Answer:
a) the acceleration is a= 2438 m/s²
b) the distance travelled during serve is d = 1.0971 m
Explanation:
a) since
v = vo + a*t ,
where v= velocity at time t , vo= velocity at time t=0 and a= acceleration
,then
a= (v-vo)/t
replacing values
a= (v-vo)/t = (73.14 m/s - 0 m/s)/( 30* 10⁻³ s) = 2438 m/s²
b) the distance travelled d is
v² = vo² + 2*a*d
then
d = (v² - vo²) /(2*a) = (73.14 m/s)² - 0²)/(2*2438 m/s²)= 1.0971 m
a) the acceleration is a= 2438 m/s²
b) the distance travelled during serve is d = 1.0971 m
What is acceleration?Acceleration represents the rate at which velocity should be changed with time, with respect to both speed and direction. Since acceleration contains both a magnitude and a direction, it is a vector quantity.
Calculation of acceleration & distance:a) since
[tex]v = vo + a\times t[/tex]
Here
v= velocity at time t ,
vo= velocity at time t=0
and a= acceleration
Now
[tex]a= (v-vo)\div t\\\\ =(73.14 m/s - 0 m/s)/( 30\times 10^{-3} s)[/tex]
= 2438 m/s²
b) Now the distance traveled d is
[tex]v^2 = vo^2 + 2\times a\times d \\\\d = (v^2 - vo^2) \div (2\timesa) \\\\=(73.14 m/s)^2 - 0^2)\div (2\times 2438 m/s^2)[/tex]
= 1.0971 m
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A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1400 N/m). The block is pulled 10.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?
To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.
From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:
[tex]KE = PE_s +W_f[/tex]
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
Where,
[tex]KE =[/tex] Kinetic Energy
[tex]PE_s =[/tex]Potential Harmonic Simple Energy
[tex]W_f =[/tex] Work made by friction.
Our values are given as,
[tex]m = 10Kg \rightarrow[/tex] mass
[tex]k = 1400N/m \rightarrow[/tex] Spring constant
[tex]A = 0.1m \rightarrow[/tex]Amplitude
[tex]f_f = 30N \rightarrow[/tex] Frictional Force
Replacing we have,
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
[tex]KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)[/tex]
[tex]KE = 4 J[/tex]
Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.
The kinetic energy of the 10-kg block as it passes through its equilibrium position is 4 Joules. This is calculated by converting the potential energy stored in the spring to kinetic energy and then subtracted the energy lost due to friction.
Explanation:The first step in solving this problem is to understand the two forces acting on the block in this question: the spring force and the frictional force. The spring potential energy when the block is pulled 10 cm to the right is given by the formula U = 1/2kx^2, where k is the force constant and x is the displacement. Substituting the given values, we have U = 1/2(1400 N/m)(0.1 m)^2 = 7 Joules. This is the initial potential energy stored in the spring. As the block passes through its equilibrium position, this potential energy is fully converted to kinetic energy.
We also need to take into account the work done against frictional force which is equal to the frictional force times the displacement, i.e., W_friction = Friction * displacement = 30N * 0.1m = 3 Joules. This is the energy lost due to friction.
Finally, we subtract the work done by the frictional force from the potential energy to achieve the kinetic energy. Therefore, the kinetic energy of the block as it passes through its equilibrium position is K = U - W_friction = 7J - 3J = 4 Joules.
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The oscillating current in an electrical circuit is as follows, where I is measured in amperes and t is measured in seconds. I = 7 sin(60πt) + cos(120πt). Find the average current for each time interval. (Round your answers to three decimal places.) (a) 0 ≤ t ≤ 1 60 amps
The average current in the given oscillating current function, I = 7 sin(60πt) + cos(120πt), can be found by integrating each component of the function over the given time interval and taking the average. Due to the complexity of the function, it is recommended to use mathematical software or a calculator with integral computation capability.
Explanation:The given function describes the oscillating current in an electrical circuit: I = 7 sin(60πt) + cos(120πt). The average current, Iave, is conceptually the net charge, ΔQ, that passes through a given cross-sectional area per unit time, Δt. Due to the complexity of this function, we must split it into two integrals to find the average currents.
For the time interval 0 ≤ t ≤ 1/60 seconds, we have two integrals for each component of the current: ∫01/607 sin(60πt) dt and ∫01/60cos(120πt) dt. Calculating these will give us the average current for this time interval. Due to the complexity of the function and the requirement of calculus to solve it, it's recommended to use mathematical software or a calculator with integral computation capability to get the numerical values, always rounding your answers to three decimal places.
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A "Barrel of Fun" consists of a large vertical cylinder that spins about its axis fast enough so that any person inside will be held against the wall. At an angular speed ω1, an upward frictional force F1 holds a person against the wall without slippingIf you double the angular speed (i.e., ω2 = 2 ω1) what is the new friction force F2?
Answer:
Friction force F₂ after doubling the angular speed is same as the friction force at angular speed ω₁
Explanation:
Consider the fig attached below. Forces acted on person are Centripetal force (-mv²/r) exerted in x direction and reversal normal force N wall exerted on person.
[tex]\sum F_{x} =0\\N+ ma_{x}\\-N=m(-\frac{v^{2}}{r})\\-N=m(-r\omega^{2})N=m(r\omega^{2})[/tex] ---(1)
In y direction there is frictional force Fs exerted in upward direction that keeps the person standing without falling which is balanced by weight of person in downward direction.
