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Brums [2.3K]
3 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]3 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]3 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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4 0
3 years ago
I will give Brainliest to whoever can answer!!!!
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Answer:

The nodes and anti nodes would reverse roles.

Explanation:

I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).

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In a collision between two unequal masses, which mass receives a greater magnitude impulse?
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<h3>Answer:  130 meters</h3>

===================================================

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represents the distance traveled. The first equation shown above is one of the four kinematics equations.

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