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3 years ago

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The elevators in the Landmark Tower in Yokohama, Japan are among the fastest in the world. After starting from rest, they reach

a maximum speed of 120 meters per minute. It takes the elevator 4.0 seconds to reach maximum speed. What is its acceleration in m/s/s (meters per second per second)?

1 answer:

MariettaO [177]3 years ago

3 0

Answer:

The acceleration is 30 m/s²

Explanation:

Given that,

Maximum speed = 120 m/s

Time = 4.0 sec

We need to calculate the acceleration

Using equation of motion

$v=u+at$

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

$120=0+a\times4.0$

$a=\dfrac{120}{4.0}$

$a=30\ m/s^2$

Hence, The acceleration is 30 m/s²

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Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would

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Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

$W = mg$

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W = Weight of the object on surface of earth

m = mass of object

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Similarly, the weight of the object on Jupiter will be given as:

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where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

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A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W

Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s$

Since, the motion is downward, final velocity must be negative. So,

$v_d=-4.95\ m/s$

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Given:

Displacement of ball (S) = 0.665 m

Initial velocity ( $v_d$ ) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s$

Since, the motion is upward, final velocity must be positive. So,

$v_{up}=3.31\ m/s$

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

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Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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Substituting the values, we get time of flight

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Answer:

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