[tex]\sum F_{y} = 0\\F_{s}-mg=0\\F_{s}=mg[/tex]----(2)
from eq 2 it can e seen that frictional force is equal to weight of person exerted in upward direction, it does not depends on angular speed ω₁. So when the angular speed is doubled i.e ω₂ = 2ω₁, frictional force Fs remains same.
You are baking and run out of an important ingredient. Rather than fire up the car, you decide to use a bicycle. The store is 1/2 mile away. Assume your car gets 30 miles to the gallon. If every driver on the planet made a similar decision once per week for a year, approximately how much gas would be saved for that year? (As this is an estimate, you may make the arithmetic easier by assuming a year has 50 weeks rather than 52 weeks. What number do you use for 'every driver on the planet'?) a. 100 million gallons b. 2 billion gallons c. 20,000 gallons d. 200,000 gallons e. 2 million gallons
Answer: this one is tough. There is a estimated 1.2 billion drivers to help best I could. I did the math the best I could and didn’t get an answer close to the choices. Just guessing I’d guess C. 20,000 because of logic.
Explanation: I can’t help much but would love to hear how to work this out.
You are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 270N .
Part A
How much work do you do on the refrigerator during its trip down the ramp?
W = ???
Final answer:
The work done on the refrigerator during its trip down the ramp is 1350 J.
Explanation:
To calculate the work done while unloading the refrigerator, we need to determine the displacement of the refrigerator and the force applied.
The displacement is given by the length of the ramp, which is 5.0 m.
The force applied is given as 270 N.
To calculate the work done, we use the formula:
Work = Force x Displacement x Cosine(angle)
In this case, the angle is 0 degrees since the force is applied horizontally.
Therefore, the work done on the refrigerator is:
Work = 270 N x 5.0 m x Cos(0°) = 1350 J.
For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equation
1/?= 2p2 mk2 e4/ h3c (Z2) (1/n2f - 1/n2i)
Using this equation, show that the ranges of wavelengths in these two series overlap.
Shortest Wavelength (m) Longest Wavelength (m)
Paschen Series
Final answer:
The Paschen and Brackett series of lines in atomic hydrogen can overlap in terms of their wavelength ranges.
Explanation:
The Paschen series of lines in atomic hydrogen occurs when nf = 3, and the Brackett series occurs when nf = 4.
The range of wavelengths in these two series can overlap because the wavelength equation 1/λ = 2π²mk²e⁴/(h³c)(Z²)(1/nf² - 1/ni²) depends on the values of nf and ni, which can be different for each series
For example, if nf = 3 in the Paschen series and nf = 4 in the Brackett series, the wavelengths in these two series can overlap depending on the values of ni.
A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp?A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp?hoop, sphere, diskhoop, disk, spheredisk, hoop, spheresphere, disk, hoopsphere, hoop, disk
Answer:
First let's write down the moment of inertia of the objects.
[tex]I_{sphere} = \frac{2}{5}mR^2\\I_{disk} = \frac{1}{2}mR^2\\I_{hoop} = mR^2[/tex]
If they all roll without slipping, then the following relation is applied to all ot them:
[tex]v = \omega R[/tex]
where v is the translational velocity and ω is the rotational velocity.
We will use the conservation of energy, because we know that their initial potential energies are the same. (Here, I will assume that all the objects have the same mass and radius. Otherwise we couldn't determine the difference. )
[tex]K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2[/tex]
For sphere:
[tex]\frac{1}{2}\frac{2}{5}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{5}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{7}{10}mv^2 = mgh\\v_{sphere} = \sqrt{\frac{10gh}{7}}[/tex]
For disk:
[tex]\frac{1}{2}\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{4}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{3}{4}mv^2 = mgh\\v_{disk} = \sqrt{\frac{4gh}{3}}[/tex]
For hoop:
[tex]\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mgh\\mv^2 = mgh\\v_{hoop}= \sqrt{gh}[/tex]
The sphere has the highest velocity, so it arrives the bottom first. Then the disk, and the hoop arrives the last.
Explanation:
The moment of inertia can be defined as the resistance to the rotation. If an object has a high moment of inertia, it resist to rotate more so its angular velocity would be lower. In the case of rolling without slipping, the angular velocity and the linear (translational) velocity are related by the radius, so the object with the highest moment of inertia would arrive the bottom the last.
The order in which a sphere, disk, and hoop reach the bottom of an incline when released from the same height is determined by their moments of inertia. The solid sphere arrives first, followed by the disk, and finally the hoop.
Explanation:In the scenario where a uniform disk, a uniform hoop, and a uniform solid sphere are released from the top of an inclined ramp and roll without slipping, the order in which they reach the bottom depends on their moments of inertia and the distribution of mass. The solid sphere has the smallest moment of inertia relative to its mass (I = 2/5 MR²), which means it will accelerate faster than the other shapes and hence get to the bottom first. The uniform disk, with a moment of inertia of I = 1/2 MR², will follow. The uniform hoop has the largest moment of inertia (I = MR²) for a given mass and radius, so it will accelerate the slowest of the three and reach the bottom last.
Therefore, the objects reach the bottom of the ramp in the following order: sphere, disk, hoop.
