Answer:
u = 14 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. On the floor just when u speeding out
Em₀ = K = ½ m v²
final point. When on top of the building, no speed
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ m v² = m g h
v = u
u = [tex]\sqrt{2 g h}[/tex]
u= [tex]\sqrt{2 \ 9.8 \ 10}[/tex]
u = 14 m / s
Which option identifies the specific knowledge that the team in the following scenario must possess?
A team of engineers is designing a space probe that will go to Saturn and collect atmospheric samples. The temperature and atmosphere on Saturn are much different from the conditions on Earth.
(A) The team must have a vast knowledge of thermodynamics.
(B) The team must have a vast knowledge of propulsion.
(C) The team must have a vast knowledge of fluid power systems.
(D) The team must have a vast knowledge of acoustics.
Answer:
The team must have a vast knowledge of thermodynamics
Explanation:
Just took the test!!!
Answer:
C. Thermodynamics
Explanation:
Galvani wrongly believed that the frog’s leg twitched during his experiment due to _____.
Answer:
nerves
Explanation:
I think, I maybe wrong.
Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please) :D
PLEASE ANSWER WITH ACTUAL ANSWER AND I WILL MARK BRAINLIEST (IF YOU GIVE ME A SCAMMY ANSWER I WILL REPORT YOU!!!)
A student wants to determine the local value of the gravitational field strength, g , in their classroom. Which of the following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured?
Select TWO answers.
A: Run a lab cart down an inclined plane; measure the length of the ramp and the time it takes the cart to reach the bottom.
B: Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.
C: Accelerate a lab cart horizontally; measure the mass of the cart and its acceleration.
D: Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.
Answer:
Most likely (B)
Explanation:
B in the passage is the most representative out of all your choices and it has evidence from the passage
Hope dis helps Jit!
Sorry i forgot to type C
B and C both measure mass while the others are calculations and are bias
The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:
Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.What is gravitational field?A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.
When a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.
When a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.
Learn more about gravitational field here:
https://brainly.com/question/26690770
#SPJ2
Help please. Question about a potential energy.
One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. The force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00 x 103 m/s, given the collision lasts 6.00 x 10-8 s is Fill input: x 106 N.
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ [tex]F=m(\frac{\Delta v}{\Delta t} )[/tex]
On substituting the given values, we get
⇒ [tex]=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )[/tex]
⇒ [tex]=6666.67 \ N[/tex]
A wave has a frequency of 67 Hz and a wavelength of 7.1 meters. What is the speed of this
wave?
Answer:
475.7 m/s
Explanation:
Given,
Frequency ( f ) = 67 Hz
Wavelength ( λ ) = 7.1 m
To find : Speed ( v ) = ?
Formula : -
v = f λ
v
= 67 x 7.1
= 475.7 m/s
Therefore,
the speed of the wave is 475.7 m/s.
A rifle can shoot a 4.00 g bullet at a speed of 998 m/s. Find the kinetic energy of the bullet. What work is done on the bullet if it starts from rest?
Answer:
1992.008J
Explanation:
Please help me!
8. Give an example of a poor blackbody radiator and explain why it is not a good blackbody radiator.
9. Does a blackbody radiator emit light waves? Explain.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Explanation:
At which point is there the most potential energy? At which point is there the most kinetic energy?
A. Potential energy A; Kinetic energy B
B. Potential energy B; Kinetic energy D
C. Potential energy A; Kinetic energy D
D. Potential energy C; Kinetic energy D
Answer:
The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy
An electron is travelling in the positive x direction. A uniform electric field is in the negative y direction. If a uniform magnetic field with the appropriate magnitude and direction also exists in the region, the total force on the electron will be zero. The appropriate direction for the magnetic field is:Group of answer choicesthe negative y directioninto the pageout of the pagethe negative x directionthe positive y direction
Answer:
into the page
Explanation:
Since the uniform electric field is in the negative y direction so its is -E and the electron is travelling in the positive x direction, it experiences an electric force F = -e × -E = + eE, so the electric force is in the positive y direction. Now since the net force on the electron is zero in the region of the magnetic field, it follows that the direction of the magnetic force is opposite to that of the electric force. Since the electric force is in the positive y direction, the magnetic force is in the negative y direction.
By the right hand rule, since the magnetic force is in the negative y direction and the electron moves in the positive x direction, it follows that the magnetic field is in the positive z direction, into the page.
The angle between reflected ray and the normal line is
Answer:
Explanation:
angle of incidence.
