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Solnce55 [7]
2 years ago
14

What is the action force and reaction force of two bumper cars collide

Physics
1 answer:
raketka [301]2 years ago
7 0
<span>action is the one car hitting the other, reaction is the other car being pushed away</span>
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a 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is intially at rest at a spotlight. The car an
harkovskaia [24]

Answer:

3.75 m/s south

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Since the car and truck stick together, v₁ = v₂.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:

(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v

-22500 kg m/s = 6000 kg v

v = -3.75 m/s

The final velocity is 3.75 m/s to the south.

4 0
2 years ago
Read 2 more answers
Giving brainiest to correct answer.
mixas84 [53]

Answer:

5.33\ m/s

Explanation:

We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s

Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s

As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s

5 0
3 years ago
How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.  

3 0
2 years ago
Read 2 more answers
An object in motion will have a speed which is a scaler , or ( blank ) which is a vector .
mylen [45]

Answer:

Velocity

Explanation:

Objects in motion usually have a speed which is scalar or velocity which is a vector.

A scalar quantity is one with magnitude but has no directional attribute.

A vector quantity is one with both magnitude and directional attribute.

Speed is a scalar quantity that describes the magnitude of motion a body accrues.

Velocity is a vector quantity that describes both the magnitude of motion and the direction of motion in a body.

3 0
3 years ago
10. Calculate the kinetic energy of a running back that has a mass of 80 kg and
EastWind [94]

Answer:

The answer is

<h2>2560 J</h2>

Explanation:

The kinetic energy of an object given it's mass and velocity can be found by using the formula

KE =  \frac{1}{2} m {v}^{2}

where

m is the mass

v is the velocity

From the question

m = 80 kg

v = 8 m/s

The kinetic energy is

KE =  \frac{1}{2}  \times 80 \times  {8}^{2}  \\  = 40 \times 64

We have the final answer as

<h3>2560 J</h3>

Hope this helps you

5 0
2 years ago
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