In an atom, an electron is confined to a space of roughly 10?10 meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty ?p in its momentum?
Answer:
[tex]5.2728\times 10^{-25}\ kgm/s[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\Delta x[/tex] = Uncertainity in position = [tex]10^{-10}\ m[/tex]
[tex]\Delta p[/tex] = Uncertainty in momentum
According to the Heisenberg uncertainity principle we have
[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta p=\dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{4\pi\times 10^{-10}}\\\Rightarrow \Delta p=5.2728\times 10^{-25}\ kgm/s[/tex]
The minimum uncertainty in its momentum is [tex]5.2728\times 10^{-25}\ kgm/s[/tex]
The free-fall acceleration on Mars is 3.7 m/s2.
(a) What length of pendulum has a period of 1.2 s on Earth? cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm An object is suspended from a spring with force constant 10 N/m.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars.
Answer:
(a) The length of the pendulum on Earth is 36.8cm
(b) The length of the pendulum on Mars is 13.5cm
(c) Mass suspended from the spring on Earth is 0.37kg
(d) Mass suspended from the spring on Mars is 0.36kg
Explanation:
Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2
( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm
(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m
(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg
(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg
The length of a pendulum with a period of 1.2 s on Earth is approximately 36.95 cm, while on Mars it is around 16.99 cm. The mass suspended from a spring that would result in a period of 1.2 s on Earth is approximately 0.722 kg, and on Mars it is approximately 0.329 kg.
Explanation:(a) What length of pendulum has a period of 1.2 s on Earth? cm
Using the equation for the period of a pendulum, T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity, we can solve for L. Rearranging the equation, we have L = (T/2π)² * g.
Given that the free-fall acceleration on Earth is approximately 9.8 m/s², substituting the values into the equation, we have:
L = (1.2/2π)² * 9.8 = 0.3695 m = 36.95 cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm
Using the same equation, L = (T/2π)² * g, we can substitute the values for the period and acceleration due to gravity on Mars:
L = (1.2/2π)² * 3.7 = 0.1699 m = 16.99 cm.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
For a spring-mass system, the period is given by T = 2π √(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we have m = (T/2π)² * k.
Given that the spring constant is 10 N/m, substituting the values into the equation, we have:
m = (1.2/2π)² * 10 = 0.722 kg.
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars. kg
Using the same equation, m = (T/2π)² * k, we can substitute the values for the period and spring constant:
m = (1.2/2π)² * 10 = 0.329 kg.
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal.
At its highest point in this trajectory, the velocityof the projectile would be what?
Answer:
169.4 m/s
Explanation:
Given that the angle of projectile is θ_1 =450°
The speed of body at maximum height is U cosθ_1 = 150 m/s
The angle in second trail is θ_2 =37°
From the given data U cosθ_1 = 150 m/s
U = 150 m/s / cosθ_1
= 150m/s / cos45°
=212.13 m/s
The velocity of the projectile at maximum height in second trail= Ucos(θ_2)
=212.13 m/s×cos37°
=169.4 m/s
The velocity of a projectile at its highest point when fired at an angle to the horizontal remains the same if the initial speed is unchanged, regardless of the angle. Hence, even when changing the angle from 45 to 37 degrees, the velocity at the highest point would still be 150 m/s.
The student's question deals with the velocity of a projectile at the highest point in its trajectory. When a projectile is fired upward at an angle, its velocity at the highest point of its trajectory is only composed of the horizontal component because the vertical component of the velocity becomes zero at that point.
In the given scenario, when the projectile is shot at 45 degrees to the horizontal, the speed at the highest point is given as 150 m/s. This speed represents the horizontal component since there's no vertical component at the highest point. When the angle is changed to 37 degrees, the horizontal component of the initial velocity is calculated using the cosine component of the initial speed, hence the velocity at the highest point remains the same as when it was fired at 45 degrees, provided that the initial speed is unchanged.
Therefore, in the second trial, with the angle at 37 degrees, the velocity of the projectile at its highest point would still be 150 m/s, because the horizontal component of the velocity is not affected by the change in angle.
A steam of hot dry nitrogen flows through a process unit that contains liquid acetone. A substantial part of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the unit at 205 ◦C and 1.1 bar, are compressed and then enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at 10 ◦C and 40 bar. The partial pressure of acetone in the feed to the condenser is 0.100 bar, and that in the effluent gas from the condenser is 0.379 bar. Assume ideal gas behavior.
(a) Calculate, for a basis of m3 of gas fed to the condenser, the mass of ace- tone condensed (kg) and the volume of gas leaving the condenser (m3).
(b) Suppose the volumetric flow rate of the gas leaving the condenser is 20.0 m3/h. Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.
The problem is solved by applying the Ideal Gas Law and Dalton's Law of Partial Pressures. First, we calculate the mass of acetone condensed using these laws and then determine the rate at which acetone is vaporized using the given volumetric flow rate.
Explanation:The given problem involves a number of gas law principles, but its main focus is on the application of the Ideal Gas Law and Dalton's Law of partial pressures. Initially, we calculate the moles of acetone in the feed using the Ideal Gas Law, and then we find out the moles of acetone in the effluent using Dalton's law. Subtracting gives us the moles of acetone condensed.
