we have that from the Question, it can be said that distance x from the left-hand end of the bar is
x=1.834m
From the Question we are told
A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position?
Generally the equation for Balanced torque is mathematically given as
T=0.3/5*3
T=0.18kg
Therefore
The clockwise torque is
T_c=0.18*1.5g
And
The anti-clockwise torque is
T=0.3/5*2g
T=0.12
Hence
0.18*1.5g=0.3/5*2g+0.9kxg
Therefore
x=1.834m
Therefore
distance x from the left-hand end of the bar is
x=1.834m
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The position of the 0.9 kg mass to balance the metal bar is 1.83 m.
The given parameters;
length of the metal bar, L = 5 mmass of the metal bar, m = 0.3 kgpivot distance = 2 mThe center of gravity of the metal bar = 2.5 m
A sketch of weight on the metal bar;
|-----------2 m-----------|--- 0.5 m-----|
0---------------------------Δ--------------------------------------------------
P ↓|-------- x----------| ↓
0.9 kg 0.3 kg
Take moment about the pivot point;
0.9x = 0.3(0.5)
0.9x = 0.15
[tex]x = \frac{0.15}{0.9} \\\\x = 0.167 \ m[/tex]
2 m - P = 0.167 m
P = 2m - 0.167 m
P = 1.83 m
Thus, the position of the 0.9 kg mass to balance the metal bar is 1.83 m.
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A piece of plastic has a net charge of +9.9 μC. How many more protons than electrons does this piece of plastic have? (e = 1.60 x 10^-19 C). (Divide your answer by 10^13, and give the result to the nearest 0.01
Answer:
The number of protons 6.19 more than electron.
Explanation:
Given that,
Charge [tex]Q=+9.9\times10^{-6}\ C[/tex]
We know that,
formula of charge
[tex]Q = nc[/tex]
Where,
Q = total charge
n = number of protons
e = charge of electron
Put the value into the formula
[tex]n = \dfrac{Q}{e}[/tex]
[tex]n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}[/tex]
[tex]n=6.1875\times10^{13}\ C[/tex]
According to statement of question
Divide the answer by [tex]10^{13}[/tex]
[tex]n=\dfrac{6.1875\times10^{13}}{10^{13}}[/tex]
[tex]n=6.19\ C[/tex]
Hence, The number of protons 6.19 more than electron.
A pitcher claims he can throw a 0.146-kg baseball with as much momentum as a 2.70-g bullet moving with a speed of 1.50 ✕ 103 m/s. (a) What must the baseball's speed be if the pitcher's claim is valid? (b) Which has greater kinetic energy, the ball or the bullet? a. The bullet has greater kinetic energy. b. The ball has greater kinetic energy. c. Both have the same kinetic energy.
(a) The pitcher must throw the ball at 27.7 m/s
The momentum of an object is given by:
[tex]p=mv[/tex]
where
m is the mass of the object
v is the object's velocity
Let's calculate the momentum of the bullet, which has a mass of
m = 2.70 g = 0.0027 kg
and a velocity of
[tex]v=1.50\cdot 10^3 m/s[/tex]
Its momentum is:
[tex]p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s[/tex]
The pitcher must throw the baseball with this same momentum. The mass of the ball is
m = 0.146 kg
So the velocity of the ball must be
[tex]v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s[/tex]
So, the pitcher must throw the ball at 27.7 m/s.
(b) a. The bullet has greater kinetic energy
The kinetic energy of an object is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where m is the mass of the object and v is its speed.
For the bullet, we have:
[tex]K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J[/tex]
For the ball:
[tex]K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J[/tex]
So, the bullet has greater kinetic energy.
Final answer:
The baseball's speed must be 27.74 m/s to match the momentum of the bullet, and the bullet has greater kinetic energy than the baseball.
Explanation:
Calculating Baseball's Speed and Comparing Kinetic Energies
To validate the pitcher's claim that a 0.146-kg baseball can have the same momentum as a 2.70-g bullet traveling at 1.50 x 10³ m/s, we must use the formula for momentum (p = mv), which gives us:
Momentum of bullet = (2.70 g) x (1.50 x 10³ m/s) = (0.0027 kg) x (1500 m/s) = 4.05 kg m/s.
Therefore, the baseball's speed v needed to have the same momentum is calculated by rearranging the formula to v = p/m, which gives us:
Baseball's speed v = 4.05 kg m/s / 0.146 kg = 27.74 m/s.
To determine which has the greater kinetic energy, we use the kinetic energy formula KE = (1/2)mv². Calculating the kinetic energy of both the bullet and the baseball:
KE of bullet = (1/2)(0.0027 kg)(1500 m/s)² = 3037.5 J.
