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Novosadov [1.4K]
3 years ago
13

2000-kg car decelerated from 10 m/s to 5 m/s. What is the work done on the car ?

Physics
1 answer:
sergey [27]3 years ago
3 0
A 2000-kg car decelerated from 10 m/s to 5 m/s. What is the work done on the car ?

Known :

Car’s mass (m) = 2000 kg

Initial speed (vo) = 10 m/s

Final speed (vt) = 5 m/s

Wanted: net work

Solution :

Net work :

Wnet= ΔEK

Wnet = ½ m (vt2 – vo2)

Wnet = ½ (2000)(52 – 102)

Wnet = (1000)(25 – 100)

Wnet = (1000)(-75)

Wnet = -75,000 Joule

The minus sign indicates that the direction of displacement is opposite with the direction of the net force.
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Which of the following has potential but not kinetic energy?
gavmur [86]

The answer would be option D "a ball sitting on a shelf." Potential energy is the amount of energy a object has while it's at rest.. (or not moving) Kinetic energy is how much energy a object is while it's moving. So in this case it's option D because a ball sitting on a shelf isn't moving therefore it has potential energy. It's not option A because thats a example of kinetic energy since how the roller coaster is moving. It's not option B because it's kinetic energy because the bike is moving. It's also not option C because it's kinetic energy because the bird is moving.


Hope this helps!

7 0
3 years ago
Whose geocentric model was accepted for 1400 years
zysi [14]
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4 years ago
Explain why a ball thrown in space could keep moving forever, while a ball thrown here on Earth will come to a stop.
Dmitriy789 [7]

Answer:

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4 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
1. Explain who is doing more work and why: a bricklayer carrying bricks and placing them on the wallof a building being
Basile [38]
Both are doing because they have chorus
7 0
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