A pendulum, comprising a light string of length L and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. P8.68). (a) Show that if the sphere is released from a height below that of the peg, it will return to this height after the string strikes the peg. (b) Show that if the pendulum is released from rest at the horizontal position (u 5 908) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5.
Answers
Answer:
the demonstrations are in the description.
Explanation:
First of all, reference is made to a picture that may be the one shown attached.
when the string hits the pin, the sphere will continue to swing to the right until a height h'.
a) Initially, the system has only potential energy because it is released with zero velocity from a height h.
if we consider that the system has no energy losses, then all that potential energy must be conserved. Thus, this energy is transformed in principle into kinetic energy and once the kinetic energy is maximum, it will be transformed again into potential energy.
The potential energy is given by:
[tex]E_p = mgh[/tex]
As we can see, potential energy only depends on mass, height and gravity. Since mass and gravity are constant, then for potential energy to be conserved, heights must be the same.
b) the statement probably has a transcription error in question b. Corrected, it would look like:
Show that if the pendulum is released from the horizontal position (∡ = 90) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5.
podemos asumir que la altura cero está dada en el punto de reposo del péndulo, por lo cual la energía inicial (que es potencial) está dada por:
[tex]E_0=MGL[/tex]
For the pendulum to swing, the tension must always be positive, and as a consequence, the centripetal force must be greater or at least equal to the weight. Mathematically
[tex]mV^2 /(L-d) = mg[/tex] (1)
Where:
[tex]mV^2 /(L-d) [/tex] It is the centripetal force that the sphere experiences.
Now, for the sphere to describe the circular motion, the minimum kinetic energy must be:
[tex]\frac{mV^2}{2} = mgL-2mg(L-d)[/tex] (2)
Then, replacing equation 1 in 2:
[tex]\frac{mg(L-d)}{2} = mgL-2mg(L-d)[/tex]
Simplifying:
[tex]L-2L+2d= \frac{L}{2} - \frac{d}{2}[/tex]
[tex]\frac{5}{2}d= \frac{3}{2}L[/tex]
Which means
[tex]d=\frac{3L}{5}[/tex]
Final answer:
When a pendulum hits a peg, it will return to its original height if released from below the peg. The minimum value of d, in order for a pendulum to swing in a complete circle centered on a peg, must be 3L/5.
Explanation:
A pendulum comprising a light string and a small sphere swings in the vertical plane. When the string hits a peg located below the point of suspension, the sphere will return to its original height if it is released from a height below that of the peg. The reason for this is that when the string hits the peg, the tension in the string abruptly changes direction, causing the sphere to swing back upwards.
If the pendulum is released from rest at the horizontal position and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5. This is because when the pendulum is at the lowest point, the tension in the string must be equal to or greater than the weight of the sphere in order to maintain circular motion. The minimum value of d ensures that this condition is satisfied.
Related Questions
A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Answers
Utilizing the principles of impulse and momentum, the magnitude of the impulsive force exerted by the steel plate on the bullet is approximately 40,576.47 N, and its direction is opposite to the bullet's initial motion.
Explanation:Impulsive Force on a Ricocheting Bullet
The problem involves finding the impulsive force exerted by the steel plate on a steel-jacketed bullet that ricochets off its surface. To determine the impulsive force, we utilize the concepts of impulse and momentum. The bullet has an initial velocity of 630 m/s and exits with a velocity of 500 m/s after ricocheting. While in contact with the plate, we are given an average speed of 600 m/s. The displacement during contact is 50 mm, equivalent to 0.050 meters.
To find the time of contact, we use the formula for distance, which is: distance = average speed * time, or time = distance / average speed. Hence, the time of contact is 0.050 meters / 600 m/s = 0.0000833 seconds (or 83.3 microseconds). Now, using the change in momentum (Δp = mass * change in velocity) and the impulse-momentum theorem (impulse = Δp), we can find the impulse. Considering the magnitude of the velocities and the mass of the bullet (26 g or 0.026 kg), the change in velocity is (500 m/s - 630 m/s) = -130 m/s. Therefore, Δp = 0.026 kg * -130 m/s = -3.38 kg*m/s. Given that impulse equals the average force times the time of the impact (Impulse = average force * time), we can find the average force as Impulse/time = -3.38 kg*m/s / 0.0000833 s = -40,576.47 N. The negative sign indicates that the force on the bullet is directed opposite to its initial motion, which is consistent with a ricochet.
The magnitude of the impulsive force exerted by the plate on the bullet is approximately 40,576.47 N, and its direction is opposite to the bullet's initial motion, demonstrating the principle of conservation of momentum during collisions.
Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black and are maintained at a temperature of 450 K. The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at 300 K. Determine the net rate of radiation heat transfer from the disks to the environment.
Answers
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:
[tex]Q = FA\sigma\Delta T^4[/tex]
Where,
F =View Factor
A = Cross sectional Area
[tex]\sigma =[/tex] Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m
[tex]L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K[/tex]
The view factor between two coaxial parallel disks would be
[tex]\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33[/tex]
[tex]\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75[/tex]
Then the view factor between base to top surface of the cylinder becomes [tex]F_{12} = 0.26[/tex]. From the summation rule
[tex]F_{13} = 1-0.26[/tex]
[tex]F_{13} = 0.74[/tex]
Then the net rate of radiation heat transfer from the disks to the environment is calculated as
[tex]\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}[/tex]
[tex]\dot{Q_3} = 2\dot{Q_{13}}[/tex]
[tex]\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)[/tex]
[tex]\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)[/tex]
[tex]\dot{Q_3} = 780.76W[/tex]
Therefore the rate heat radiation is 780.76W
The net radiation heat transfer from the disks to the environment is computed by applying the Stefan-Boltzmann law for the radiation heat transfer of black bodies. The area of one side of the disk and the given temperatures are substituted into the law's formula to obtain the desired value.
Explanation:The physical concept relevant to the question is the Stefan-Boltzmann law related to radiation heat transfer. Since both disks are black, they are considered perfect black bodies with an emissivity (e) of 1.
Firstly, we calculate the area (A) of one disk as A = π(D/2)² (because the back sides are insulated on both disks, we only need to consider the radiation from one side of each disk). Then, using Stefan-Boltzmann law formula: Qnet = 2σeA(T₁⁴ - T₂⁴) (the factor 2 is due to having two disks), where T₁ = 450K (temperature of the disks) and T₂ = 300K (temperature of the surrounding environment) is used to find the desired rate of heat transfer (Qnet). The Stefan-Boltzmann constant (σ) is known to be 5.67 × 10⁻⁸ J/s.m².K⁴.
With the value of A calculated from the given diameter and the above values substituted, we can calculate Qnet.
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You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experiments
1. Shining the green laser through grating A you see the first maximum 1 meter away from the center
2. Shining the red laser through grating B, you see the first maximum 1 meter away from the center
In both cases, the gratings are the same distance from the screen.
(a) What can you deduce about the gratings?
(b) What would you observe if you shone the green laser through grating B:?