KE of baseball = (1/2)(0.146 kg)(27.74 m/s)² = 55.90 J.
Comparing the results shows that the bullet has greater kinetic energy than the baseball.
Give both a written definition and a mathematical equation for torque.
Answer:
Explanation:
When two equal and opposite force acting at a line of action, they form a couple. It always gives a turning effect to the body.
The turning force is said to be torque .
Torque is the product of either force and the perpendicular distance between the forces.
Torque = force x perpendicular distance
Its SI unit is Nm. It is a vector quantity.
Torque is a physical quantity that measures the effectiveness of a force in changing or accelerating a rotation. The equation for the magnitude of torque is t = rF sin θ.
Explanation:Torque is a physical quantity that measures the effectiveness of a force in changing or accelerating a rotation. It incorporates the magnitude, direction, and point of application of the force. The equation for the magnitude of torque is given by t = rF sin θ, where t is torque, r is the distance from the pivot point to the point of force application, F is the magnitude of the force, and θ is the angle between the force and the lever arm.
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A boat that travels with constant speed of 6.10 m/s in still water is to go directly across a river. The current in the river flows at 1.95 m/s. (a) At what angle must the boat be steered?
Explanation:
The boat's velocity relative to the water is 6.10 m/s, and the water's velocity relative to the shore is 1.95 m/s. We want the boat's velocity relative to the shore to be perpendicular to the shore.
If we say that θ is the angle the boat makes with the shore, then:
6.10 cos θ - 1.95 = 0
6.10 cos θ = 1.95
cos θ = 1.95 / 6.10
θ = 71.4°
The boat should be steered 71.4° relative to the shore, or 18.6° relative to the vertical.
The angle at which the boat must be steered can be determined by using the trigonometric function sine (sin), taking the river current speed as the opposite side and the boat's speed in still water as the hypotenuse in a right triangle. Calculate θ = arcsin(1.95 m/s / 6.10 m/s) to find the angle.
Explanation:To find the angle the boat must be steered, we need to consider the velocity of both the boat and the river current. Firstly, consider that the boat's velocity (6.10 m/s) and the river current velocity (1.95 m/s) form a right triangle. The speed in still water is the hypotenuse while the river current speed forms the opposite side of the triangle.
To find the angle θ which is the angle between the boat's direction and the river current, we use the trigonometric function sine (sin), defined as the length of the opposite side divided by the length of the hypotenuse. Therefore, sin(θ) = (river current speed) / (boat speed in still water), sin(θ) = 1.95 m/s / 6.10 m/s. To find the angle θ, find the inverse of sin(θ).
Therefore, θ = arcsin(1.95 m/s / 6.10 m/s), which can be calculated using a scientific calculator to get an accurate degree measurement for the angle.
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A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an amplitude of 5,000 mm/s with a fre- quency of 10 Hz. Compute the maximum displacement the machine undergoes during this oscillation.
[tex]a=5000\dfrac{mm}{s}=5\dfrac{m}{s}[/tex]
[tex]f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s[/tex]
[tex]a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}
[/tex]
Hope this helps.
r3t40
The maximum displacement or amplitude of the oscillating machine, given the acceleration of 5,000 mm/s² and a frequency of 10 Hz, is calculated using principles of simple harmonic motion as approximately 13 mm.
Explanation:The question you asked involves the principles of Physics, in particular, the topics of simple harmonic motion and amplitude are applied. The machine's oscillation modelled as a simple spring-mass system operates under the principles of simple harmonic motion. The maximum displacement the machine undergoes, also known as the amplitude, can be calculated using the given data of acceleration and frequency.
Acceleration in a simple harmonic system is given by the formula a = -ω²x, where ω represents angular frequency, x represents displacement, and a represents acceleration. The maximum displacement, or amplitude, can be described as A = a_max/ω².
To find ω, we use the given frequency and recall that ω=2πf, so substituting 10 Hz for f, we get ω=20π rad/sec. To find the acceleration, we need to convert 5000 mm/s² to m/s² which is 5 m/s².
Substituting these values back into the amplitude equation A = a_max/ω², we get A = 5/400π² which calculates approximately as 0.013 m or 13 mm.
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Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a speed of 5.10 106 m/s kg · m/s (b) a 15.0-g bullet moving with a speed of 480 m/s kg · m/s (c) a 72.5-kg sprinter running with a speed of 10.5 m/s kg · m/s (d) the Earth (mass = 5.98 1024 kg) moving with an orbital speed equal to 2.98 104 m/s. kg · m/s
The magnitude of linear momentum can be calculated using the equation p=mv. The momentum for the different objects given are: (a) Proton: 8.517 * 10^-21 kg*m/s, (b) Bullet: 7.2 kg*m/s, (c) Sprinter: 761.25 kg*m/s, and (d) Earth: 1.78324 * 10^29 kg*m/s.
