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Brums [2.3K]
2 years ago
7

How high has a 45 g golf ball been driven if it has a potential energy of 2.06 J?

Physics
2 answers:
Mice21 [21]2 years ago
7 0
PE=mgh
2.06 J = 2.06 kg·m²/s² = (0.045 kg)·(9.8 m/s²)·h
where h is the height the ball has reached
Solve for h:
h = 2.06/(0.045·9.8) = 4.67 to the nearest hundredth
h = 4.67 m

nydimaria [60]2 years ago
4 0
Here, U = mgh

We have: m = 45 g = 0.045 Kg
U = 2.06 J, g = 9.8 m/s², h = ?

Substitute the values in the formula:
2.06 = 0.045 × 9.8 × h
2.06 = 0.441 × h
h = 2.06 / 0.441
h = 4.67 m

In short, your final answer would be 4.67 meters

Hope this helps!

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Answer:

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Explanation:

The first sentence of this question is not explanatory enough. However, I'll assume the force to be 15N

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Required

Solve for the x and y components

Since the given angle is to the horizontal, the x and y coordinates are calculated using the following illustrations.

Sin\theta = \frac{y}{Force} ---- y component

Cos\theta = \frac{x}{Force} ---- x component

Calculating the y component.

Substitute 15 for Force and 36.7 for \theta

Sin\theta = \frac{y}{Force} becomes

Sin(36.7) = \frac{y}{15}

Make y the subject

y = 15 * Sin(36.7)

y = 15 * 0.5976

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Calculating the x component.

Substitute 15 for Force and 36.7 for \theta

Cos\theta = \frac{x}{Force} becomes

Cos(36.7) = \frac{x}{15}

Make y the subject

x = 15 * Cos(36.7)

x = 15 * 0.8018

x = 12.027N

<em>Hence, the x and y components of the force are: 8.964N and 12.027N respectively.</em>

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