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3 years ago

2. Does the bowling ball have more potential energy or kinetic energy as it is half way through its fall

Physics

2 answers:

lesya [120]3 years ago

3 0

Answer:

Equal

Explanation:

Recall that KE = 1/2(m)(v²) and PE = mgh, where m is mass, g is gravity, v is velocity, and h is height.

When the bowling ball is first dropped, it has a maximum potential energy but minimum kinetic energy. The height is max, so the potential energy will be greatest. Velocity is 0, so kinetic energy will be 0.

When the bowling ball is half way through its fall, the height will be half the initial height and the velocity will be half of the final velocity.

When the bowling ball is at the bottom and reaches the ground, the height is 0 so potential energy is 0 while the kinetic energy is max because velocity is the greatest.

Zinaida [17]3 years ago

3 0

Answer:

I think that it is kinetic because it is gaining speed through out it's fall.

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bagirrra123 [75]

Answer:

The group that is exposed to the variable to be tested is called as experimental group.

Explanation:

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4 years ago

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Natali5045456 [20]

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3 years ago

An electron is held at a distance from a stationary, positively charged object.

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Answer:

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Explanation:

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3 years ago

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) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th

Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

Now, we will substitute 0 for both $z_{1}$ and $z_{2}$ , 0 for w, 334.9 kJ/kg for $h_{1}$ , 2726.5 kJ/kg for $h_{2}$ , 5 m/s for $V_{1}$ and 220 m/s for $V_{2}$ .

Putting the given values into the above formula as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

$1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0$

q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

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