Explanation:The magnitude of the linear momentum of an object is calculated using the equation p=mv, where 'p' is the momentum, 'm' is the mass of the object, and 'v' is its velocity.
(a) Using the given values, m = 1.67 * 10^-27 kg and v = 5.10 * 10^6 m/s, the momentum of the proton is p = m*v = (1.67 * 10^-27 kg)*(5.10 * 10^6 m/s) = 8.517 * 10^-21 kg*m/s.
(b) The bullet's mass is 15.0 g which is 0.015 kg. Its velocity is 480 m/s. Thus, its momentum is p = m*v = (0.015kg)*(480m/s) = 7.2 kg*m/s.
(c) The sprinter's mass is 72.5 kg and velocity is 10.5 m/s. So, his momentum is p = m*v = (72.5kg)*(10.5m/s) = 761.25 kg*m/s.
(d) The Earth's mass is 5.98 * 10^24 kg and its orbital speed is 2.98 * 10^4 m/s. Hence, its momentum is p = m*v = (5.98 * 10^24 kg)*(2.98 * 10^4 m/s) = 1.78324 * 10^29 kg*m/s.
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The magnitude of the linear momentum for a proton, a bullet, a sprinter, and the Earth is approximately 8.52 * 10^-21 kg*m/s, 7.2 kg*m/s, 760.5 kg*m/s, and 1.78 * 10^29 kg*m/s, respectively.
Explanation:To calculate the magnitude of the linear momentum, we need to multiply the mass of the object by its velocity, which follows the formula: momentum = mass * velocity.
For the given cases:
Proton (1.67 * 10^-27 kg, 5.10 * 10^6 m/s): Using the momentum formula, you multiply these values together to get an approximate momentum of 8.52 * 10^-21 kg*m/s.Bullet (15.0 * 10^-3 kg, 480 m/s): Similarly, multiplying these entities gives us a momentum of approximately 7.2 kg*m/s.Sprinter (72.5 kg, 10.5 m/s): For the sprinter, the momentum will be approximately 760.5 kg*m/s.Earth (5.98 * 10^24 kg, 2.98 * 10^4 m/s): The Earth's momentum is a whopping 1.78 * 10^29 kg*m/s!Learn more about Linear Momentum here:https://brainly.com/question/34645079
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A long non-conducting hollow cylinder with inner radius, ri and outer radius, r2 is charged with a uniform positive charge with a charge density ?. (a) Use Gauss's Law in differential form for regions outside the cylinder (r >r2) to show that the electric field decreases as 1/r (note that no marks will be awarded for using Gauss's Law in integral form).
Final answer:
Gauss' Law in differential form can be used to show that the electric field outside a long non-conducting hollow cylinder decreases as 1/r.
Explanation:
Gauss' Law in differential form can be used to show that the electric field outside a long non-conducting hollow cylinder decreases as 1/r. To do this, we need to consider a Gaussian surface in the form of a cylindrical shell with radius r and length L. The charge enclosed by this surface is the charge on a length L of the cylinder, which is given by Qenc = λL, where λ is the linear charge density. According to Gauss' Law in differential form, the electric field through the Gaussian surface is given by E⋅dA = λenc/ε₀, where E is the electric field, dA is the area vector of the Gaussian surface, λenc is the charge enclosed by the surface, and ε₀ is the permittivity of free space. Since the Gaussian surface is a cylindrical shell, the area vector is perpendicular to the electric field and its magnitude is equal to the area of the cylindrical shell, which is 2πrL. Substituting these values in the equation, we get E(2πrL) = λL/ε₀. Solving for E, we find that the electric field outside the cylinder is given by E = λ/(2πε₀r). Since λ is constant for a long hollow cylinder, we can replace it with the charge density ρ multiplied by the length of the cylinder, L. Therefore, the electric field outside the cylinder decreases as 1/r, where r is the distance from the center of the cylinder.
An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.
Answer:
a) Current = 11 mA
b) Energy = 66 mJ
c) Power = 101.54 W
Explanation:
a) Voltage, V = IR
Voltage, V = 6 V, Resistance, R = 550 Ω
Current, I [tex]=\frac{6}{550}=0.011A=11mA[/tex]
b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ
c) [tex]\texttt{Power=}\frac{Energy}{Time}=\frac{0.066}{0.65\times 10^{-3}}=101.54W[/tex]
A ramp leading to the entrance of a building is inclied upward at an angle of 7 degree. A suitcase is to be pulled up the ramp by a handle that makes an angle of 35 degree with the horizontal. How mch force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs?