(a) Using Ideal Gas Law, we have PV=nRT. Hence, n (acetone, feed) = P (acetone, feed) * V(feed) /RT(feed). To find the moles of acetone in the effluent, we use Dalton's law and the Ideal Gas Law to get n (acetone, effluent) = P(acetone, effluent) * V (effluent) / RT (effluent). Subtracting moles in effluent from moles in feed gives moles condensed. Multiplying by the molar mass of acetone, we get mass of acetone condensed.
(b) The question tells us the volumetric flow rate of the gas leaving the condenser. Therefore, number of moles of acetone vaporized per hour can be calculated using the Ideal Gas Law and then can be converted into mass by using the molar mass of acetone. Hence, rate at which acetone is vaporized = moles (acetone, vaporized per hour) * molar mass (acetone).
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An electromagnetic wave is transporting energy in the negative y direction. At one point and one instant the magnetic field is in the positive x direction. The electric field at that point and instant is:Why is (D) the answer? I thought it's (C)A) positive y directionB) negative y directionC) positive z directionD) negative z direction
Answer:negative Z direction
Explanation:
It is given that EM wave is travelling in negative y direction i.e. -[tex]\hat{j}[/tex]
Magnetic Field is in positive x direction i.e. [tex]\hat{i}[/tex]
We know that Electric field and magnetic field are perpendicular to each other and they are mutually perpendicular to direction of wave propagation.
Mathematically if Electric field is in negative z direction it will yield the direction of wave propagation
[tex]=(-\hat{k})\times \hat{i}[/tex]
[tex]=-\hat{j}[/tex]
What is the object’s velocity, in meters per second, at time t = 2.9? Calculate the object’s acceleration, in meters per second squared, at time t = 2.9. What is the magnitude of the object’s maximum acceleration, in meters per second squared? What is the magnitude of the object’s maximum velocity, in meters per second?
Answer:
the question is incomplete, below is the complete question
"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.
a.What is the object's velocity, in meters per second, at time t = 2.9?
b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.
c. What is the magnitude of the object's maximum acceleration, in meters per second squared?
d.What is the magnitude of the object's maximum velocity, in meters per second?"
a.[tex]v(t)==24.1m/s[/tex]
b.[tex]a(t)=3.79m/s^{2}[/tex]
c.[tex]a_{max}=106.48m/s^{2}[/tex]
d.[tex]v_{max}=24.2m/s[/tex]
Explanation:
the gneral expression for the displacement of object in simple harmonic motion is represented by
[tex]x(t)=Acos(wt- \alpha)\\[/tex]
while the velocity is express as
[tex]v(t)=-Awcos(4.4t-1.8)\\[/tex]
and the acceleration is
[tex]a(t)=-aw^{2}cos(wt- \alpha )\\[/tex]
Note: the angle is in radians
The expression for the displacement from the question is [tex]x(t)=5.5cos(4.4t-1.8)\\[/tex]
comparing, A=5.5, w=4.4,α=1.8
a.To determine the object velocity at t=2.9secs,
we substitute for t in the velocity equation
[tex]v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\[/tex]
[tex]v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s[/tex]
b.To determine the object acceleration at t=2.9secs,
we substitute for t in the acceleration equation
[tex]a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\[/tex]
[tex]a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}[/tex]
c. The acceleration is maximum when the displacement equals the amplitude. hence magnitude of the object acceleration is
[tex]a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}[/tex]
d.The maximum velocity is expressed as
[tex]v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s[/tex]
A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictionless surface. Let the initial position of the object be defined as x-0. The object is pulled to position x = A and then released, after which it undergoes simple harmonic motion. In one full cycle of its motion, the total distance traveled by the object is 2 4 O 4A
Answer:
The total distance traveled by an object attached to a spring that is pulled to position x=A and then released is 4 A.
Explanation:
We have an small object attached to a relaxed spring whose opposite end is fixed. The spring rests on a frictionless surface. This means the only force acting on the object is the elastic force of the spring, a conservative force, since the weight and the normal force compensate between them.
The initial position of the object is x=0 then is pulled to position x=A and released. After which it undergoes a simple harmonic motion with an amplitude A. From the position x=A to the equilibrium position in x=0 the object travels a distance A. From the equilibrium position x=0 to maximum negative position in x= -A the object travels again a distance A. Then to return to the original position the object should travel a distance 2 A in reverse direction.
In one full cycle of its motion the object travels a distance 4 A.
All Houston Methodist buildings system wide have an emergency power generator that turns on to supply emergency power after normal power shuts down within:
A. 5 seconds
B. 30 seconds
C. 60 seconds
D. 10 seconds
Answer:
(D) 10 seconds
Explanation:
The Houston Methodist Hospital automated the emergency power supply system (EPSS) testing processes in order to ensure the safety of patients during an outage
There is an emergency power generator that turns on to supply emergency power after normal power shuts down within 10 seconds.
Answer:
D. 10 seconds
Explanation:
The Houston Methodist Hospital has eleven diesen powered generators that turn on at most 10 seconds after the normal power shuts down.
Considering that it is a hospital, this system is really important.
So the correct answer is:
D. 10 seconds
The hydrofoil boat has an A-36 steel propeller shaft that is 100ft long. It is connected to an inline diesel engine that delivers a maximum power of 2500 hp and causes the shaft to rotate at 1700 rpm. If the outer diamater of the shaft is 8 in. and the wall thickness is 3/8 in., determine the maximum sheer stress developed in the shaft. Also, what is the "wind up," or angle of twist in the shaft at full power?
