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____ [38]
3 years ago
7

A driver in a 2144 kg car traveling at 15 m/s hits the brakes, coming to a stop in 67 meters. How far would it take the car to s

top if it was instead traveling at 45 m/s?
Physics
1 answer:
Komok [63]3 years ago
4 0
1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
2aS = v_f^2-v_i^2 = -v_i^2
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
a= \frac{-v_i^2}{2S}= \frac{-(15m/s)^2}{2\cdot 67 m}=-1.68 m/s^2

2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
F=ma
This force will be constant, and since m is always the same, then a is the same even in the second situation.

3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
2aS=-v_i^2
S= \frac{-v_i^2}{2a} = \frac{-(45 m/s)^2}{2\cdot (-1.68 m/s^2)}=603 m
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Answer:

A. from a low-temperature areas to high-temperature areas

Explanation:

it goes from low-temperature to high-temperature because it's reacting

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2 years ago
The ramp has an efficiency of 80%. If you perform 600 J of work, how much useful work does the ramp do
PIT_PIT [208]

The ramp does 480J of useful work with an efficiency of 80% .

<h3>What is efficiency of work done ?</h3>
  • Efficiency is the ratio of the useful energy released by a system to the input energy .
  • Mathematically, efficiency of energy = out put energy/ input energy

<h3>What is the useful work done by the ramp having efficiency 80% and an input work done 600J?</h3>
  • The efficiency =output work done/ input work done
  • 80% =output work done/ 600J
  • output work done =( 80×600)/100

=480J

Thus, we can conclude that the useful work done by the ramp is 480J.

Learn more about efficiency of energy here:

brainly.com/question/14280607

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5 0
2 years ago
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the
Paul [167]

Answer:

<em>a) 3344 N</em>

<em>b) 3344 N</em>

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = <em>3344 N</em>

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is <em>3344 N </em>

7 0
3 years ago
Calculate the mechanical advantage of a pulley system if only 10 lb is required to lift a 500 lb block
ExtremeBDS [4]

As we know that mechanical advantage is defined as the ratio of output force and input force for the pulley system

Here we know that it required 10 lb of input force in order to lift the block of 500 lb so here output force is 500 lb

now we have

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here we have

F_{out} = 500 lb

F_{in} = 10 lb

now from above formula we have

MA = \frac{500}{10}

MA = 50

so mechanical advantage of this system is 50

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The latent heat of fusion, Lf. This is the heat per kilogram needed to make the change between the solid and liquid phases, as when water turns to ice or ice turns to water.

The latent heat of vaporization, Lv.This is the heat per kilogram needed to make the change between the liquid and gas phases, as when water boils or when steam condenses into water.

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Thanks!
4 0
3 years ago
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