Explanation:
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / F
F = 50 / cos 28°
F = 56.6 lbs
56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change in momentum.
As given in the problem if a ramp leading to the entrance of a building is inclined upward at an angle of 7 degrees. A suitcase is to be pulled up the ramp by a handle that makes an angle of 35 degrees horizontal.
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / Force
Force = 50 / cos 28°
Force = 56.6 lbs
Thus, the 56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs
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A118 kg defensive lineman is running at a 74 kg running back at a speed of 1 m/s. If the running back is running at the lineman with a velocity of 6 m/s (assume this one is negative) and the two collide (assume inelastic collision), what is the final velocity of the lineman/running back combo? (One decimal place and no spaces between units and answer)
Answer:
Final velocity will be 1.7 m/s in the initial direction of running back
Explanation:
Since there is no external force on the system of two line man so here we can say that momentum of the two person will remain constant
So we can say
initial momentum of the two persons = final momentum of them
[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2) v[/tex]
now from the above equation we will have
[tex]v = \frac{m_1v_{1i} + m_2v_{2i}}{m_1 + m_2}[/tex]
now plug in all the values in the above equation
[tex]v = \frac{118(1) + 74(-6)}{118 + 74}[/tex]
[tex]v = - 1.7 m/s[/tex]
In Millikan’s experiment, the oil droplets acquire one or more negative charges by combining with the negative charges that are produced from the ionization of air by X rays. By measuring the charges on the oil droplets, he calculated the charge on a single electron as −1.60×10−19 C. The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electron.If Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons on it, what charge would he have measured on the droplet?
The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C with 6 negatively charged electrons on it.
If the charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electron [tex]10^{-19[/tex] C. and Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons.
Given that :
Number of electrons = 6
Given charge = −1.60× [tex]10^{-19}[/tex]C
The total charge can be calculated as:
Total charge = Number of electrons × Fundamental charge
Total charge = 6 × −1.60× [tex]10^{-19}[/tex]C
= 9.6 ×[tex]10^{-19}[/tex]
The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C.
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Using the findings of Millikan's oil drop experiment, the charge on an oil droplet with 6 negatively charged electrons would be -9.60×10^−19 Coulombs.
Explanation:According to the data from Millikan’s oil drop experiment, the charge of a single electron is identified to be −1.60×10−19 C. It was found that the charge on any negatively charged oil droplet is always a whole-number multiple of this fundamental charge. Therefore, if Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons on it, he would calculate the charge as
-1.60×10−19 C (charge of a single electron) multiplied by 6 (number of electrons) resulting in a total charge of -9.60×10^−19 C.
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A wire with a weight per unit length of 0.071 N/m is suspended directly above a second wire. The top wire carries a current of 72.8 A, and the bottom wire carries a current of 72.7 A. The permeability of free space is 4π × 10−7 T · m/A . Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. Answer in units of mm.
Answer:
d = 15 mm
Explanation:
Force of repulsion between two current carrying wire is given by
[tex]F = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]
now this force of repulsion is counterbalanced by the weight of the wire
so we have
[tex]mg = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]
now we have
[tex]d = \frac{\mu_0 i_1 i_2 L}{2\pi mg}[/tex]
so here we can say that
[tex]d = \frac{\mu_0 i_1 i_2}{2\pi (m/L)g}[/tex]
now plug in all values in it
[tex]d = \frac{4\pi \times 10^{-7} (72.8)(72.7)}{2\pi (0.071)}[/tex]
[tex]d = 0.015 m[/tex]
d = 15 mm
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 3°C, the resistance of the carbon resistor is 217.6 2 Ω. What is the temperature on a spring day when the resistance is 215.3 2 Ω? (Take the temperature coefficient of resistivity for carbon to be a-5.00 x 10^-4°C^-1) (Give your answer in decimal using "degreeC" °C) as unit)
Answer:
24.14 ⁰C
Explanation:
T₀ = initial temperature = 3 °C
R₀ = initial resistance of thermometer at initial temperature = 217.62 Ω
R = Final resistance of thermometer at final temperature = 215.32 Ω
T = final temperature = ?
α = temperature coefficient of resistivity for carbon = - 5.00 x 10⁻⁴ C⁻¹
Final resistance of thermometer at final temperature is given as
R = R₀ (1 + α (T - T₀ ))
Inserting the values
215.32 = 217.62 (1 + (- 5.00 x 10⁻⁴) (T - 3))
T = 24.14 ⁰C
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 310 m/s at 50.0° above the horizontal. It explodes on the mountainside 39.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?