Final answer:
The maximum shear stress developed in the shaft is approximately 7.57 MPa. The angle of twist in the shaft at full power is approximately 3.00°.
Explanation:
To determine the maximum shear stress developed in the shaft, we can use the formula:
Shear stress (τ) = Torque (T) / Polar Moment of Inertia (J)
To find the torque, we can use the formula:
Torque (T) = Power (P) / Angular Velocity (ω)
We are given that the power is 2500 hp and the angular velocity is 1700 rpm. Converting these values to W and rad/s respectively, we have:
Power (P) = 2500 hp * 745.7 W/hp ≈ 1,864,250 W
Angular Velocity (ω) = 1700 rpm * 2π rad/minute ≈ 17897.37 rad/s
Substituting these values into the torque formula, we have:
Torque (T) = 1,864,250 W / 17897.37 rad/s ≈ 103.98 N*m
Next, we need to find the polar moment of inertia (J) of the shaft. The polar moment of inertia for a solid shaft can be calculated using the formula:
J = (π/2) * (outer diameter^4 - inner diameter^4)
Converting the diameter to meters, we have:
Diameter (d) = 8 in * 0.0254 m/in = 0.2032 m
Substituting this value into the polar moment of inertia formula, we have:
J = (π/2) * (0.2032^4 - (0.2032 - 2 * 3/8)^4)
= (π/2) * (0.2032^4 - 0.1926^4)
0.013743 m^4
Finally, we can calculate the maximum shear stress using the formula:
Shear stress (τ) = 103.98 N*m / 0.013743 m^4 ≈ 7.57 MPa
As for the angle of twist, we can use the formula:
Angle of twist (θ) = T * L / (G * Given that the length of the shaft is 100 ft and Young's modulus (G) for A-36 steel is approximately 200 GPa, we can calculate the angle of twist as follows:
Angle of twist (θ) = 103.98 N*m * 100 ft * 0.3048 m/ft / (200 GPa * 0.013743 m^4)
≈ 0.0524 radians or 3.00°
(FLUID MECHANICS)
The apparent weight of an object when a system is not in equilibrium is represented by:
A. Fg (object)
B. Fnet (system)
C. Fb
D. rhoV
Answer:
option B
Explanation:
When a body is immersed in liquid there will be two force is acting on the body.
First one force acting downward due to weight of the body.
And the second force acting on the object will be buoyant force.
If the object is not in equilibrium the apparent weight will be equal to net force acting on the object.
[tex]F_{net} = W - F_b[/tex]
W is the weight of the object acting downward
Fb is the buoyancy force acting upward on the object.
Hence, the correct answer is option B
A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder and, using a triple beam balance, finds the mass to be 3.21 kg. The density of this particular alloy of brass is 8.62 g/cm 3 . The student ties a massless string to one end of the cylinder and submerges it into a tank of water where there is an apparent reduction in the weight of the cylinder. With this information, calculate the volume, V , of the cylinder and the tension, T , in the string when it is submerged in the tank of water. The density of water is 1.00 g/cm 3 , and the acceleration due to gravity is gVolume of the cylinder: _____ cm3Tension in the string: ______ N
Answer:
V = 0.3724 m³
T = 27.836 N
Explanation:
Given :
m = 3.21 kg , W= 3.21 * 9.81 m / s² = 31.4901 N
ρ = 8.62 g / cm ³ = 8620 kg / m³
V = m / ρ = 3.21 kg / 8620 kg / m³
V = 0.3724 m³
when submerged the weight of brass cylinder is equal to the tension in string:
F = (0.3724m³) * (1000 kg / m³) * (9.81 m/s²²) = 3.653 ≈ 3.65 N
T = 31.4901 N - 3.65 N
T = 27.836 N
For a new TV series "Stupidity Factor contestants are dropped into the ocean (p 1030 kg/ms) along with a Styrofoam soo kg/ms block that is mm x 3 mm X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.
Answer:
Explanation:
Heres the correct and full question:
For a new TV series "Stupidity Factor contestants are dropped into the ocean (p=1030 kg/m³) along with a Styrofoam p=3oo kg/m³ block that is 2m x 3 m X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.
answer:
volume of styrofoam block=V=2m x 3m x 0.20m =1.2m³
density of styrofoam=ρs=300kg/m³
mass of styrofoam=ms=v(ρs)=1.2 x 300=360kg
weight of styrofoam = ws=(ms)g=360 x 9.8=3528N
consider number of contestants= n
toatal weight of ccontestants=W=n(70 x 9.8)=n(686N)
since, styrofoam fully submerged into water, then bat force,
B=ρ(vs)g=1030 x 1.2 x 9.8 = 12112.8N
At equilibrium,
B - W - Ws = 0
12112.8 - 686n - 3528 = 0
[tex]n=\frac{12112.8-3528}{686}=12.5[/tex]
n=12 person(contestants)
A simple piping system consisting of a small pipe (diameter 6 cm) connected to a larger pipe (diameter 30 cm) is used to transfer and magnify force from one end to the other. The system is full of water and completely enclosed by two pistons, one on each end. If a force 11.2 N is exerted (additional to the forces already present) on the small piston, then how much additional force is exerted at the larger piston
Answer:
280N
Explanation:
Pascal's law states that the pressure in a fluid is transmitted across every point in the fluid system.