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
b. (a) grating B has more lines/mm; (b) the first maximum less than 1 meter away from the center
c.(a) grating B has more lines/mm; (b) the first maximum more than 1 meter away from the center
d. (a) grating B has more lines/mm: (b) the first maximum 1 meter away from the center
e. (a) grating A has more lines/mm; (b) the first maximum more than 1 meter away from the center
f. (a) grating A has more lines/mm; (b) the first maximum 1 meter away from the center
Answers
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
Final answer:
Comparison of the diffraction patterns for two lasers and two gratings reveals that Grating B must have more lines per millimeter, and if the green laser were shone through Grating B, the first maximum would be more than 1 meter away from the center.
Explanation:
To analyze this problem, we need to apply our knowledge of diffraction gratings and wavelength of light. The position of the maxima on the screen depends on the grating spacing (number of lines per millimeter) and the wavelength of the light. The formula for the angle of the maxima for a diffraction grating is:
nλ = d sin θ,
where:
When comparing two diffraction gratings with different lasers:
A green laser (shorter wavelength) producing a first maximum at 1 meter suggests that the spacing between lines (d) in grating A supports this particular maximum for that wavelength.A red laser (longer wavelength) producing a first maximum at the same distance suggests that grating B must have a smaller d (more lines per mm) to compensate for its longer wavelength to produce a maximum at the same distance.Hence, the answer is (c):
a) Grating B has more lines/mm; because it compensates for the longer wavelength of the red light to still create a maximum at the same position as the green light with grating A.
b) If the green laser (shorter wavelength) were shone through grating B (more lines/mm), the first maximum would be more than 1 meter away from the center, since a grating with more lines per millimeter spreads the maxima further apart for the same wavelength, compared to a grating with fewer lines per millimeter.
A disk is rotating with angular speed ω1=2.0 rad/s about axle. The moment of inertia of disk & axle is 0.47 kg m2. A second disk of moment of inertia 0.31 kg m2 is dropped onto first. If dropped disk is rotating in opposite direction of ω1 with angular velocity ω2=1.0 rad/s, find magnitude of angular velocity of combination of two disks. Express your answer in rad/s.
Answers
Answer:
w = 0.808 rad / s
Explanation:
As indicated by the moment of inertia t the angular velocity of the disks we use the concept of conservation of the angular momentum, for this we define the system as formed by the two discs, therefore the torque during the crash is internal and the angular momentum is conserved
Let's write in angular momentum
Initial. Before impact
L₀ = I₁ w₁ + I₂ w₂
Final. After the rock has stuck
[tex]L_{f}[/tex] = (I₁ + I₂) w
The two discs are rotating in opposite directions, we consider the rotation of the first positive disc, so the angular velocity of the second is negative
L₀ =[tex]L_{f}[/tex]
I₁ w₁ - I₂ w₂ = (I₁ + I₂) w
w = (I₁ w₁ - I₂ w₂) / (I₁ + I₂)
Let's calculate
w = (0.47 2.0 - 0.31 1.0) / (0.47+ 0.31)
w = 0.63 / 0.78
w = 0.808 rad / s
in the direction of disc rotation 1
An 80 kg astronaut has gone outside his space capsule to do some repair worlc Unfortunately, he forgot to lock his safetytether in place, and he has drifted 5.0 m away from the capsule. Fortunately, he has a 1000 w portable laser with freshbatteries that will operate it for 1.0 hr His only chance is to accelerate himself toward the space capsule by firing the laser inthe opposite direction. He has a 10-h supply of oxygen. How long will it take him to reach safety?
Answers
Final answer:
The question involves using the principles of conservation of momentum and kinematics in space to rescue an astronaut adrift from their space capsule using a laser for propulsion. However, the time required for the astronaut to reach safety cannot be calculated without the thrust or force specifics of the laser.
Explanation:
The scenario described involves the principles of conservation of momentum and the astronaut's ability to utilize a laser as a propulsion device. The astronaut's mass is 80 kg, and they need to use the laser to accelerate towards the space capsule. Without any external forces, momentum is conserved, and hence, the astronaut can generate thrust in space by expelling photons in the opposite direction of the desired movement, albeit very weak thrust. To determine how long it will take him to reach safety, we would need to calculate the actual propulsion force of the laser and the resulting acceleration. This can be derived from the conservation of momentum and Newton's second law of motion. However, with the current information provided, it is impossible to provide an accurate estimate without knowing the momentum or thrust provided by the laser.
Without such crucial information, assuming an ideal scenario and if the astronaut could somehow generate a constant force to obtain a tangible acceleration, we can use kinematics equations to estimate travel time once the acceleration is known. Yet in this specific situation, it's not feasible to calculate the time required for the astronaut to reach the space capsule without additional details about the actual force the laser can exert on the astronaut.
A glass column is filled with mercury and inverted in a pool of mercury. The mercury column stabilizes at a height of 735 mm above the pool of mercury. What is the pressure of the atmosphere?
Answers
Answer:
[tex] 735 mm Hg = 0.967 atm= 97991.940 Pa=97.992Pa[/tex]
Explanation:
Previous concepts
Atmospheric pressure is defined as "the force per unit area exerted against a surface by the weight of the air above that surface".
Torricelli shows that we can calculate the atmosphric pressure with a glass tube inside of a tank and with the height we can find the pressure with the relation
[tex]P_{atm}=\rho_{Hg} g h[/tex]
Solution to the problem
For this case we can use the following conversion factor:
[tex]1 atm = 760 mm Hg[/tex]
And if we convert the 735 mm Hg to atm we got this:
[tex]735 mm Hg * \frac{1atm}{760 mm Hg}=0.967 atm[/tex]
And also we can convert this value to Pa and Kpa since we have this conversion factor:
[tex] 1 atm =101325 Pa[/tex]
And if we apply the conversion we got:
[tex]0.967 atm *\frac{101325 Pa}{1 atm}=97991.940 Pa[/tex]
And that correspond to 97.99 Kpa.
Finally we can express the atmospheric pressure on different units for this case :
[tex] 735 mm Hg = 0.967 atm= 97991.940 Pa=97.992Pa[/tex]
The atmospheric pressure (atm) resulting from the mercury column is 0.967 atm.
The given parameters;
the height of the mercury column, h = 735 mm HgThe atmospheric pressure (atm) resulting from the mercury column is calculated as follows;
760 mmHg ------- 1 atm
735 mmHg -------- ?
[tex]= \frac{735 \ \times \ 1 atm}{760} \\\\= 0.967 \ atm[/tex]
Thus, the atmospheric pressure (atm) resulting from the mercury column is 0.967 atm.
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A horizontal tube consists of a 7.0-cm-diameter pipe that narrows to a 2.0-cm-diameter throat. In the pipe, the water pressure is twice atmospheric pressure and the water flows with a speed of 0.40 m/s. What is the pressure in the throat, assuming that the water behaves like an ideal fluid? The density of water is 1000 kg/m3 , and atmospheric pressure is 1.01 × 105 Pa.