The shell has horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] at time [tex]t[/tex] according to
[tex]x=\left(310\dfrac{\rm m}{\rm s}\right)\cos50.0^\circ t[/tex]
[tex]y=\left(310\dfrac{\rm m}{\rm s}\right)\sin50.0^\circ t-\dfrac g2t^2[/tex]
where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
The shell explodes after 39.0 s, at which point its coordinates are
[tex]x=7770\,\mathrm m[/tex]
[tex]y=1810\,\mathrm m[/tex]
Calculate the magnitude of the force between two point charges where q1 = +5.30 ?C and Q2 = +11.2 11C where the separation between the charges is R = 0.22 m. O A. 11.0 Newtons O B. 18.3 Newtons ° C. 27.4 Newtons 0 D. 41.5 Newtons
Answer:
Magnitude of force, F = 11 Newtons
Explanation:
Charge 1, [tex]q_1=5.3\ \mu C=5.3\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=11.2\ \mu C=11.2\times 10^{-6}\ C[/tex]
The separation between the charges is, r = 0.22 m
We have to find the magnitude of the force between two point charges. It can be calculated using the formula as :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{5.3\times 10^{-6}\ C\times 11.2\times 10^{-6}\ C}{(0.22\ m)^2}[/tex]
F = 11.03 N
or
F = 11 Newtons
Hence, this is the required solution.
A large stick is pivoted about one end and allowed to swing back and forth with no friction as a physical pendulum. The mass of the stick is 4.8 kg and its center of gravity (found by finding its balance point) is 1.4 m from the pivot. If the period of the swinging stick is 9 seconds, what is the moment of inertia of the stick about an axis through the pivot
Answer:
[tex]I = 94.33 kg m^2[/tex]
Explanation:
Let say the rod is slightly pulled away from its equilibrium position
So here net torque on the rod due to its weight is given as
[tex]\tau = mg dsin\theta[/tex]
since rod is pivoted at distance of 1.4 m from centre of gravity
so its moment of inertia about pivot point is given as
[tex]Inertia = I[/tex]
now we have
[tex]I \alpha = mg d sin\theta[/tex]
now for small angular displacement we will have
[tex]\alpha = \frac{mgd}{I}\theta[/tex]
so angular frequency of SHM is given as
[tex]\omega = \sqrt{\frac{mgd}{I}}[/tex]
now we will have
[tex]\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}[/tex]
[tex]I = 94.33 kg m^2[/tex]
Final answer:
The moment of inertia from the pivot is approximately [tex]72.4 kg\(\cdot\)m2.[/tex]
Explanation:
To generate an accurate answer for the moment of inertia of the stick about the pivot when it is used as a physical pendulum, we can use the formula for the period of a physical pendulum, which is:
[tex]T = 2\(\pi\)\(\sqrt{I/(mgd)}\),[/tex]
where T is the period, I is the moment of inertia, m is the mass of the rod, g is the acceleration due to gravity (approximately 9.8 m/s2), and d is the distance from the pivot to the center of mass. Given that T = 9 seconds, m = 4.8 kg, and d = 1.4 m, we can rearrange the formula to solve for I:
[tex]I = T2mgd/(4\(\pi\)2).[/tex]
Plugging the given values into this equation:
[tex]I = 92 * 4.8 kg * 9.8 m/s^2 * 1.4 m / (4 * \(\pi\)2).[/tex]
Calculating this yields the moment of inertia I:
[tex]I \(\approx\) 72.4 kg\(\cdot\)m2.[/tex]
An experiment is designed to test the relationship between the initial height of a basketball before it is dropped to the height of its rebound bounce. The height of the rebound bounce is measured using a scale positioned behind the ball. In the above experiment, which condition would not be controlled?
A). the starting height of the ball
B). the surface the ball bounces on
C). the type of ball
D). the method to measure the rebound height of the ball
Answer:
A. The starting height of the ball
Explanation:
When we talk about controlled variables, we refer to the variable that should be kept the same throughout the experiment. The reason why we do this, is to limit anything else that is not being tested, that may affect the results of the experiment.
In the scenario given, the experiment is to see the relationship between the initial height of a basketball and the height of its rebound bounce.
So you the starting height of the ball should vary, meaning it is NOT controlled.
Answer:
Starting height of the ball
Explanation:
A spring with spring constant 20 N/m has a mass of 5.0 kg attached. The spring is compressed and released. What is the period of oscillation?
Answer:
The period of oscillation is 3.14 sec.