Hence, the pressure in both tubes must remain same;
So, pressure = F1/A1 = F2/A2
where F1 = initial force on small pipe, A1 = area of small pipe
F2 = force on larger piston, A2 = area of larger piston
Given:
F1 = 11.2N
D1 = diameter of smaller pipe = 6 cm
D2 = diameter of larger piston = 15 cm
A1 = π*(r1)² = π*(6/2)² =π*9 (As radius, r = diameter/2)
A2 = π*(r2)² = π*(30/2)² =π*225
Hence 11.2/π*9 = F2/π*225
solving, we have
F2 = 280N
Billiard ball A of mass mA = 0.117 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.135 kg . As a result of the collision, ball A is deflected off at an angle of θ′A = 30.0∘ with a speed v′A = 2.10 m/s, and ball B moves with a speed v′B at an angle of θ′B to original direction of motion of ball A.Part A Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the x direction.A. 0=mAv′Asinθ′A−mBv′Bsinθ′BB. mAvA=mAv′Acosθ′A+mBv′Bcosθ′BC. mAvA=mAv′Acosθ′A−mBv′Bsinθ′BD. 0=(mAvA+mBv′B)sinθ′BPart B Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the y direction.A. mAvA=mAv′Acosθ′A−mBv′Bsinθ′BB. 0=(mAvA+mBv′B)sinθ′BC. 0=mAv′Asinθ′A−mBv′Bsinθ′BD. mAvA=mAv′Acosθ′A+mBv′Bcosθ′BPart C Solve these equations for the angle, θ′B , of ball B after the collision. Do not assume the collision is elastic.Part D Solve these equations for the speed, v′B , of ball B after the collision. Do not assume the collision is elastic.
Answer:
Part A:
B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
Part B:
C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
since vAy = 0 m/s
Part C:
θ′B = tan⁻¹(1.0699) = 46.94°
Part D:
v′B = 1.246 m/s
Explanation:
Given:
mA = 0.117 kg
vA = vAx = 2.80 m/s
mB = 0.135 kg
vB = 0 m/s
θ′A = 30.0°
v′A = 2.10 m/s
Part A: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the x direction is
B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
Part B: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the y direction is
C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
since vAy = 0 m/s
Part C: Solving these equations for the angle, θ′B , of ball B after the collision and assuming that the collision is not elastic:
mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
⇒ (0.117)(2.80) = (0.117)(2.10)Cos 30° + (0.135)*v′B*cosθ′B
⇒ v′B*cosθ′B = 0.8505 ⇒ v′B = 0.8505/cosθ′B
then
0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
⇒ 0 = (0.117)(2.10)Sin 30° - (0.135)*v′B*sinθ′B
⇒ v′B*sinθ′B = 0.91 ⇒ v′B = 0.91/sinθ′B
if we apply
0.8505/cosθ′B = 0.91/sinθ′B
⇒ tanθ′B = 0.91/0.8505 = 1.0699
⇒ θ′B = tan⁻¹(1.0699) = 46.94°
Part D: Solving these equations for the speed, v′B , of ball B after the collision and assuming that the collision is not elastic:
if v′B = 0.91/sinθ′B
⇒ v′B = 0.91/sin 46.94°
⇒ v′B = 1.246 m/s
Conservation of momentum can be used to solve the equations for the angle and the speed of ball B after the collision, using the initial conditions, final conditions, and trigonometric identities. This involves the application of physics concepts, combined with the mathematics of trigonometry.
Explanation:Part A: The correct equation expressing the conservation of momentum for the components in the x direction would be option B, mAvA=mAv′Acosθ′A+mBv′Bcosθ′B. This equation shows that the initial momentum of ball A (mAvA) equals the sum of the momentum of ball A and ball B after the collision in the x direction.
Part B: For the components in the y direction, the right answer is C, 0=mAv′Asinθ′A−mBv′Bsinθ′B. Since ball B was initially at rest and ball A was moving along the x-axis, there was no momentum in the y direction before the collision. Therefore, the total momentum in the y direction after the collision should also be 0.
Part C and D: To solve these equations for the angle and the speed of ball B after the collision, you need to use these equations in combination with the conservation of kinetic energy formula (1/2*m*v^2) and the trigonometric identities. Detailed solution steps require knowledge of the involved mathematics.
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A 1.6-kg block is attached to the end of a 2.0-m string to form a pendulum. The pendulum is released from rest when the string is horizontal. At the lowest point of its swing when it is moving horizontally, the block is hit by a 10-g bullet moving horizontally in the opposite direction. The bullet remains in the block and causes the block to come to rest at the low point of its swing. What was the magnitude of the bullet's velocity just before hitting the block?