Answers
Answer:
[tex]1.9\times10^{5} Pa[/tex]
Explanation:
[tex]d_{p}[/tex] = diameter of the pipe = 7 cm
[tex]v_{p}[/tex] = speed of water in the pipe = 0.40 m/s
[tex]A_{p}[/tex] = Area of cross-section of pipe = [tex](0.25)\pi d_{p}^{2}[/tex]
[tex]d_{t}[/tex] = diameter of the throat = 2 cm
[tex]v_{t}[/tex] = speed of water in the throat
[tex]A_{t}[/tex] = Area of cross-section of throat = [tex](0.25)\pi d_{t}^{2}[/tex]
Using equation of continuity
[tex]A_{p} v_{p} = A_{t} v_{t} \\(0.25)\pi d_{p}^{2} v_{p} = (0.25)\pi d_{t}^{2} v_{t} \\(7)^{2} (0.40) = (2)^{2} v_{t}\\v_{t} = 4.9 ms^{-1}[/tex]
[tex]P_{o}[/tex] = atmospheric pressure = 1.01 x 10⁵ Pa
[tex]P_{p}[/tex] = Pressure in the pipe = [tex]2 P_{o}[/tex] = 2.02 x 10⁵ Pa
[tex]P_{t}[/tex] = Pressure in the throat
Using Bernoulli's theorem
[tex]P_{t} + (0.5)\rho v_{t}^{2} = P_{p} + (0.5)\rho v_{p}^{2}\\P_{t} + (0.5)(1000) (4.9)^{2} = 2.02\times10^{5} + (0.5)(1000) (0.40)^{2}\\P_{t} + 12005 = 202080\\P_{t} = 190075 Pa\\P_{t} = 1.9\times10^{5} Pa[/tex]
The pressure in the throat of the pipe, assuming that the water behaves like an ideal fluid, can be obtained using the continuity equation and Bernoulli's equation. The velocity at the throat is first calculated using the continuity equation and then Bernoulli's equation is used to find the pressure. The pressure comes out to be 1.94 x 105 Pa.
Explanation:This problem can be resolved using the continuity equation and Bernoulli's equation, both of which refer to the conservation laws of mass and energy respectively in fluid dynamics. Bernoulli's equation states that, along a streamline, the sum of all forms of fluid energy is constant with respect to any increase in height. The continuity equation states that the mass flow rate must remain constant in the pipe i.e., the velocity of fluid at the wide end of the pipe multiple by its area will be equal to the velocity at the narrow end multiplied by its area.
The velocity at the throat can be found using the continuity equation: (7)^2 x (0.40) = (2)^2 x V2. Solving this gives V2 = 2.45 m/s.
Next, we'll use Bernoulli's equation, P1 + 0.5 x rho x (V1)^2 = P2 + 0.5x rho(V2)^2, where P1 is the pressure at the wide end (which is given as twice of atmospheric pressure), rho is density of water, V1 is velocity at the wide end, P2 is the pressure at the narrow end (which we are to find), and V2 is the velocity at the narrow end.
Substituting the given values, we have (2 x 1.01e5) + 0.5 x 1000 x (0.4)^2 = P2 + 0.5*1000*(2.45)^2. Solving this equation gives P2 = 1.94 x 105 Pa.
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A person is trying to judge whether a picture (mass = 1.05 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.720. What is the minimum amount of pressing force that must be used?
Answers
Answer:
the minimum amount of pressing force P will be 14.29 N
Explanation:
the friction force Fr will be
Fr = μ*N
where μ= coefficient of static friction , N= force normal to the plane
then N=P (force applied by the person)
from Newton's first law
net force = F = 0
Fr - weight = 0
μ*P - m*g =0
P = m*g/μ = 1.05 kg*9.8 m/s² / 0.720 = 14.29 N
P = 14.29 N
Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for problems like this one. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.130 m, an angular speed of 78.0 rad/s, and a moment of inertia of 1.25 kg·m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of 6 during a time of 3.00 s. (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.
Answers
Answer:
21.67 rad/s²
208.36538 N
Explanation:
[tex]\omega_f[/tex] = Final angular velocity = [tex]\dfrac{1}{6}78=13\ rad/s[/tex]
[tex]\omega_i[/tex] = Initial angular velocity = 78 rad/s
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
t = Time taken
r = Radius = 0.13
I = Moment of inertia = 1.25 kgm²
From equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2[/tex]
The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875[/tex]
Frictional force is given by
[tex]F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N[/tex]
The magnitude of the force of friction applied by the brake shoe is 208.36538 N
The angular deceleration of the cylinder is 46.0 rad/s².
The force of friction applied by the brake shoe is 1180 N.
Here's how we can approach it:
(a) Angular Deceleration:
Initial Angular Speed (ω₀): 92.0 rad/s
Final Angular Speed (ωf): ω₀/2 = 92.0 rad/s / 2 = 46.0 rad/s
Time (Δt): 4.00 s
We can use the following equation to find the angular deceleration (α):
α = (ωf - ω₀) / Δt
Substituting the values:
α = (46.0 rad/s - 92.0 rad/s) / 4.00 s
α = -46.0 rad/s² (negative sign indicates deceleration)
Therefore, the magnitude of the angular deceleration of the cylinder is 46.0 rad/s².
(b) Force of Friction:
Moment of Inertia (I): 1.36 kg·m²
Angular Deceleration (α): 46.0 rad/s²
The net torque (τ) acting on the cylinder is equal to the product of its moment of inertia and angular deceleration:
τ = I * α
The frictional force (F) applied by the brake shoe creates a torque that opposes the cylinder's rotation. This torque is equal to the force multiplied by the radius of the cylinder (r):
τ = F * r
Since the net torque is caused solely by the frictional force, we can equate the two torque equations:
I * α = F * r
Solving for the force of friction:
F = I * α / r
Substituting the values:
F = 1.36 kg·m² * 46.0 rad/s² / 0.0530 m
F = 1180 N
Therefore, the magnitude of the force of friction applied by the brake shoe is 1180 N.
9. Would the maximin criterion achieve perfect income equality? a. Yes. There would be no way to reallocate resources to raise the utility of the poor. b. Yes. The maximin criterion would eliminate poverty. c. No. It is impossible for complete equality to benefit the worst-off people in society. d. No. Complete equality would reduce incentives to work, which would reduce total income, which would reduce the incomes of the worst-off people in society.
Answers
Answer:
C
Explanation:
Although the maximin criterion emphasizes the worst-off person in society and it's targeted towards equalizing of the distribution of income by transferring income from the rich to the poor, it will not lead to a complete egalitarian society. Because this will make the people not to have incentive to work hard and the societal total income will substantially fall off and the least fortunate person will be worse off. Thus, this rule still allows disparities in income.
Bored, a boy shoots his pellet gun at a piece of cheese that sits, keeping cool for dinner guests, on a massive block of ice. On one particular shot, his 1.3 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is 73 m/s and the cheese has a mass of 109 g, what is the coefficient of friction between the cheese and ice?
Answers
To calculate the coefficient of friction between the cheese and ice, we can use the work-energy theorem. By calculating the work done on the cheese by the friction force, we can determine the coefficient of friction. By plugging in the given values, the coefficient of friction is found to be 0.047.
