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3 years ago

At what point on the track does a roller coaster have the greates kinetic energy?

Physics

1 answer:

vredina [299]3 years ago

3 0

At the bottom of the first and biggest drop. I hope I helped ^u^

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2. What will be the extension of this spring if the load is a) 4N and b) 75 g?

Furkat [3]

Answer:

Explanation:

just add

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2 years ago

_________ is a layer of the earth that is classified not by composition

Sidana [21]

Answer:

Asthenosphere

Explanation:

The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.

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3 years ago

A ball is thrown up into the air. The time that it takes to go up equals

hammer [34]

Answer:

The time it takes the ball to rise equals the time it takes to fall.

Explanation:

because what goes up at some point must come down

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3 years ago

A soccer ball is kicked and left the ground at angle 45° above the horizantal, moving at 25 m/s.

Tcecarenko [31]

Answer:

a) 3.6 sec

b) 17.7 m/s

c) 17.7 m/s

d) 63.6 m

Explanation:

y = (25sin45)t - 4.9t^2 =17.7t - 4.9t^2

a) to find hang time set y = 0

4.9t^2 = 17.7t or t = 3.6 sec

b) vx = 25cos45 = 17.7 m/s

c) vy = 25sin45 = 17.7 m/s

d) To find the range R, use the results in (a) and (b) into the equation R = vxt:

R = (17.7 m/s)(3.6 sec) = 63.6 m

8 0

3 years ago

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m². The father

natulia [17]

Answer:

Part a)

$KE = 101.4 J$

Part b)

$N = 0.043 revolution$

Part c)

F = 2.7 N

Explanation:

Part a)

As we know that the rotational kinetic energy of the merry go round is given as

$KE = \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}84.4(\omega^2)$

here we know that

$\omega = 2\pi(\frac{14.8}{60})$

$\omega = 1.55 rad/s$

Now we have

$KE = \frac{1}{2}(84.4)(1.55^2)$

$KE = 101.4 J$

Part b)

Now we know that work done due to torque = change in kinetic energy

$W = KE_f - KE_i$

$\tau (2N\pi) = 101.4 - 0$

$375(2\pi N) = 101.4$

$N = 0.043 revolution$

Part c)

In order to stop it in four revolutions we have

$\tau(2\pi N) = \Delta KE$

$FR(2\pi N) = 101.4$

$F(1.5)(2\pi \times 4) = 101.4$

F = 2.7 N

3 0

2 years ago