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grigory [225]
3 years ago
5

Suppose that a ball decelerates from 8.0 m/s to a stop as it rolls up a hill, losing 10% of its kinetic energy to friction. Dete

rmine how far vertically up the hill the ball reaches when it stops. Show your work.(2 points)
Physics
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

The maximum height is 0.33 m.

Explanation:

initial velocity, u = 8 m/s

final velocity, v = 0 m/s

10% of  kinetic energy is lost in friction.

The kinetic energy used to move up the top,

KE = 10 % of 0.5 mv^2

KE = 0.1 x 0.5 x m x 8 x 8 = 3.2 m

Let the maximum height is h.

Use conservation of energy

KE at the bottom = PE at the top

3.2 m = m x 9.8 x h

h = 0.33 m  

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A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.
jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = <u>21 m  </u>

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>

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3 years ago
During a blackout, you are trapped in a tall building. you want to call rescuers on your cell phone, but you can't remember whic
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Let the height where we are trapped is H

now to find the time to reach the key at the bottom is given as

y = v_i t + \frac{1}{2}at^2

now we have

H = \frac{1}{2}gt^2

t = \sqrt{\frac{2H}{g}}

now if the speed of sound is considered to be 340 m/s then time taken by the sound to reach at the top is given as

t = \frac{H}{340}

now the total time is given as

\sqrt{\frac{2H}{g}} + \frac{H}{340} = 3.27

now by solving above equation we have

H = 48 m

now height of one floor is 3 m

so our position must be

N = \frac{48}{3} = 16 th \:floor

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Answer:

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2 years ago
Read 2 more answers
An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas
Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

F_f=\mu mg

where

\mu is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

\mu=0.500 (coefficient of friction)

m = 1260 kg (mass of the car)

g=9.8 m/s^2

Therefore, the force of friction is

F_f=(0.500)(1260)(9.8)=6174 N

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

F=m\frac{v^2}{r}

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

v=54.1 km/h =15.0 m/s is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

F=(1260)\frac{15.0^2}{41.6}=6814 N

Therefore, the force of friction is not enough to keep the car in the curve, since F_f

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3 years ago
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