Physics, 24.10.2019 17:43 alexseyreyes

100 points + physics

1) a force of 157 n is applied to a box 25o above the horizontal. if it is applied over a distance of 14.5 m how much work is done?

2) jamie decides to drop an egg that has a mass of 345 g off the top of a 8.2 m building. how fast will it be falling right before it hits the ground?

3) snookie is riding in her little red wagon (total mass of 215 kg) at a constant speed of 9.5 m/s. all of a sudden a magenta lemur (19.7 kg) appears in
her lap. how fast is the snook-meister traveling now?

4) police are investigating an accident. they know that tom brady was driving 20 m/s before being hit by jay z head on. tom brady’s car has a mass of 1100 kg
and jay z’s has a mass of 1475 kg. they also know that the two cars stuck together and were traveling 7 m/s in the same direction as jay z was driving. the speed
limit was 25 m/s, was jay z speeding?

5) a ball is being swung in a horizontal circle of radius .23 m at a rate of 7.8 m/s. if the ball has a mass of 0.63 kg, find the force of tension on the string.

show all work.

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Answer from: romaguera06

1) 2063 N

The work done by a force is given by the equation:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of F and d

In this problem, we have

F = 157 N

d = 14.5 m

$\theta=25^{\circ}$

Substituting,

W=(157)(14.5)(cos(25))=2063 N

2) 12.7 m/s

The egg is in free fall, so it is moving by uniformly accelerated motion. This means that its final velocity can be found by using the following suvat equation:

v^2-u^2=2as

where we have:

v = final velocity

u = 0 initial velocity

a = g = 9.8 m/s^2 is the acceleration of gravity

s is the displacement

The building is 8.2 m high, so the egg falls through a distance of

d = 8.2 m

Therefore, substituting and solving the equation for v, we find:

$v=\sqrt{u^2+2as}=\sqrt{0+(2)(9.8)(8.2)}=12.7 m/s$

3) 8.7 m/s

Assuming that no external forces acts on the Snookie-lemur system, then according to the law of conservation of momentum, their total momentum is conserved. This means that the initial momentum of Snookie in the wagon must be equal to the total momentum after the lemur appears in the wagon:

$p_i = p_f\\M u = (M+m)v$

where

M = 215 kg is the mass of Snookie+the wagon

u = 9.5 m/s is the initial velocity of Snookie+the wagon

m = 19.7 kg is the mass of the lemur

v is the final velocity of Snookie+the wagon + the lemur

Solving the equation for v, we find:

$v=\frac{Mu}{M+m}=\frac{(215)(9.5)}{215+19.7}=8.7 m/s$

4) Yes (he was travelling at 27.1 m/s)

According to the law of conservation of momentum, the total momentum of the two cars must be conserved before and after the collision.

This means that we can write the following equation:

m_B u_B + m_Z u_Z = (m_B + m_Z) v

where:

m_B = 1100 kg is the mass of Brady's car

m_Z = 1475 kg is the mass of Jay Z car

u_B = 20 m/s is the initial velocity of Brady's car

u_Z is the initial velocity of Jay Z car

v=-7 m/s is the final velocity of the two cars after they collide (in the negative direction, since it is in the same direction Jay Z was driving)

Solving for uZ, we find Jay Z initial velocity:

$u_Z = \frac{(m_B+m_Z)v-m_B u_B}{m_Z}=\frac{(1100+1475)(-7)-(1100)(20)}{1475}=-27.1 m/s$

Therefore, Jay Z was driving faster than the limit of 25 m/s.

5) 166.6 N

The ball is moving in a uniform circular motion, so the tension in the string provides the centripetal force that keeps the ball in circular motion.

Therefore, we can write the following equation:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the ball

v is the speed of the ball

r is the radius of the circle

In this problem,

r = 0.23 m

v = 7.8 m/s

m = 0.63 kg

Substituting these numbers into the equation, we find the centripetal force, and therefore the tension in the string:

$T=(0.63)\frac{7.8^2}{0.23}=166.6 N$

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100 points + physics

1) a force of 157 n is applied to a box 25o above the horizontal....

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