Explanation:To find the coefficient of friction between the cheese and ice, we can use the concept of work-energy theorem. The work done on the cheese by the friction force is equal to the change in kinetic energy of the cheese. The work done by friction is given by the equation:
Work = Force x Distance
In this case, the force is the friction force and the distance is the distance the cheese slid. We can express the friction force as:
Friction Force = coefficient of friction x Normal Force
Since the cheese is in contact with the ice, the normal force exerted on the cheese is equal to its weight:
Normal Force = mass of cheese x acceleration due to gravity
Substituting the expressions for friction force and normal force into the work equation, we get:
Work = (coefficient of friction x mass of cheese x acceleration due to gravity) x distance
Since the work done is equal to the change in kinetic energy of the cheese, we have:
0.5 x mass of cheese x final velocity^2 - 0.5 x mass of cheese x initial velocity^2 = (coefficient of friction x mass of cheese x acceleration due to gravity) x distance
Simplifying the equation, we can solve for the coefficient of friction:
coefficient of friction = (0.5 x mass of cheese x (final velocity^2 - initial velocity^2)) / (mass of cheese x acceleration due to gravity x distance)
Plugging in the given values, we find that the coefficient of friction between the cheese and ice is 0.047.
The answer is [tex]\mu \[/tex]approx 0.0153.
The coefficient of friction between the cheese and the ice can be determined by analyzing the conservation of momentum and the work-energy principle.
First, let's calculate the initial momentum of the pellet:
[tex]\[ p_{pellet} = m_{pellet} \cdot v_{pellet} \][/tex]
[tex]\[ p_{pellet} = 0.0013 \, \text{kg} \cdot 73 \, \text{m/s} \][/tex]
[tex]\[ p_{pellet} = 0.0949 \, \text{kg} \cdot \text{m/s} \][/tex]
When the pellet gets stuck in the cheese, the momentum is transferred to the cheese-pellet system. By conservation of momentum:
[tex]\[ m_{pellet} \cdot v_{pellet} = (m_{pellet} + m_{cheese}) \cdot v_{cheese} \][/tex]
[tex]\[ 0.0013 \, \text{kg} \cdot 73 \, \text{m/s} = (0.0013 \, \text{kg} + 0.109 \, \text{kg}) \cdot v_{cheese} \][/tex]
[tex]\[ v_{cheese} = \frac{0.0013 \, \text{kg} \cdot 73 \, \text{m/s}}{0.1103 \, \text{kg}} \][/tex]
[tex]\[ v_{cheese} = \frac{0.0949 \, \text{kg} \cdot \text{m/s}}{0.1103 \, \text{kg}} \][/tex]
[tex]\[ v_{cheese} \approx 0.8608 \, \text{m/s} \][/tex]
Now, we use the work-energy principle to find the coefficient of friction. The work done by friction is equal to the kinetic energy lost by the cheese as it slides to a stop:
[tex]\[ W_{friction} = f \cdot d \][/tex]
[tex]\[ f = \mu \cdot N \][/tex]
[tex]\[ N = m_{cheese} \cdot g \][/tex]
[tex]\[ W_{friction} = \mu \cdot m_{cheese} \cdot g \cdot d \][/tex]
The kinetic energy lost by the cheese is equal to its initial kinetic energy:
[tex]\[ KE_{initial} = \frac{1}{2} m_{cheese} \cdot v_{cheese}^2 \][/tex]
[tex]\[ KE_{initial} = \frac{1}{2} \cdot 0.109 \, \text{kg} \cdot (0.8608 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{initial} \approx 0.0396 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ KE_{initial} \approx 0.0396 \, \text{J} \][/tex]
Setting the work done by friction equal to the kinetic energy lost:
[tex]\[ \mu \cdot m_{cheese} \cdot g \cdot d = KE_{initial} \][/tex]
[tex]\[ \mu \cdot 0.109 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 0.25 \[/tex], [tex]\text{m} = 0.0396 \, \text{J} \][/tex]
[tex]\[ \mu \cdot 2.5963 \, \text{kg} \cdot \text{m/s}^2 = 0.0396 \[/tex], [tex]\text{J} \][/tex]
[tex]\[ \mu \approx \frac{0.0396 \, \text{J}}{2.5963 \, \text{kg} \cdot \text{m/s}^2} \] \[ \mu \approx 0.0153 \][/tex]
The correct format for the answer is:
[tex]\[ \boxed{\mu \approx 0.0153} \][/tex]
A refrigerator is being pulled up a ramp with a horizontal force P, which acts at the top corner. The refrigerator has a mass of 75 kg, acting through point G. The ramp is inclined at 20º, and the coefficient of static friction is 0.3 between the refrigerator and the ramp.
(a) Find the force P required to move the refrigerator.
(b) Does the refrigerator tip or slide?
Answers
Answer:
(a) P = 459.055 N.
(b) the refrigerator tips.
Explanation:
Given, the angle of ramp is 20°.
When the weight of refrigerator is resolved in directions parallel and perpendicular to ramp, 75×g×sin(20°) and 75×g×cos(20°).
⇒ normal contact force is 75×g×cos(20°).
⇒ frictional force is 0.3×75×g×cos(20°) = 207.414 N
so, total opposite force is 207.414 + 75×g×sin(20°) = 459.055 N.
so, the force needed is P = 459.055 N
And as the moment due to both opposite force and P force are in same direction the refrigerator tips rather than just sliding.
I take 1.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium is reached, I have 2.0 kg of ice at 0°C. The water was originally at 0°C.
The specific heat of water = 1.00 kcal/kg⋅°C, the specific heat of ice = 0.50 kcal/kg⋅°C, and the latent heat of fusion of water is 80 kcal/kg.
The original temperature of the ice was:
a. one or two degrees below 0°C.b. −80°C.c. −160°C.d. The whole experiment is impossible.
Answers
Answer:
.c. −160°C
Explanation:
In the whole process one kg of water at 0°C loses heat to form one kg of ice and heat lost by them is taken up by ice at −160°C . Now see whether heat lost is equal to heat gained or not.
heat lost by 1 kg of water at 0°C
= mass x latent heat
= 1 x 80000 cals
= 80000 cals
heat gained by ice at −160°C to form ice at 0°C
= mass x specific heat of ice x rise in temperature
= 1 x .5 x 1000 x 160
= 80000 cals
so , heat lost = heat gained.
The original temperature of the ice was a. one or two degrees below 0°C.
Explanation:Heat transfer is the process by which thermal energy is exchanged between systems. It occurs through conduction, where heat moves through materials, convection, involving the movement of fluids, and radiation, which involves electromagnetic waves. Understanding heat transfer is essential in fields like physics, engineering, and environmental science.
To find the original temperature of the ice, we need to calculate the heat transferred during the process. First, we need to bring the ice up to 0°C and melt it. This requires a heat transfer of 4.74 kcal. This will lower the temperature of the water by 23.15°C. After the ice has melted and the system reaches equilibrium, the final temperature of the water is 20.6°C. Therefore, the original temperature of the ice was one or two degrees below 0°C.
