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2 years ago

A 3 kg rock sits on a 0.8 meter ledge. If it is pushed off, how fast will it be going at the bottom?

1 answer:

andrey2020 [161]2 years ago

6 0

As long as it sits on the shelf, its potential energy
relative to the floor is . . .

 Potential energy = (mass) x (gravity) x (height) =

 (3 kg) x (9.8 m/s²) x (0.8m) = 23.52 joules .

If it falls from the shelf and lands on the floor, then it has exactly that
same amount of energy when it hits the floor, only now the 23.52 joules
has changed to kinetic energy.

 Kinetic energy = (1/2) x (mass) x (speed)²

 23.52 joules = (1/2) x (3 kg) x (speed)²

Divide each side by 1.5 kg : 23.52 m²/s² = speed²

Take the square root of each side: speed = √(23.52 m²/s²) = 4.85 m/s  (rounded)

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If wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:

GenaCL600 [577]

The frequency of the wheel is given by:
$f= \frac{N}{t}$
where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:
$f= \frac{100}{10 s}=10 s^{-1}$

And now we can find the angular speed of the wheel, which is related to the frequency by:
$\omega=2 \pi f=2 \pi (10 s^{-1})=62.8 s^{-1}$

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3 years ago

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Answer:

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Explanation:

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2 years ago

Jamie had a solid, angular crystal and noticed that it was very hard to break. Which model could represent the crystal?

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Explanation:

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2 years ago

If you drop a 50 gram piece of metal that has a temperature of 110°Celsius into 1000 grams of water at 25°Celsius, what best des

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If you drop a 50 gram piece of metal that has a temperature of 110°Celsius into 1000 grams of water at 25°Celsius, D.)The water and the metal’s temperature will reach the same temperature. In any system undergoing heat transfer, the objects involved will eventually reach the same temperature, signifying thermal equilibrium.

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2 years ago

Read 2 more answers

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connec

juin [17]

Answer:

Explanation:

Given that:

angular frequency = 11.3 rad/s

Spring constant (k) = $= \omega^2 \times m$

k = (11.3)² m

k = 127.7 m

where;

$x_1$ = 0.065 m

$x_2$ = 0.048 m

According to the conservation of energies;

$E_1=E_2$

∴

$\Big(\dfrac{1}{2} \Big) kx_1^2 =\Big(\dfrac{1}{2} \Big) mv_2^2 + \Big(\dfrac{1}{2} \Big) kx_2^2$

$kx_1^2 = mv_2^2 + kx_2^2$

$(127.7 \ m) \times 0.065^2 = v_2^2 + (127.7 \ m) \times 0.048^2$

$0.5395325 = v_2^2 +0.2942208 \\ \\ 0.5395325 - 0.2942208 = v_2^2 \\ \\ v_2^2 = 0.2453117 \\ \\ v_2 = \sqrt{0.2453117} \\ \\ \mathbf{ v_2 \simeq0.50 \ m/s}$

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2 years ago