Explanation:
Given that,
Spring constant = 20 N/m
Mass = 5.0 kg
We need to calculate the time period
Using formula of time period
[tex]T =2\pi\sqrt{\dfrac{m}{k}}[/tex]
m= mass
k = spring constant
[tex]T =2\times3.14\sqrt{\dfrac{5.0}{20}}[/tex]
[tex]T =3.14\ sec[/tex]
Hence, The period of oscillation is 3.14 sec.
A solid plastic cube with uniform density (side length 0.5 m) of mass 100 kg is placed in a vat of fluid whose specific gravity is 1.2. What fraction of the cube's volume floats above the surface of the fluid? O C. 2/3 O 1.45
Answer:
0.042 m^3
Explanation:
side, a = 0.5 m, m = 100 kg
density of fluid, d = 1.2 g/cm^3 = 1200 kg/m^3
Volume of cube, V = side^3 = (0.5)^3 = 0.125 m^3
density of solid = mass of solid / volume of solid cube
density of solid, D = 100 / 0.125 = 800 kg/m^3
Let v be the volume of cube immersed in fluid.
According to the principle of flotation
Weight of cube = Buoyant force acting on the cube
V x D x g = v x d x g
0.125 x 800 = v x 1200
v = 0.083 m^3
Volume above the level of fluid = V - v = 0.125 - 0.083 = 0.042 m^3
A student weighing 200N climbs a flight of stairs 7.0m high in 8s. What power is required to perform this task?
Answer:
Power required, P = 175 watts
Explanation:
It is given that,
Weight of a student, F = mg = 200 N
The student climbs a flight of stairs of height, h = 7 m
Time taken, t = 8 s
We have to find the power required to perform this task. Work done per unit time is called the power required. Mathematically, it is given by :
[tex]P=\dfrac{W}{t}[/tex]
W = work done
t = time taken
[tex]P=\dfrac{mgh}{t}[/tex]
[tex]P=\dfrac{200\ N\times 7\ m}{8\ s}[/tex]
P = 175 watts
Hence, the power required to complete this task is 175 watts.
Using coulombs inverse square law, if the charges of two similarly charged spheres increases, what would happen to the value of A? (In this experiment we moved a charged sphere closed to second one and observed how this second one moved away) A is fromt the equation Fcoul= A/r 2
Answer:
The value of A will increase as you increase the charges.
Explanation:
A is actually kq1*q2. So if you increase the charges, A will also increase. The big increase comes if you make r smaller.
Increased charges is a direct variation. As the charge increases the value of A increases. As the charges decrease, the value of A decreases.
r is an inverse variation. As r increases the force decreases. As r decreases the force increases. R does not affect a.
2. Calculate the dipole moment of 5 & 3 columb charges separated by
a. 10m
b. 1micron
c. 0.005 nanometres
A large cruise ship of mass 6.40 ✕ 107 kg has a speed of 11.6 m/s at some instant. (a) What is the ship's kinetic energy at this time? J (b) How much work is required to stop it? (Give the work done on the ship. Include the sign of the value in your answer.) J (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 3.20 km? N
(a) [tex]4.3\cdot 10^9 J[/tex]
The kinetic energy of an object is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
For the ship in the problem, we have
[tex]m=6.40\cdot 10^7 kg[/tex] is the mass
[tex]v=11.6 m/s[/tex] is the speed
So its kinetic energy is
[tex]K=\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=4.3\cdot 10^9 J[/tex]
(b) [tex]-4.3\cdot 10^9 J[/tex]
According to the work-kinetic energy theorem, the work done on an object is equal to the change in kinetic energy of the object:
[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]
where
W is the work done
m is the mass of the object
v is the final speed of the object
u is the initial speed of the object
Here we want to find the work done to stop the ship, so the final speed of the ship is v=0, while the initial speed is u=11.6 m/s. So the work done will be
[tex]W= 0 -\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=-4.3\cdot 10^9 J[/tex]
(c) [tex]1.3\cdot 10^6 N[/tex]
The work done on an object can be also written as follows
[tex]W=Fd[/tex]
where
F is the magnitude of the force
d is the displacement of the object
Here we know:
[tex]W=-4.3\cdot 10^9 J[/tex] is the work done
d = 3.20 km = 3200 m is the displacement of the ship
So we can solve the formula to find F, the force exerted on the ship to stop it:
[tex]F=\frac{W}{d}=\frac{-4.3\cdot 10^9 J}{3200 m}=-1.3\cdot 10^6 N[/tex]
and the negative sign simply means that the force is opposite to the displacement of the ship (in fact, the force acts against the motion of the ship).
The ship's kinetic energy is 5.77 × 10^9 J. The work required to stop it is -5.77 × 10^9 J. The magnitude of the force required to stop it is 1.80 × 10^3 N.