In 1995 a research group led by Eric Cornell and Carl Wiemann at the University of Colorado successfully cooled Rubidium atoms to the 20-200 nK temperature range. Assuming (incorrectly) that the Rubidium atoms behave like particles of a classical ideal gas, calculate the RMS speed of a Rubidium atom at a temperature of 85.0 nK. In the experiments one particular isotope of Rubidium was used, Rubidium-87. The molar mass of this isotope is 86.91 g/mol.
Answers
Answer:
0.00493 m/s
Explanation:
T = Temperature of the isotope = 85 nK
R = Gas constant = 8.341 J/mol K
M = Molar mass of isotope = 86.91 g/mol
Root Mean Square speed is given by
[tex]v_r=\sqrt{\dfrac{3RT}{M}}\\\Rightarrow v_r=\sqrt{\dfrac{3\times 8.314\times 85\times 10^{-9}}{86.91\times 10^{-3}}}\\\Rightarrow v_r=0.00493\ m/s[/tex]
The Root Mean Square speed is 0.00493 m/s
Final answer:
The RMS speed of Rubidium-87 atoms at 85 nK can be estimated using the ideal gas approximation and the formula 'Urms = √(3kBT/M)'. The molar mass is converted to kg/mol and the temperature to kelvins before calculation.
Explanation:
To calculate the root-mean-square (RMS speed) of Rubidium-87 atoms at a temperature of 85.0 nK assuming classical ideal gas behavior, we can use the formula:
Urms = √(3kBT/M)
Where Urms is the root-mean-square speed, kB is Boltzmann's constant (1.38 × 10-23 J/K), T is the absolute temperature in kelvins, and M is the molar mass in kilograms per mole. First, we convert the molar mass of Rubidium-87 from grams per mole to kilograms per mole by dividing it by 1000:
M = 86.91 g/mol ∖ 0.08691 kg/mol
Next, we convert the temperature from nanokelvins to kelvins:
T = 85.0 nK = 85.0 × 10-9 K
Substituting the values into the RMS speed equation gives us:
Urms = √(3 × (1.38 × 10-23 J/K) × (85.0 × 10-9 K) / 0.08691 kg/mol)
After calculating, we find that the RMS speed of Rubidium-87 atoms at 85.0 nK is:
Urms ≈ ... m/s
Note that the calculation here assumes that the Rubidium atoms behave as a classical ideal gas, which is not an accurate assumption for atoms at such low temperatures.
A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg. (a) What is the angular velocity of the arm immediately after catching the ball as measured from the arm socket? (b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?
Answers
To solve the problem, we require an understanding of physics concepts like angular velocity, moment of inertia, and torque. The catcher catching the ball changes its angular momentum, resulting in an angular velocity. The torque experienced when the arm stops the rotation can be computed using known equations.
Explanation:This question involves concepts of physics like angular velocity, moment of inertia, and torque. Initially, with the catcher's arm at the ready position, the system (arm and ball) has zero angular velocity. Then when the catcher catches the ball, he applies an impulse to it and changes not just the linear momentum but the angular momentum about the shoulder as well.
The change in angular momentum (angular impulse) will be equal to the product of the mass of the baseball, its velocity, and the arm's length, i.e., 0.145kg × 40m/s × 0.5m= 2.9 kg m²/s. This change in angular momentum over time will induce an angular velocity, which can be calculated by dividing the change in angular momentum by the moment of inertia of the system (arm and ball).
For part (b), the torque experienced by the arm when it stops the rotation can be computed from the known equation Torque = (Moment of Inertia × Angular Acceleration). The angular acceleration is determined by the change in angular velocity divided by the time taken which in this case is 0.3 seconds. Taking all these physics concepts into account will yield the correct numerical solutions for parts (a) and (b).
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To find the angular velocity of the arm after catching the ball, conservation of angular momentum was used, resulting in an angular velocity of 7.85 rad/s. The torque needed to stop the arm's rotation in 0.3 seconds is -9.66 N - m, calculated using the angular deceleration. The answer involves concepts of angular momentum and torque.
A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg.
(a) What is the angular velocity of the arm immediately after catching the ball?
To find the angular velocity of the arm immediately after catching the ball, we need to use the principle of conservation of angular momentum. The initial angular momentum of the ball can be calculated using:
L_initial = m_ball * v_ball * r_arm
where m_ball = 0.145 kg, v_ball = 40 m/s, and r_arm = 0.5 m.
L_initial = 0.145 kg * 40 m/s * 0.5 m = 2.9 kg·m²/s
The moment of inertia of the arm plus the ball (approximated as point mass at the end) is:
I_total = I_arm + m_ball * r_arm²
Using the formula for the moment of inertia of a rod about one end: I_arm = (1/3) * m_arm * (r_arm)², where m_arm = 4.0 kg and r_arm = 0.5 m:
I_arm = (1/3) * 4.0 kg * (0.5 m)² = 0.333 kg·m²
Adding the moment of inertia of the ball:
I_total = 0.333 kg·m² + 0.145 kg * (0.5 m)² = 0.333 kg·m² + 0.03625 kg·m² = 0.36925 kg·m²
Since angular momentum is conserved, L_initial = I_total * ω, where ω is the angular velocity:
ω = L_initial / I_total = 2.9 kg·m²/s / 0.36925 kg·m² = 7.85 rad/s
(b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?
Torque (τ) can be calculated using the relationship between torque, angular deceleration (α), and moment of inertia (I):
τ = I_total * α
First, we find the angular deceleration. The arm stops, meaning final angular velocity is 0.
Using the angular kinematic equation: ω_final = ω_initial + α * t, where ω_final = 0 and t = 0.3 s:
0 = 7.85 rad/s + α * 0.3 s
α = -7.85 rad/s / 0.3 s = -26.17 rad/s²
Now, calculate torque:
τ = 0.36925 kg·m² * (-26.17 rad/s²) = -9.66 N·m
The negative sign indicates that the torque is in the direction opposite to the rotation.
wo parallel-plate capacitors have the same dimensions, but the space between the plates is filled with air in capacitor 1 and with plastic in capacitor 2. Each capacitor is connected to an identical battery, such that the potential difference between the plates is the same in both capacitors. Compare the magnitudes of the electric fields between the plates, ????1 and ????2, and the magnitudes of the free charges on the plates, ????1 and ????2.
Answers
Final answer:
The electric field between the plates will be the same in both capacitors, but the charge on the plates of capacitor 2 with plastic between the plates will be greater than the charge on the plates of capacitor 1 with air between the plates.
Explanation:
When comparing the magnitudes of the electric fields between the plates, we can use the formula E = Q / (ε0 * A), where E is the electric field, Q is the charge on the plates, ε0 is the permittivity of free space, and A is the area of the plates. In this case, since the capacitors have the same dimensions and the same charge, the electric field between the plates in both capacitors will be the same.
Regarding the magnitudes of the free charges on the plates, we know that Q = C * V, where Q is the charge, C is the capacitance, and V is the potential difference. Since the capacitance is directly proportional to the dielectric constant of the material between the plates, and the charge is the product of the capacitance and potential difference, the charge on the plates of capacitor 2 with plastic between the plates will be greater than the charge on the plates of capacitor 1 with air between the plates, because the dielectric constant of plastic is greater than 1.
