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goldfiish [28.3K]

3 years ago

7

A Viking ship roller coaster at the fair has a mass of 36,000 kg. If at its peak it reaches a height of 20 m off the ground, how

much gravitational potential energy does it have?

1 answer:

adell [148]3 years ago

8 0

Answer:

7056 kJ

Explanation:

Given that,

Mass of a ship roller coaster is 36,000 kg.

It reaches a height of 20 m off the ground

We need to find the gravitational potential energy does it have. The formula for the gravitational potential energy ios given by :

E = mgh

g is acceleration due to gravity

E = 36,000 kg × 9.8 m/s² × 20 m

= 7056000 J

or

E = 7056 kJ

So, it will have 7056 kJ of gravitational potential energy.

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A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$a_{max} = 0.00246 \ m/s^2$

b

$k =722.2 \ N/m$

Explanation:

From the question we are told that

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Generally this test weight is mathematically represented as

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7 0

3 years ago

A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational mome

Leya [2.2K]

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

$L=I\omega$

$L=\dfrac{1}{2}mr^2\omega$

Put the value into the formula

$L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320$

$L=12.96\ kg m^2/s$

(B). Rotation momentum of sphere is same rotational momentum of the flywheel

We need to calculate the magnitude of the rotational speed of sphere

Using formula of rotational momentum

$L_{sphere}=L_{flywheel}$

$I\omega_{sphere}=I\omega_{flywheel}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}$

Put the value into the formula

$\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}$

$\omega_{sphere}=400\ rad/s$

Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

4 0

3 years ago