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Black_prince [1.1K]
3 years ago
10

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after

the objects collide and bounce apart?
Physics
2 answers:
Bezzdna [24]3 years ago
7 0

Answer:

On edg it should be the second one

Explanation:

I took the tst

s2008m [1.1K]3 years ago
6 0

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

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Answer:

Gravity :)

Explanation:

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How much heat energy is required to raise the temperature of 0.368kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of cop
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The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.

Heat required = mCΔT
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3 years ago
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For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci
MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

z = \frac{V_o}{i_o}

now we will have

z = \frac{125}{3.20} = 39.06 ohm

Part b)

Now we also know that

\frac{R}{z} = cos\theta

\theta = 0.982 rad = 56.3 degree

now we have

\frac{R}{39.06} = cos56.3

R = 21.7 ohm

5 0
3 years ago
if a pressure of 70.kPa on a volume of 80.cm cubed is reduced to 10.kPa, by what factor does the pressure change
boyakko [2]

70-10/70 x 100  percentage change ....

60/70, 6/7 fract change


6 0
3 years ago
An engine has an energy input of 125j and 35 j of that energy is transformed into useful energy
Tatiana [17]
The engine's efficiency is (35J)/125J) = 28% .

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5 0
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