Question:The mechanical properties of cobalt may be improved by incorporating fine particles of tungsten carbide (WC). Given that the moduli of elasticity of these materials are, respectively, 200 GPa and 700 GPa, plot modulus of elasticity vs. the volume percent of WC in Co from 0 to 100 vol% using both upper- and lower-bound expressions to form a performance envelope into which the material will fall. Remember that the isostrain case is the upper-bound case and the isostress is the lower-bound case.Mechanical Properties:Mechanical properties help to identify and classify the materials. The most common Mechanical properties are ductility, tenacity, strength, impact resistance, hardness. The mechanical property does not depend on the quantity of the material.
Answers
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Final answer:
To explore how the incorporation of tungsten carbide (WC) into cobalt (Co) affects the composite material's modulus of elasticity, the moduli values are calculated for a range of WC volume percentages under both isostrain and isostress conditions, illustrating the performance envelope and guiding material design in engineering applications.
Explanation:
To plot the modulus of elasticity vs. the volume percent of WC in Co from 0 to 100 vol%, we use both upper- and lower-bound expressions to form a performance envelope. For the isostrain (upper-bound) case, the mixture's modulus of elasticity (Ecomposite, isostrain) can be calculated using the rule of mixtures formula: Ecomposite, isostrain = ECoVCo + EWCVWC, where E represents modulus of elasticity, and V represents volume fraction. For the isostress (lower-bound) case, the inverse of the composite's modulus of elasticity can be expressed as 1/Ecomposite, isostress = VCo/ECo + VWC/EWC. Here, ECo = 200 GPa is the modulus of elasticity for cobalt, and EWC = 700 GPa is for tungsten carbide.
As the volume percent of WC increases from 0 to 100%, the calculated Ecomposite, isostrain values will show a trend of increasing modulus due to the higher modulus of WC compared to Co. Conversely, the Ecomposite, isostress values will also follow an upward trend but at a different rate, illustrating the material's variation within the performance envelope depending on the volume fraction of WC. This graphical representation will help to visualize how incorporating tungsten carbide into cobalt affects the composite's mechanical performance, underlining the importance of material design and engineering applications.
A slice of bread contains about 100 kcal. If specific heat of a person were 1.00 kcal/kg·°C, by how many °C would the temperature of a 70.0-kg person increase if all the energy in the bread were converted to heat?a. 2.25°Cb. 1.86°Cc. 1.43°Cd. 1.00°C
Answers
Answer:
(c) 1.43°C
Explanation:
If the energy in the bread are converted to heat.
Then, The heat transferred from the bread to person = 100 kcal.
From specific heat capacity,
Q = cmΔT............................ equation 1
Where Q = quantity of heat, m = mass of the person, c = specific heat capacity of the person, Δ = increase in temperature.
Making ΔT the subject the equation 1,
ΔT = Q/cm........................ equation 2
Where Q = 100 kcal, c= 1.00 kcal/kg.°C, m = 70.0 kg
Substituting these values into equation 2,
ΔT = 100/(1×70)
ΔT = 100/70
ΔT = 1.428
ΔT ≈ 1.43°C
The increase in temperature of the body is = 1.43°C
The right option is (c) 1.43°C
The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of the sun. (Take e = 1, and σ = 5.67 × 10−8W/m2⋅K4).
a. 3.95 × 1026 W
b. 5.17 × 1027 W
c. 9.62 × 1028 W
d. 6.96 × 1030 W
Answers
Answer:
a. [tex]3.95\times10^{26} [/tex]W
Explanation:
[tex]T[/tex] = temperature of the surface of sun = 5800 K
[tex]r[/tex] = Radius of the Sun = 7 x 10⁸ m
[tex]A[/tex] = Surface area of the Sun
Surface area of the sun is given as
[tex]A = 4\pi r^{2} \\A = 4(3.14) (7\times10^{8})^{2}\\A = 6.2\times10^{18} m^{2}[/tex]
[tex]e[/tex] = Emissivity = 1
[tex]\sigma[/tex] = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴
Using Stefan's law, Power output of the sun is given as
[tex]P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (1) (6.2\times10^{18}) (5800)^{4}\\P = 3.95\times10^{26} W[/tex]
An oil film with index of refraction 1.64 is trapped between two pieces of glass with index of refraction 1.43. No light is reflected by such a film when 578 nm light falls on it at normal incidence. What is the nonzero minimum thickness of the oil film that will satisfy these conditions? Answer in units of nm.
Answers
Final answer:
The minimum nonzero thickness of an oil film with index of refraction 1.64, between glass with index 1.43 to achieve no reflection for 578 nm light at normal incidence, is approximately 88 nm. This results from the condition for destructive interference.
Explanation:
When light of wavelength 578 nm hits the film at normal incidence and no light is reflected, it means that the reflected light waves from the top and bottom surfaces of the oil film are exactly out of phase, causing destructive interference.
For destructive interference to occur, the path difference between the two reflected waves must be an odd multiple of half the wavelength in the medium, which is given by the formula:
Path Difference (in the medium) = 2 × n × thickness of the film = [tex](m + \(\frac{1}{2}\)) \times \(\frac{\lambda}{n}\)[/tex], where m = 0, 1, 2, ...
Considering that there is a [tex]\(\frac{\lambda}{2}\)[/tex] phase shift when light reflects off a medium with a lower index of refraction to a higher one, and the film's index of refraction is greater than that of the surrounding glass, we take m = 0 for the smallest non-zero thickness. Thus, the minimum thickness of the oil film is:
Thickness = [tex]\(\frac{\lambda}{4 * n}\)[/tex] = [tex]\(\frac{578 nm}{4 \times 1.64}\)[/tex] = 88.16 nm
The nonzero minimum thickness of the oil film that will satisfy the conditions of no reflected light is therefore approximately 88 nm.
The emission of x rays can be described as an inverse photoelectric effect.
What is the potential difference through which an electron accelerates to produce x rays with a wavelength of 0.10 nm?
Answers
Answer:
The potential difference through which an electron accelerates to produce x rays is [tex]1.24\times 10^4\ volts[/tex].
Explanation:
It is given that,
Wavelength of the x -rays, [tex]\lambda=0.1\ nm=0.1\times 10^{-9}\ m[/tex]
The energy of the x- rays is given by :
[tex]E=\dfrac{hc}{\lambda}[/tex]
The energy of an electron in terms of potential difference is given by :
[tex]E=eV[/tex]
So,
[tex]\dfrac{hc}{\lambda}=eV[/tex]
V is the potential difference
e is the charge on electron
[tex]V=\dfrac{hc}{e\lambda}[/tex]
[tex]V=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times 0.1\times 10^{-9}}[/tex]
V = 12431.25 volts
or
[tex]V=1.24\times 10^4\ volts[/tex]
So, the potential difference through which an electron accelerates to produce x rays is [tex]1.24\times 10^4\ volts[/tex]. hence, this is the required solution.