To increase the potential energy of the system, what did you have to do?
Answer:
You can use work to add kinetic energy to a system or to increase potential energy in the system.
Explanation:
Potential energy stored in any system can be released as kinetic energy. Kinetic energy can be transformed to do work or to increase potential energy.
hope this helped
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.70 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a) v = 1.1 m/s
b) A = 0.315 m
c) v = 1.1 m/s A= 0.15 m
Explanation:
a)
In any travelling wave, there exists a fixed relationship between the propagation speed, the wavelength and the frequency, as follows:[tex]v = \lambda * f (1)[/tex]
If the wave crests are spaced a horizontal distance of 5.7 m apart, this means that the wavelength of the wave is just the same, i.e., 5.70 m.Regarding the frequency, we know that the frequency is just the inverse of the period, i.e., the time needed to complete one oscillation.If it takes a time of 2.60 s to go from the highest point to the lowest, the time needed to complete an oscillation (the period T) will be just double of this time:⇒ T = 2.60 s * 2 = 5.20 s (2)Since we have now T, we can find the frequency f as follows:[tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]
Replacing f and λ in (1) we get:[tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]
b)
The amplitude of the wave is just the amount that the water aparts from its equilibrium level, which is just the half of the distance between its highest point and the lowest one, as follows:[tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]
c)
Part a) will not be affected by the new amplitude, because we have showed that the speed is independent of the amplitude, so v can be written as follows:v = 1.10 m/s (6)
Part b) will change , due to the amplitude changes. If the total vertical distance traveled by the boat is 0.30 m, by the same token as explained in b), the new amplitude will be just half of this, as follows:[tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]
NEED TO SUBMIT THIS IN 10 MINS, PLS HELP!!!!
Answer:
Your answer is B
because it's on sneel's law.
that is sin of incident ray / sin of refracted ray is refractive index
the atom of an element x has 21protrons and 23neutrons. What is the
(a) Electron number
(b) Mass number
(c) Neutron number
A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and
rebounds with a velocity of 22 m/s. Is the change in momentum of the cart
Explanation:
mass, m = 5kg
initial velocity, u = 16m/s
final velocuty, v = -22m/s
change in momentum, ∆p = ?
∆p = m (v-u)
5(-22-16)
5(38)
∆p = 190kgm/s
check the calculations!
If 10 Coulombs flow through a circuit every 2 seconds, what is the current?
A. Not enough info
B. 5 A
C. 10 A
D. 1 A
Answer:
not enought info
Explanation:
tbh I just know it's not 5 10 or 1
Answer:
B. 5 A
Explanation:
10/2= 5
Educere
Easy physics question help.!!!
Answer: This is not easy lol
Explanation:
When you cool a gas, how does this affect the de Broglie wavelength of the gas atoms? When you cool a gas, how does this affect the de Broglie wavelength of the gas atoms? Being cooled, the gas atoms slow down so that their de Broglie wavelength will increase. Being cooled, the gas atoms slow down so that their de Broglie wavelength will decrease. The de Broglie wavelength will remain the same because it does not depend on temperature.
Answer:
The de Broglie wavelength will remain the same because it does not depend on temperature.
Explanation:
de Broglie wavelength of a particle is independent of the temperature and hence the properties of emitted particle such as photoelectric effect, radioactive radiation etc. does not depend on the temperature.
Also, until unless the kinetic energy of a moving particle is not driven by the
thermal energy, the de Broglie wavelength is independent of the temperature
What fuel does a main-sequence star use for nuclear fusion?
oxygen (0)
petroleum
helium (He)
hydrogen (H)
Answer:
A main sequence star is powered by fusion of hydrogen into helium in its core
Explanation:
Explain why it is not advisable to be in a garage when the car engine is being
heated.
Answer:
You can breathe in too much carbon monoxide, which will eliminate the flow of oxygen to your bloodstream and can kill you.
Explanation:
An artificial satellite circling the Earth completes each orbit in 126 minutes. (a) Find the altitude of the satellite.
Answer:
Explanation:
Time period of rotation
T = 2πR/ V where R is radius of orbit and V is orbital velocity
Orbital velocity V = √ ( GM/R ) , m is mass of the earth .