Explanation:(a) The kinetic energy of the ship can be calculated using the formula:
Ek = (1/2)mv2
where m is the mass of the ship and v is its velocity. Plugging in the given values, we get a kinetic energy of 5.77 × 109 J.
(b) To stop the ship, work needs to be done to bring it to a complete stop. The work done is equal to the change in kinetic energy of the ship. Initially, the ship had a kinetic energy of 5.77 × 109 J. When it comes to a stop, its kinetic energy becomes zero. Therefore, the work required to stop the ship is -5.77 × 109 J (negative since work is done on the ship).
(c) The magnitude of the constant force required to stop the ship can be calculated using the work-energy theorem:
Work = Force × Distance
Given that the displacement of the ship is 3.20 km, or 3.20 × 106 m, and the work done on the ship is -5.77 × 109 J, we can solve for the force:
Force = Work / Distance = -5.77 × 109 J / (3.20 × 106 m) = -1.80 × 103 N
Learn more about kinetic energy here:https://brainly.com/question/26472013
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Car A has twice the mass of car B , and car A has the half the kinetic energy of car B . How do their speeds compare?
Answer:
Speed of car A is half the speed of car B.
Explanation:
Kinetic energy, [tex]KE=\frac{1}{2}mv^2[/tex]
Car A has twice the mass of car B.
[tex]m_A=2m_B[/tex]
Car A has the half the kinetic energy of car B
[tex]KE_A=\frac{1}{2}KE_B[/tex]
We have
[tex]KE_A=\frac{1}{2}KE_B\\\\\frac{1}{2}m_Av_A^2=\frac{1}{2}\times \frac{1}{2}m_Bv_B^2\\\\\frac{1}{2}\times 2m_Bv_A^2=\frac{1}{2}\times \frac{1}{2}m_Bv_B^2\\\\v_A^2=\frac{v_B^2}{4}\\\\v_A=\frac{v_B}{2}[/tex]
Speed of car A is half the speed of car B.
A bullet is shot at an angle of 32° above the horizontal on a level surface. It travels in the air for 6.4 seconds before it strikes the ground 92m from the shooter. What was the maximum height reached by the bullet? Round to one decimal place and include units.
Answer:
Maximum height reached by the bullet = 4.01 m
Explanation:
Horizontal displacement = 92 m
Time taken = 6.4 s
Horizontal velocity
[tex]=\frac{92}{6.4}=14.375m/s[/tex]
We have angle of projection = 32°
Horizontal velocity = u cos 32 = 14.375
u = 16.95 m/s
Vertical velocity = u sin θ = 16.95 x sin 32 = 8.98 m/s
Time of flight till it reaches maximum height = 0.5 x 6.4 = 3.2s
Now we have vertical motion of bullet
S = ut + 0.5 at²
Vertical velocity = u = 16.95 m/s
a = -9.81 m/s²
t = 3.2s
Substituting
S = 16.95 x 3.2 - 0.5 x 9.81 x 3.2² = 4.01 m
Maximum height reached by the bullet = 4.01 m
A mass attached to a spring oscillates and completes 53 full cycles in 28 s. What is the time period (in s) and frequency (in Hz) of this system? period s frequency Hz
Explanation:
It is given that, a mass attached to a spring oscillates and completes 53 full cycles in 28 s. Frequency of the system is given by the number of oscillations or vibrations per second. Here, the frequency of this system is given by :
[tex]f=\dfrac{no\ of\ oscillations}{time}[/tex]
[tex]f=\dfrac{53}{28}[/tex]
f = 1.89 Hertz
The relationship between the frequency and the time period of the spring is inverse i.e.
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{1.89\ Hz}[/tex]
T = 0.52 seconds
Hence, this is the required solution.
What is the speed of a 2 HZ wave with a wavelength of 10 m
Explanation:
The speed [tex]v[/tex] of a wave is given by the following equation:
[tex]v=\frac{\lambda}{T}[/tex] (1)
Where [tex]\lambda[/tex] is the wavelength and [tex]T[/tex] the period of the wave.
In addition we know there is an inverse relation between the period of a wave and its frequency [tex]f[/tex] :
[tex]f=\frac{1}{T}[/tex] (2)
Substituting (2) in (1):
[tex]v=\lambda.f[/tex] (3)
Now we can find the velocity of the wave with the known given values applying equation (3):
[tex]v=(10m)(2Hz)[/tex] (4)
Finally:
[tex]v=20 \frac{m}{s}[/tex]
A playground slide is 8.80 ft long and makes an angle of 25.0° with the horizontal. A 63.0-kg child, initially at the top, slides all the way down to the bottom of the slide. (a) Choosing the bottom of the slide as the reference configuration, what is the system's potential energy when the child is at the top and at the bottom of the slide? What is the change in potential energy as the child slides from the top to the bottom of the slide? (Include the sign of the value in your answer.)