A seaside cliff is 30 m above the ocean surface, and Sam is standing at the edge of the cliff. Sam has three identical stones. The first stone he throws off the cliff at 30° above the horizontal. The second stone he throws vertically downward into the ocean. The third stone he drops into the ocean.
1. In terms of magnitude, which stone has the largest change in its velocity over a one second time interval after its release? (Sam’s throwing speed is 10 m/s.)
Answers
Answer:
In terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.
Explanation:
Stone 1:
vi = 10 m/s
vix = vi*Cos ∅ = (10 m/s)*Cos 30° = 8.66 m/s = vx
viy = vi*Sin ∅ = (10 m/s)*Sin 30° = 5 m/s
vy = viy - g*t = (5 m/s) - (9.8m/s²)*(1 s) = -4.8
then
v = √(vx²+vy²) = √((8.66)²+(-4.8)²) = 9.90 m/s
Δv = v - vi = 9.902 m/s - 10 m/s
⇒ Δv = -0.098 m/s
Stone 2:
vi = 10 m/s
v = vi + g*t = (10 m/s) + (9.8m/s²)*(1 s) = 19.8 m/s
Δv = v - vi = (19.8 m/s) - (10 m/s)
⇒ Δv = 9.8 m/s
Stone 3:
vi = 0 m/s
v = g*t = (9.8m/s²)*(1 s) = 9.8 m/s
Δv = v - vi = (9.8 m/s) - (0 m/s)
⇒ Δv = 9.8 m/s
Finally, in terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.
1. A 70-kg swimmer dives horizontally off a 500-kg raft. The diver's speed immediately after leaving the raft is 6.0 m/s. A micro-sensor system attached to the edge of the raft measures the time interval during which the diver applies an impulse to the raft just prior to leaving the raft surface. If the time interval is read as 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?
Answers
To solve this problem it is necessary to apply the concepts related to momentum theorem.
The equation for impulse is given as
[tex]I = Ft[/tex]
Where
I = Force
t = Time
At the same time we have the equation for momentum is given as
[tex]p = mv[/tex]
The impulse momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Therefore
I = p
Ft = mv
Solving to find the force
[tex]F = \frac{mv}{t}[/tex]
[tex]F = \frac{(70)(6)}{0.25}[/tex]
[tex]F = 1680N[/tex]
Therefore the magnitude of the average horizontal force by diver on the raft is 1680N
The magnitude of the average horizontal force exerted by the diver on the raft is 1680 N.
Explanation:To find the magnitude of the average horizontal force exerted by the diver on the raft, we need to start by calculating the change in momentum of the diver. The momentum of an object is given by the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. The change in momentum is equal to the impulse, which is given by the equation J = Δp = mΔv.
Since the swimmer dives horizontally, the change in velocity is equal to the initial velocity of the swimmer. Therefore, Δv = 6.0 m/s. Substituting the values, we get J = (70 kg)(6.0 m/s) = 420 kg·m/s.
The impulse is equal to the average force multiplied by the time interval, so we can rearrange the equation to solve for the average force. F = J / Δt = 420 kg·m/s / 0.25 s = 1680 N.
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what consistent physiological pattern is more common in men's' teeth? A. Men have more room in their mouths for their molars. B. Men generally have stronger enamel. C. Men have more teeth. D. Men generally have larger teeth.
Answers
Answer:
B.
Explanation:
Men generally have stronger enamel than the women.
Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 ∘ C and 721 mmHg ?
Answers
Answer:
The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%
Explanation:
From General gas equation.
PV = nRT............................... Equation 1
Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.
n = mass/molar mass .................. Equation 2
substituting equation 2 into equation 1.
PV = (mass/molar mass)RT
⇒ Mass/molar mass = PV/RT..................... Equation 3
But mass = Density × Volume
⇒ M = D × V.................... Equation 4
Where D = density, M = mass
Substituting equation 4 into equation 3
DV/molar mass = PV/RT............ Equation 5
Dividing both side of the equation by Volume (V) in Equation 5
D/molar mass = P/RT .............. Equation 6
Cross multiplying equation 6
D × RT = P × molar mass
∴ Molar mass = (D × RT)/P.................. Equation 7
Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,
T = 25°C = 25 + 273 = 298 K,
P =721 mmHg = (721/760) atm= 0.949 atm
Substituting these values into equation 7
Molar mass = (0.518 × 0.0821 × 298)/0.949
Molar mass = 13.35 g/mole
The molar mass of the mixture is =13.35 g/mole
Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.
∴ 13.35 = 4(y) + 32(1-y)
13.35 = 4y + 32 - 32y
Collecting like terms in the equation,
32y - 4y = 32 - 13.35
28y = 18.65
y = 18.65/28
y =0.666
y = 0.666 × 100 = 66.6%
∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%
The percent (by moles) of He in the helium-oxygen mixture with a density of 0.518 g/L at 25 ∘C and 721 mmHg is 13.6%.
Explanation:To determine the percent (by moles) of He in the helium-oxygen mixture, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P is the pressure (721 mmHg)V is the volume (0.518 L)n is the number of moles of the gas we're interested in (He in this case)R is the ideal gas constant (0.0821 L·atm/mol·K)T is the temperature in Kelvin (25 + 273 = 298 K)Rearranging the equation to solve for n:
n = PV / RT
Substituting the values:
n = (721 mmHg * 0.518 L) / (0.0821 L·atm/mol·K * 298 K)
n = 0.136 mol
Since the total number of moles in the mixture is 1 (as it is a binary mixture), the percent (by moles) of He can be calculated as:
Percent of He = (0.136 mol of He / 1) * 100%
Percent of He = 13.6%
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At high noon, the sun delivers 1 000 W to each square meter of a blacktop road.
What is the equilibrium temperature of the hot asphalt, assuming its emissivity e = 1? (σ = 5.67 × 10−8W/m2⋅K4) .
a. 75°Cb. 84°Cc. 91°Cd. 99°C
Answers
Answer:
The correct answer is c, T = 91.3°C
Explanation:
For this exercise let's use Stefan's equation on the emission of a black body
P = σ A e T⁴
Where σ is the Stefan-Boltzmann constant, A the area, and 'e' emissivity and T the absolute temperature
In this case give the absorbed power is 1000W per square meter, let's clear the temperature equation
T⁴ = (P / A) 1/σ e
Let's calculate
T⁴ = 1000 1 / (5.67 10⁻⁸ 1)
T⁴ = 176.37 10⁸
T =[tex]\sqrt[4]{176.37 10^8}[/tex]
T = 3.6442 10² K
Let's reduce to degrees Celsius
T = 364.42 -273.15
T = 91.3 ° C
The correct answer is c
The equilibrium temperature of the hot asphalt, assuming it behaves as a perfect blackbody with emissivity of 1, can be calculated using the Stefan-Boltzmann law, resulting in approximately 91°C, making choice (c) the correct answer.
To determine the equilibrium temperature of the hot asphalt, we can use the concept of blackbody radiation.
Since the emissivity (e) is 1, the asphalt behaves as a perfect blackbody, which means it absorbs and emits radiation efficiently.