T = 2πR √R / GM
T² = 4π²R³ / GM
Putting the values
( 126 x 60 )² = 4 x 3.14² x R³ / 6.67 x 10⁻¹¹ x 5.97 x 10²⁴
57.15 x 10⁶ = 39.44 x R³ / 39.82 x 10¹³
R³ = 577 X 10¹⁸
R = 8.325 x 10⁶ m
= 8325 km
Radius of earth = 6400 km
height of satellite = 8325- 6400 = 1925 km .
A mass of 3 kg stretches a spring 9m. The mass is acted on by an external force of 2 AND. The Mass moves in a medium that imparts a viscous force of 1 N when the speed of the mass is 4m/sec The mass is pulled down 8 cm below its equilibrium position, and then set in motion inthe upward direction with a velocity of 5 m/sec. State the initial value problem describing the motion of the mass. DO NOT SOLVE.
Answer:
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
Explanation:
In this exercise, you are asked to state the problem you are posing. We are going to find the equation of motion for this exercise. Let's start with Newton's second law
Let's set a reference system with the y-axis in a vertical and positive direction upwards.
We have four forces: an external downward force, negative in sign, the but that goes down and is negative, the Hook force that goes up and is positive and the friction force that opposes the movement, in this case it goes down being negative
let's write Newton's second law
F_e -F -fr - W = m a
where
F_e = -kDy = - k y
fr = - b v = -b dy / dt
W = mg
we substitute for the specific case, that is, using the signs
k y -b [tex]\frac{dy}{dt}[/tex] - m g - F = m [tex]\frac{d^2y}{dt^2}[/tex]
In the initial condition of the problem, before starting the movement, the friction force is zero and the acceleration is also zero
k y - m g - F = 0
from this equation you can find the spring constant, y= 9m and F=2 N
It is not clear if when the movement starts this external force becomes zero, but since it balances the weight we can eliminate the two forces that have the same magnitude and opposite direction, so the equation remains
k y - b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
therefore, to initiate the movement, a small external force F 'is applied that moves the system to a new equilibrium position and this small force F' is made zero, thus initiating an oscillatory movement, described by the equation.
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
This is a differential equation of the second degree, therefore it needs two initial conditions for its complete solution
The initial amount of displacement corresponds to the amplitude of movement A = 0.08 m
Blue light (450 nm) and orange light
(625 nm) pass through a diffraction
grating with d = 2.88 x 10-6 m. What is
the angular separation between them
for m = 1?
Answer:
3.54
Explanation:
some nerd thing I found it on Yahoo answers
Answer:
3.54º
Explanation:
Find the blue θ first
sin⁻¹(540x10⁻⁹/2.88x10⁻⁶)=8.99°
Then find the orange θ
sin⁻¹(625x10⁻⁹/2.88x10⁻⁶)=12.53°
Take the differences and subtract
12.53°－8.99°=3.54°
A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.6 A runs through the wire, in the x direction. Along 0.17 m of the length of the wire there is a magnetic field of 0.52 tesla in the y direction, due to a large magnet nearby. At other locations in the circuit, the magnetic field due to external sources is negligible. What is the magnitude of the magnetic force on the wire
Answer:
the magnitude of the magnetic force on the wire is 0.2298 N
Explanation:
Given the data in the question;
we know that, the magnitude of magnetic force is given as;
|F[tex]_{mg}^>[/tex] | = I([tex]B^>[/tex] × [tex]L^>[/tex] )
given that
I = 2.6 A
[tex]B^>[/tex] = 0.17
[tex]L^>[/tex] = 0.52
so we substitute
|F[tex]_{mg}^>[/tex] | = 2.6( 0.17i" × 0.52j" )
|F[tex]_{mg}^>[/tex] | = 0.2298 N
Therefore, the magnitude of the magnetic force on the wire is 0.2298 N
Each of the two grinding wheels has a diameter of 6 in., a thickness of 3/4 in., and a specific weight of 425 lb/ft3. When switched on, the machine accelerates from rest to its operating speed of 3450 rev/min in 5 sec. When switched off, it comes to rest in 35 sec. Determine the motor torque and frictional moment, assuming that each is constant. Neglect the effects of the inertia of the rotating motor armature.
Answer:
[tex]0.842\ \text{lb ft}[/tex]
[tex]0.1052\ \text{lb ft}[/tex]
Explanation:
d = Diameter of wheel = 6 in
r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]
t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]
w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]
[tex]t_2[/tex] = Time taken to slow down = 35 s
[tex]t_1[/tex] = Time taken to reach operating speed = 5 s
[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
Weight is given by
[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]
Mass is given by
[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]
Moment of inertia is given by
[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]
Angular acceleration while slowing down is given by
[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]
Frictional moment is given
[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]
Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]
Angular acceleration while speeding up is given by
[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]
Motor torque is given by
[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]
Motor torque is [tex]0.842\ \text{lb ft}[/tex].
