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3 years ago

7

If you drop an object from a height of 1.9 m, it will hit the ground in 0.62 s. if you throw a baseball horizontally with an ini

tial speed of 34 m/s from the same height, how long will it take the ball to hit the ground?

1 answer:

faltersainse [42]3 years ago

6 0

That's the cool thing about free fall. The amount of time it takes to fall remains the same.

In this case, a ball that is simply dropped from rest will fall at the same rate as a ball that had some umph in the horizontal direction.

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What is the kinetic energy of a 620.0 kg roller coaster moving with a velocity of 9.00 m/s?

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Ek = 1/2 x m x v ²

1/2 x 620.0 x 9.00 ² = 25,100

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An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

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$W = 0.44\ kJ$

c) Power output

t = 0.296 s

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3 years ago

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2 years ago

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A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

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3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

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The force exerted by the ring to push her is equal to 422.36 N.

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2 years ago