The power per unit area absorbed by the asphalt is:
P = 1000 W/m²According to the Stefan-Boltzmann law, the power radiated per unit area by a blackbody is given by:
P = σeT⁴where
σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²·K⁴), and T is the equilibrium temperature in KelvinGiven that e = 1 for a perfect blackbody, we can set the absorbed power equal to the emitted power:
1000 = 5.67 × 10⁻⁸ T⁴Solving for T:
T⁴ = 1000 / 5.67 × 10⁻⁸T⁴ ≈ 1.76 × 10¹⁰T ≈ (1.76 x 10¹⁰)∧1/4T ≈ 278.8 KConverting to Celsius:
T ≈ 278.8 K - 273.15 ≈ 5.65°CThis result does not match any answer choices, which suggests a potential issue. The correct calculations should yield a higher temperature due to an error in an earlier assumption or value misunderstanding. Revisiting the calculations correctly:
Solving again for higher accuracy:
T ≈ (1000 / 5.67 x 10⁻⁸)∧1/4 ≈ 364 K ≈ 91°CTherefore, the correct equilibrium temperature is 91°C, making the correct choice: (c) 91°C
Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis oriented at 0°. A second polarizer has its transmission axis oriented at 45° and a third polarizer oriented with its axis at 90°. Determine the fraction of light intensity exiting the third sheet with and without the second sheet present.
Answers
Answer:
Explanation:
Given
Initial Intensity of light is S
when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.
When it is passed through a second Polarizer with its transmission axis [tex]\theat =45^{\circ}[/tex]
[tex]S_1=S_0\cos ^2\theta [/tex]
here [tex]S_0=\frac{S}{2}[/tex]
[tex]S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}[/tex]
[tex]S_1=\frac{S}{4}[/tex]
When it is passed through third Polarizer with its axis [tex]90^{\circ}[/tex] to first but [tex]\theta =45^{\circ}[/tex] to second thus [tex]S_2[/tex]
[tex]S_2=S_0\cos ^2\theta [/tex]
[tex]S_2=\frac{S}{4}\times \frac{1}{2}[/tex]
[tex]S_2=\frac{S}{8}[/tex]
When middle sheet is absent then Final Intensity will be zero
Final answer:
The first polarizing sheet reduces the intensity of unpolarized light to 50%. The second polarizing sheet further reduces the intensity to 25%. The third polarizing sheet, with the second sheet present, does not allow any light to pass through.
Explanation:
When unpolarized light passes through a polarizing sheet, the intensity of the light reduces by half. The first polarizing sheet reduces the intensity to 50% of the original intensity. The second polarizing sheet, oriented at an angle of 45° to the first sheet, further reduces the intensity by 50%. So, the intensity exiting the second sheet is 25% of the original intensity (50% x 50% = 25%).
However, the third polarizing sheet, oriented at an angle of 90° to the first, does not allow any light to pass through because the transmission axis of the third sheet is perpendicular to the polarization direction of the light. Therefore, the fraction of light intensity exiting the third sheet, with the second sheet present, is 0%.
Blood flows through a section of a horizontal artery that is partially blocked by a deposit along the artery wall. As a hemoglobin molecule moves from the narrow region into the wider region, its speed changes from v2 = 0.800 m/s to v1 = 0.630 m/s. What is the change in pressure, P1 - P2, that it experiences? The density of blood is 1060 kg/m3.
Answers
Answer:
[tex]\Delta P=128.843\ Pa[/tex]
Explanation:
given,
speed of blood = v₂ = 0.800 m/s
v₁ = 0.630 m/s
density of blood = 1060 kg/m³
Atmospheric pressure = 8.89 ✕ 10⁴ N/m²
Using Bernoulli equation
[tex]P_1 - P_2 = \dfrac{1}{2}\rho (v_2^2-v_1^2)[/tex]
[tex]\Delta P= \dfrac{1}{2}\rho (0.8^2-0.63^2)[/tex]
[tex]\Delta P= \dfrac{1}{2}\times 1060\times (0.8^2-0.63^2)[/tex]
[tex]\Delta P=128.843\ Pa[/tex]
the change of pressure is equal to [tex]\Delta P=128.843\ Pa[/tex]
When the daughter nucleus produced in a radioactive decay is itself unstable, it will eventually decay and form its own daughter nucleus. If the newly formed daughter nucleus is also unstable, another decay will occur, and the process will continue until a nonradioactive nucleus is formed. Such a series of radioactive decays is called a decay chain.
A good example of a decay chain is provided by 232 90Th, a naturally occurring isotope of thorium.
What is the energy Q released in the first step of the thorium-232 decay chain? The atomic mass of 232 90Th is 232.038054 u and the atomic mass of 228 88Ra is 228.0301069 u.
Answer in (MeV) and show your work
Answers
Answer:
4.981 MeV
Explanation:
The quantity of energy Q can be calculated using the formula
Q = (mass before - mass after) × c²
Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass between the thorium and radium ( 232 - 228) and ( 90 - 88) show α particle was emitted.
1 u = 931.494 Mev/c²
Q = (mass before - mass after) × c²
Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²
Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²
Q = 0.0053471 u × c²
replace 1 u = 931.494 MeV/ c²
Q = 0.0053471 × c² × (931.494 MeV / c²)
cancel c² from the equation
Q = 0.0053471 × 931.494 MeV = 4.981 MeV
A team of dogs drags a 53.9 kg sled 1.62 km over a horizontal surface at a constant speed. The coefficient of friction between the sled and the snow is 0.234. The acceleration of gravity is 9.8 m/s 2 . Find the work done by the dogs. Answer in units of kJ.
Answers
Answer:
200.24 kJ
Explanation:
[tex]m[/tex] = mass of sled = 53.9 kg
[tex]d[/tex] = distance traveled by the sled = 1.62 km = 1620 m
[tex]\mu[/tex] = Coefficient of friction between sled and snow = 0.234
frictional force acting on the sled is given as
[tex]f = \mu mg[/tex]
[tex]F[/tex] = Applied force by the dogs on the sled
Since the sled moves at constant speed, the force equation for the motion of the sled is given as
[tex]F = f \\F = \mu mg[/tex]
[tex]W[/tex] = Work done by the dogs on the sled
Work done by the dogs on the sled is given as
[tex]W = F d\\W = \mu mg d\\W = (0.234) (53.9) (9.8) (1620)\\W = 200237.64 J\\W = 200.24 kJ[/tex]
The work done by the dogs is 200.238 kJ.
The work done by the dog is equal to the work done to move the sled through the distance and the work done against friction.
Formula:
W = dma+μmgd............. Equation 1Where:
W = work done by the dogsm = mass of the sleda = acceleration of the sledg = acceleration due to gravityμ = coefficient of frictionFrom the question,
Given:
m = 53.9 kga = 0 m/s² (move with constant speed)d = 1.62 km = 1620 mg = 9.8 m/s²μ = 0.234Substitute these values into equation 1
W = (53.9×0×1620)+(53.9×9.8×0.234×1620)W = 2002378 JW = 200.238 kJHence, The work done by the dogs is 200.238 kJ.
Learn more about work done here: https://brainly.com/question/8119756