Answer:
initial: 1654.6 J, final: 0 J, change: -1654.6 J
Explanation:
The length of the slide is
L = 8.80 ft = 2.68 m
So the height of the child when he is at the top of the slide is (with respect to the ground)
[tex]h = L sin \theta = (2.68 m)sin 25.0^{\circ}=1.13 m[/tex]
The potential energy of the child at the top is given by:
[tex]U = mgh[/tex]
where
m = 63.0 kg is the mass of the child
g = 9.8 m/s^2 is the acceleration due to gravity
h = 1.13 m
Substituting,
[tex]U=(63.0 kg)(9.8 m/s^2)(2.68 m)=1654.6 J[/tex]
At the bottom instead, the height is zero:
h = 0
So the potential energy is also zero: U = 0 J.
This means that the change in potential energy as the child slides down is
[tex]\Delta U = 0 J - (1654.6 J) = -1654.6 J[/tex]
The potential energy at the top is 700.9 J. At the bottom is 0 J. Change is -700.9 J (loss).
The length of the slide [tex]\( L = 8.80 \, \text{ft} \)[/tex] and the angle with the horizontal [tex]\( \theta = 25.0^\circ \)[/tex], we can use the sine function to find the vertical height h from the top to the bottom of the slide:
[tex]\[ h = L \sin \theta \][/tex]
[tex]\[ h = 8.80 \, \text{ft} \times \sin 25.0^\circ \][/tex]
First, calculate [tex]\( \sin 25.0^\circ \)[/tex]:
[tex]\[ \sin 25.0^\circ \approx 0.4226 \][/tex]
Now, calculate h:
[tex]\[ h = 8.80 \times 0.4226 \approx 3.719 \, \text{ft} \][/tex]
Potential energy PE is given by:
[tex]\[ PE = mgh \][/tex]
where [tex]\( m = 63.0 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \)[/tex] , and h is the height
Convert h from feet to meters (1 ft = 0.3048 m):
[tex]\[ h = 3.719 \, \text{ft} \times 0.3048 \, \text{m/ft} = 1.133 \, \text{m} \][/tex]
The potential energy at the top:
[tex]\[ PE_{\text{top}} = 63.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.133 \, \text{m} \][/tex]
[tex]\[ PE_{\text{top}} \approx 63.0 \times 9.81 \times 1.133 \approx 700.9 \, \text{J} \][/tex]
Since the child is at the bottom of the slide, where h = 0, the potential energy at the bottom [tex](\( PE_{\text{bottom}} \))[/tex] is 0 Joules.
The change in potential energy
The change in potential energy [tex](\( \Delta PE \))[/tex] is:
[tex]\[ \Delta PE = PE_{\text{bottom}} - PE_{\text{top}} \][/tex]
[tex]\[ \Delta PE = 0 - 700.9 \, \text{J} \][/tex]
[tex]\[ \Delta PE = -700.9 \, \text{J} \][/tex]
A cannon is shot at a 30 degree angle above the horizontal with a velocity of 240 m/s. What is the range of this shot?
Answer:
Range, R = 5090.1 meters
Explanation:
It is given that,
Velocity of cannon, v = 240 m/s
A cannon is shot at a 30 degree angle above the horizontal. The range of this shot is given by the formula as follows :
[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]
g = acceleration due to gravity
[tex]R=\dfrac{(240\ m/s)^2\ sin2(30)}{9.8\ m/s^2}[/tex]
R = 5090.1 meters
So, the range of this shot is 5090.1 meters. Hence, this is the required solution.
A composite material is to be designed with epoxy (Em 3.5 GPa) and unidirectional fibers. The longitudinal elastic modulus of the designed composite material is to be at least E 320 GPa. If the fiber volume fraction is 70%, calculate the minimum elastic modulus requirement of the fiber material.
Answer:
Minimum elastic modulus of fiber = 455.64 GPa
Explanation:
Contents of composite material = Epoxy and Unidirectional fibers
Elastic modulus of epoxy = 3.5 GPa
Elastic modulus of composite material = 320 GPa
Volume fraction of fiber = 70 %
Volume fraction of epoxy = 100 - 70 = 30%
Elastic modulus of composite material = 3.5 x 0.3 + Elastic modulus of fiber x 0.7 = 320
0.7 x Elastic modulus of fiber = 320 - 1.05 = 318.95
Elastic modulus of fiber = 455.64 GPa
Minimum elastic modulus of fiber = 455.64 GPa