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8090 [49]
2 years ago
11

What are the things being compared in the following argument?

Physics
1 answer:
White raven [17]2 years ago
3 0

Answer:

C

Explanation:

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Answer:

Empirical formula

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3 years ago
A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at
Rasek [7]

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

W_{all}=Change\ in\ kinetic\ energy\ of\ system

We know that kinetic energy of moving mass given as

KE=\dfrac{1}{2}mv^2

Now by pitting the values

KE=\dfrac{1}{2}mv^2

KE=\dfrac{1}{2}\times 0.011\times 29^2

KE= 4.25 J

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3 years ago
Which of the following would be a good question that could be scientifically investigated? What is the best advice a parent can
Nesterboy [21]
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4 0
3 years ago
Read 2 more answers
Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli
Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

Explanation:

Given:

Mass of  Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = v_{f}=?

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}

Substituting the values we get

2\times 14 + 5\times 0 =(2+5)\times v_{f}

v_{f}=\dfrac{28}{7}=4\ m/s

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

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3 years ago
Asap plz Which statement describes a controlled experiment?
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