Explain why your image never disappears and never flips over as you bring the convex mirror
close to your eye.
Explanation:
When you get closer to the mirror than the focal point a virtual image is formed behind the mirror and this image is not inverted. That's why the image flips as you get closer. ... With a virtual image the light rays never come to a focus so there is no place you can put a piece of paper to see the image.
Please help I will mark you brainliest
I believe the answer is a
A wooden cylinder (in the form of a thin disk) of uniform density and a steel hoop are set side by side, released from rest at the same moment, and roll down an inclined plane towards a wall at the bottom. The cylinder has a larger radius than the hoop, but the hoop weighs more than the cylinder.
Required:
Who reaches the bottom first and why?
Answer:
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
Explanation:
a. Who reaches the bottom first
The kinetic energy of the objects is given by
K = 1/2mv² + 1/2Iω² where m = mass of object, v = velocity of object, I = moment of inertia and ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy, K = 1/2mv² + 1/2(mr²/2)(v/r)²
K = 1/2mv² + 1/4mv²
K = 3mv²/4
For the steel hoop, I' = mr'² where m' = mass of steel hoop and r' = radius of steel hoop and v' = velocity of steel hoop
So, its kinetic energy, K' = 1/2m'v'² + 1/2(m'r'²)(v'/r')²
K' = 1/2m'v'² + 1/2m'v'²
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
3mv²/4 = m'v'²
v²/v'² = 4m/3m'
v²/v'² = 4/3(m/m')
v/v' = √[4/3(m/m')]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
b. Why
Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
K = K₀ + 1/2Iω² where K₀ = translational kinetic energy of wooden cylinder
K - 1/2Iω² = K₀
3/4mv² - 1/2(mr²/2)(v/r)² = K₀
3/4mv² - 1/4mv² = K₀
K₀ = 1/2mv²
For the steel hoop,
K' = K₁ + 1/2I'ω'² where K₁ = translational kinetic energy of steel hoop
K' - 1/2I'ω'² = K₁
m'v'² - 1/2(m'r'²)(v'/r')² = K₁
m'v'² - 1/2m'v'² = K₁
K₁ = 1/2m'v'²
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
What is Kinetic energy?
The energy of the body due to its movement in a particular direction under the influence of a force like a free-falling body due to gravitaional force is called Kinetic energy.
The kinetic energy of the objects is given by
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}Iw^2[/tex]
where
m = mass of object,
v = velocity of object,
I = moment of inertia and
ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy,
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{mr^2}{2})\dfrac{v}{r}^2[/tex]
[tex]K = \dfrac{3mv^2}{4}[/tex]
For the steel hoop,
I' = mr'²
where
m' = mass of steel hoop and
r' = radius of steel hoop and
v' = velocity of steel hoop
So, its kinetic energy,
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}(m'r'^2)\dfrac{v'}{r'}^2[/tex]
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}m'v'^2[/tex]
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
[tex]\dfrac{3mv^2}{4 }= m'v'^2[/tex]
[tex]\dfrac{v^2}{v'^2} =\dfrac{ 4m}{3m'}[/tex]
[tex]\dfrac{v^2}{v'^2} = \dfrac{4}{3}(\dfrac{m}{m'})[/tex]
[tex]\dfrac{v}{v'} = \sqrt{[\dfrac{4}{3}(\dfrac{m}{m'})][/tex]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
(b) Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
[tex]K = K_o + \dfrac{1}{2}Iw^2[/tex]
where
K₀ = translational kinetic energy of wooden cylinder
[tex]K - \dfrac{1}{2}Iw^2 = K_o[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{2}(\dfrac{mr^2}{2})(\dfrac{v}{r})^2 = K_a[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{4}mv^2 = K_o[/tex]
[tex]K_o = \dfrac{1}{2}mv^2[/tex]
For the steel hoop,
[tex]K' = K_1 + \dfrac{1}{2}I'w'^2[/tex]
where
K₁ = translational kinetic energy of steel hoop
[tex]K' - \dfrac{1}{2}I'w'^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}(m'r'^2)(\dfrac{v'}{r'})^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}m'v'^2 = K_1[/tex]
[tex]K_1= \dfrac{1}{2}m'v'^2[/tex]
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
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