Answer:
1002.2688 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
h = The length of a string = 2 m
m = Mass of block = 1.6 kg
[tex]m_2[/tex] = Mass of bullet = 0.01 kg
Here, the potential energy of the fall will balance the kinetic energy of the bullet
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 2}\\\Rightarrow v=6.26418\ m/s[/tex]
Velocity of block is 6.26418 m/s
As the momentum of system is conserved we have
[tex]mv=m_2u\\\Rightarrow u=\dfrac{mv}{m_2}\\\Rightarrow u=\dfrac{1.6\times 6.26418}{0.01}\\\Rightarrow u=1002.2688\ m/s[/tex]
The magnitude of velocity just before hitting the block is 1002.2688 m/s
A car of mass 900 kg is traveling at 20 m/s when the brakes are applied. The car then comes to a complete stop in 5 s. What is the average power that the brakes produce in stopping the car?
A. 36,000 W
B. 7200 W
C. 3600 W
D. 1800 W
Answer:
A. 36,000 W
Explanation:
[tex]m[/tex] = mass of the car = 900 kg
[tex]v_{o}[/tex] = Initial speed of the car = 20 m/s
[tex]v_{f}[/tex] = Final speed of the car = 0 m/s
[tex]W[/tex] = Work done by the brakes on the car
Magnitude of work done on the car by the brakes is same as the change in kinetic energy of the car.hence
[tex]W = (0.5) m (v_{o}^{2} - v_{f}^{2})\\W = (0.5) (900) ((20)^{2} - (0)^{2})\\W = 180000 J[/tex]
[tex]t[/tex] = time taken by the car to come to stop = 5 s
[tex]P[/tex] = Average power produced by the car
Average power produced by the car is given as
[tex]P = \frac{W}{t} =\frac{180000}{5} \\P = 36000 W[/tex]
You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.1 s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward.
(a) Use the momentum principle to determine the speed that the ball had just AFTER it left your hand.
vinitial = ?? m/s
(b) Use the Energy Principle to determine the maximum height above your hand reached by the ball.
h = ?? m
Answer:
10.791 m/s
5.93505 m
Explanation:
m = Mass of ball
[tex]v_f[/tex] = Final velocity
[tex]v_i[/tex] = Initial velocity
[tex]t_f[/tex] = Final time
[tex]t_i[/tex] = Initial time
g = Acceleration due to gravity = 9.81 m/s²
From the momentum principle we have
[tex]\Delta P=F\Delta t[/tex]
Force
[tex]F=mg[/tex]
So,
[tex]m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s[/tex]
The speed that the ball had just after it left the hand is 10.791 m/s
As the energy of the system is conserved
[tex]K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m[/tex]
The maximum height above your hand reached by the ball is 5.93505 m
The speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.
What is momentum principal?When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the principal of momentum.
The net force using the momentum principle can be given as,
[tex]F_{net}=\dfrac{\Delta P}{\Delta t}[/tex]
Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. Therefore, the above equation for initial and final velocity and time can be written as,
[tex]F_{net}=\dfrac{m(v_f-v_i)}{(t_f-t_i)}[/tex]
(a) The speed that the ball had just AFTER it left your hand-The mass of the ball is 1 kg, and it takes 2.2 s to go up and down. The net force acting on the body is mass times gravitational force.
Therefore, the above equation can be written as,
[tex]mg=\dfrac{m(v_f-v_i)}{(t_f-t_i)}\\g=\dfrac{(v_f-v_i)}{(t_f-t_i)}\\[/tex]
As the final velocity is zero and initial time is also zero. Therefore,
[tex](-9.81)=\dfrac{(0-v_i)}{(1.1-0)}\\v_i=10.791 \rm m/s[/tex]
(b) The maximum height above your hand reached by the ball-Using the energy principle, we can equate the kinetic energy of the system to the potential energy of the system as,
[tex]\dfrac{1}{2}mv_i^2=mgh\\\dfrac{1}{2}(10.791)^2=(9.81)h\\h=5.94\rm m[/tex]
Thus, the speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.
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Question:
Jane, looking for Tarzan, is running at top speed (4.5 m/s) and grabs a vine hanging vertically from a tall tree in the jungle.
How high can she swing upward?
Energy Conservation
According to the conservation of the mechanical energy the total energy remains constant of the system, therefore the kinetic energy of the system will get converted into the potential energy of the system.
Answer:
She can swing 1.0 m high.
Explanation:
Hi there!
The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).
The kinetic energy is calculated as follows:
KE = 1/2 · m · v²
And the potential energy:
PE = m · g · h
Where:
m = mass of Jane.
v = velocity.
g = acceleration due to gravity (9.8 m/s²).
h = height.
Then:
ME = KE + PE
Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:
ME = KE + PE (PE = 0)
ME = KE
ME = 1/2 · m · (4.5 m/s)²
ME = m · 10.125 m²/s²
When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:
ME = KE + PE (KE = 0)
ME = PE
ME = m · 9.8 m/s² · h
Then, equallizing both expressions of ME and solving for h:
m · 10.125 m²/s² = m · 9.8 m/s² · h
10.125 m²/s² / 9.8 m/s² = h
h = 1.0 m
She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).
