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zloy xaker [14]

3 years ago

12

A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti

on between the ramp surface and crate, what is the velocity of the crate at the bottom of the ramp? (g = 9.8 m/s2) Group of answer choices

1 answer:

Alenkinab [10]3 years ago

6 0

Answer:

$v_f = 3.45 m/s$

Explanation:

As we know that box was initially at rest

so here work done by all forces on the box = change in its kinetic energy

so we will have

$mg sin\theta L = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 10 kg

$\theta = 10^0$

L = 3.5 m

so we will have

$10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)$

$2(9.81) sin10 \times 3.5 = v_f^2$

so final speed is given as

$v_f = 3.45 m/s$

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago

Which example demonstrates constant speed with changing direction?

Darya [45]

Ffffffddddffffffffffc

3 0

3 years ago

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, i

Lapatulllka [165]

Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

Explanation:

a. KE =1/2 (MV^2) where the M is mass of electron

b. E = V/d

c. V= 0 V (momentarily the pd changes to zero)

d KE= 300*3600 v = 1.08 MeV

6 0

3 years ago

?????Help please?????

Sindrei [870]

The correct selections are C, C, B, D, A, B, and A .

6 0

3 years ago

11) A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the sa

lesya692 [45]

15mph

If the man turns and runs toward point A, he will cover

3/8 of the length of the bridge in the time that it takes

the train to reach A.

If the man runs forward toward point B, what part of the bridge

will he cover before the train reaches A? Well, he will cover

3/8 of the bridge, only heading forward toward B. This will put

him 3/8 + 3/8 = 6/8 = 3/4 of the way across the bridge by the

time the train reaches A.

since we know that the man and the train will meet at B, this

means that in the time it takes the man to run the remaining

1/4 of the bridge, the train will cover the entire length of

the bridge.

If it takes the man the same time to cover 1/4 of the bridge

that it takes the train to cover the whole bridge, then the train

must be going four times as fast as the man. Another way of saying

this is that the man runs at 1/4 the speed of the train.

Since the train's speed is known to be 60 mph, this means that

the man runs at (1/4) 60 = 15 mph.

7 0

3 years ago