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Deffense [45]
3 years ago
11

Please help ASAP. A 1550 kg car is initially at rest on level ground when the engine does 450 000 J of work on it. After the car

reaches its top speed, the engine cuts out and the car begins to roll up a hill. If the car continues to roll until its velocity is zero, what final height will it reach?
Physics
1 answer:
AlekseyPX3 years ago
5 0

Answer:

29.6 m

Explanation:

Work is change in KE, and at the bottom of the hill, the car only has KE. Therefore, the car's total mechanical energy (ME) at the bottom of the hill must be equal to its KE (450 000 J). At the top of the hill, the car is no longer moving, so its total energy must be all gravitational potential energy. Following convervation of ME:

ME_{i} = ME_{f}

KE_{i} = GPE_{f}

450 000 J = mgh_{f}

h_{f} = \frac{450 000 J}{m * g}

h_{f} = \frac{450 000 J}{1550 kg * 9.81 m/s^{2} }  = 29.6 m

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A leftward force of 7N is applied to a box while the force of friction is 3N. The weight of the box is 4N as is the normal force
Flauer [41]

Answer:

0N

Explanation:

When you draw a free body diagram, You have a 7N arrow pointing left, and a 3N arrow pointing right. When you subtract the two, you still have a force of 4N on the left. However, seeing as how gravity is in play at a force of 4N, the two forces cancel out and you get 0N.

8 0
3 years ago
8. A car initially has 100 J of total energy. After driving down the road, the car's
sveta [45]

The work done on the car is -20 J.

Work done on the car is negative, meaning that the car actually does work on the external system.

<h3>Energy and law of conservation of energy</h3>
  • Energy is the ability to do work
  • the law of conservation of energy states that the total energy in a system is conserved

From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.

  • Initial energy = 100 J
  • Initial energy = Final energy - work done on car
  • Final Energy = Work done on car + initial energy

80J = Work done on car + 100 J

Work done on car = 80 - 100J

Work done on car = -20 J

Hence, the work done on the car is -20 J

Work done on car is negative.

Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.

Learn more about energy and work at: brainly.com/question/13387946

8 0
2 years ago
What is shot-curciting​
rjkz [21]

Answer:

A path that allows most of the current in an electric circuit to flow around or away from the principal elements or devices in the circuit.

3 0
3 years ago
A battery of voltage V delivers power P to a resistor of resistance R connected to it. By what factor will the power delivered t
Anettt [7]

Answer:

Explanation:

Power P = V² / R

a ) The resistance is changed to 2.90R

Power will become 1 / 2.9 times .

b )The voltage of the battery is now 2.90V, but the resistance is R

P = (2.9V)² / R

= 8.41 x V² / R

So power becomes 8.41 times

c )The resistance is 2.90R and voltage is 2.90V

Power P = (2.9V)² / 2.9 R

= 2.9 V²/R

So power becomes 2.9 times

d ) The resistance is 2.90R and the voltage is V/2.90

Power P = ( V/2.90)² x 1 / 2.90R

1 / ( 2.9 )³ x V² / R

= 1 / 24.389 x V² / R

So power becomes  1 / 24.389 times .

4 0
3 years ago
A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a heigh
dlinn [17]

Answer:

a) t = H/v0

b) H = -(v0)²/g

Explanation:

Hi there!

a)The position of the balls can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t.

y0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity.

t = time.

For the ball that is thrown upwards, the initial height is zero, then, the equation can be written as follows:

y = v0 · t + 1/2 · g · t²

The second ball is initially at a height H and the initial velocity is zero. The equation of height for the second ball will be:

y = H + 1/2 · g · t²

When the two balls collide, their height is the same. Then, equalizing both equations we can obtain the time at which they collide:

v0 · t + 1/2 · g · t² = H + 1/2 · g · t²

v0 · t = H

t = H/v0

b) When the first ball is at the highest point its velocity is zero. Using the equation of velocity we can find the time at which the ball is at that point. The equation of velocity is the following:

v = v0 + g · t

At the highest point v = 0.

0 = v0 + g · t

Solving for t:

-v0/g = t

The time at which the first ball is at the highest point is t = -v0/g

The time at which both balls collide was calculated above:

t = H/v0

Then, equalizing both times and solving for H:

H/v0 = -v0/g

H = -v0/g · v0

H = -(v0)²/g

3 0
3 years ago
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