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Keith_Richards [23]
3 years ago
9

How does reducing the mass of a moving object by half (1/2) change its kinetic energy?​

Physics
1 answer:
allochka39001 [22]3 years ago
7 0
Kinetic energy = 1/2 m v²
If we reduce the mass by half > m/2
Kinetic energy = 1/2 m/2 v²
We should know that 1/2 × 1/2 = 1/4
So kinetic energy will be :
1/4 × m × v²
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3 years ago
AP Physics I, shouldn't be too hard.
Nana76 [90]

Answer:

The correct option is;

D. The kinetic energy decreases by 3·m₀·v₀²

Explanation:

The given parameters are;

The mass of object X = m₀

The initial velocity of object X = v₀

The mass of object Y = 2·m₀

The initial velocity of object Y = -2·v₀

By conservation of linear momentum, we have;

The total initial momentum = The total final momentum

Therefore, we have;

The total initial momentum = m₀·v₀ - 2·m₀·2·v₀ = The total final momentum

∴ The total final momentum = -3·m₀·v₀

The total mass of the two object after sticking together = 2·m₀ + m₀ = 3·m₀

Therefore, the velocity of the two objects after collision = (The total final momentum)/(Total mass) = -3·m₀·v₀/(3·m₀) = -v₀

The kinetic energy = 1/2 × Mass × (Velocity)²

Therefore, the kinetic energy after collision = 1/2 × (3·m₀) × v₀² = 3·m₀·v₀²/2

The kinetic energy before collision = 1/2 × m₀ × v₀² + 1/2 × (2·m₀) × (2·v₀)² = (1/2 + 4) × (m₀·v₀²)

∴ The kinetic energy before collision =  9·(m₀·v₀²)/2

The change in kinetic energy = The kinetic energy after collision - The kinetic energy before collision = 3·m₀·v₀²/2 - 9·(m₀·v₀²)/2 = -3·m₀·v₀²

Therefore, the kinetic energy decreases by 3·m₀·v₀².

5 0
3 years ago
How many ohms of resistance must be present in a circuit that has 50 volts and a current of 2 amps.
Lady_Fox [76]

Answer:

25 ohms

Explanation:

Hi there!

Ohm's law states that V=IR where V is voltage, I is current and R is resistance.

Plug the given information into Ohm's law (V=50, I=2) and solve for R

V=IR\\50=2R

Divide both sides by 2 to isolate R

\frac{50}{2}= \frac{2R}{2}  \\25=R

Therefore, there are 25 ohms of resistance present in this circuit.

I hope this helps!

5 0
3 years ago
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