Which of the following statements concerning the nuclear force is false? O The nuclear force is attractive and not repulsive. O The nuclear is one of only four known types of forces in nature. O The nuclear force is very short-ranged. O The nuclear force acts on both protons and neutrons. O The nuclear force is very weak and much smaller in relative magnitude than the electrostatic and gravitational forces. Submit Request
Answer:
The nuclear force is attractive and not repulsive.The nuclear force is very weak and much smaller in relative magnitude than the electrostatic and gravitational forces.Explanation:
Nuclear force is the strongest existing force in the nature. It has the shortest range.Its main function is to hold the subatomic particles together in nature.The nuclear force is created by the exchange of pi mesons between the nucleons of an atom, but for this exchange to happen the particles must be close to one another of the order of few femtometer.At about 1 femtometer the nuclear force is very strongly attractive in nature but at distance greater than 2.5 femtometer it fades away. The force becomes repulsive in nature at distance less than 0.7 femtometer.This force holds the likely charged protons together in the nucleus.How much heat energy is required to convert 93.4 g of solid ethanol at − 114.5 ° C to gasesous ethanol at 149.8 ° C ? The molar heat of fusion of ethanol is 4.60 kJ/mol , and its molar heat of vaporization is 38.56 kJ/mol . Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C . The specific heat capacity of liquid ethanol is 2.45 J / g ⋅ ° C , and that of gaseous ethanol is 1.43 J / g ⋅ ° C .
Answer:
Q' = 140.859 kJ
Explanation:
Given that, 93.4 g of solid ethanol at − 114.5 °C is converted to gasesous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol , and its molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Lets solve this step wise ;
Given 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given = [tex]\frac{93.4}{46.06}[/tex] = 2.02 moles
step 1: solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ heat required = 4.60 × 2.02 = 9.292 kJ.
step 2: liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q = m×S×ΔT ; Q = heat required
m = mass of the substance
S = specific heat of the substance
ΔT = change in temperature
Here S = S(l);
⇒ Q = 93.4×2.45×(78.4-(-114.5))
= 44.141 kJ
step 3: liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
⇒ required heat = 38.56×2.02 = 77.89 kJ
step 4: gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q = m×S×ΔT
Here, S = S(g)
Q = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
⇒ Q' = 140.859 kJ
The total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
Given data:,
93.4 g of solid ethanol at − 114.5 °C is converted to gaseous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol.
Molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Since, 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given is
⇒93.04/46.06 = 2.02 moles
Heat required for conversion of solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ Q = 4.60 × 2.02 = 9.292 kJ.
Heat required to convert liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q' = m×s×ΔT
Here,
m = mass of the substance
s = specific heat of the substance
ΔT = change in temperature
Solving as,
Q' = 93.4×2.45×(78.4-(-114.5))
Q' = 44.141 kJ
Heat required to convert liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
Q'' = 38.56×2.02 = 77.89 kJ
Heat required to convert gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q''' = m×S×ΔT
Q''' = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
Thus, we can conclude that the total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
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Interactive Solution 9.63 illustrates one way of solving a problem similar to this one. A thin rod has a length of 0.620 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.185 rad/s and a moment of inertia of 1.43 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?
Answer:
w = 7.89 10⁻² rad/s
Explanation:
We will solve this exercise with the conservation of the annular moment, let's write it in two moments
Initial. With the insect in the center
L₀ = I w₀
End with the bug on the edge
[tex]L_{f}[/tex]= I w + [tex]I_{bug}[/tex] w
The moments of inertia are
For a rod
I = 1/3 M L²
For the insect, taken as a particle
I = m L²
The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved
L₀ = [tex]L_{f}[/tex]
I w₀ = I w + [tex]I_{bug}[/tex] w
w = I / (I + [tex]I_{bug}[/tex]) w₀
w = I / (I + m L²) w₀
Let's calculate
w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)² 0.185
w = 1.43 10⁻³ / 3.352 10³ 0.185
w = 7.89 10⁻² rad/s
An object is 12 m long, 0.65 m wide, and 13 cm high. Calculate the volume of this object.
The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.
Mathematically the volume of this body is given as
[tex]V = lWh[/tex]
Where,
L = Length
W = Width
H = High
[tex]V = (12)(0.65)(13*10^{-2})[/tex]
[tex]V = 1.014m^3[/tex]
Note: The value given for the height was in centimeters, so it was transformed to meters.
A spring that is stretched 23 cm from its equilibrium point experiences a force of 103 N.a.) How much energy is stored in the spring?
Answer:
U = 11.85 J
Explanation:
given,
spring is stretched = 23 cm
x = 0.23 m
Force experiences = 103 N
we know,
F = k x
where k is the spring constant
[tex]k =\dfrac{F}{x}[/tex]
[tex]k =\dfrac{103}{0.23}[/tex]
k = 447.83 N/m
energy stored in the spring
[tex]U =\dfrac{1}{2}kx^2[/tex]
[tex]U =\dfrac{1}{2}\times 447.83 \times 0.23^2[/tex]
U = 11.85 J
hence, energy stored in the spring is equal to U = 11.85 J
The elastic potential energy stored in the spring when stretched to a length of 23 cm, given a spring constant of 4 N/cm, is 0.18 J.
Explanation:The question pertains to how much energy is stored in a spring when it is stretched 23 cm from its equilibrium point, where it experiences a force of 103 N. The potential energy stored in the spring can be calculated using the formula U = 1/2kx². Given that the spring constant is 4 N/cm and the displacement of the spring from its unstretched length is 3 cm (since the unstretched length is 20 cm), the potential energy can be calculated as follows: U = 1/2 * 4 N/cm * (3 cm)² = 0.18 J. Thus, the elastic potential energy contributed by the spring when it is stretched to a length of 23 cm is 0.18 J.