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3 years ago

A crate of 890 kg gets raised to a height of 2.1m. Calculate the crate's gravitational potential energy.

Physics

2 answers:

Licemer1 [7]3 years ago

8 0

Gravitational potential energy = mass x height x gravity

So you would do 890 x 2.1 x gravity (which is 9.8 on earth)

mars1129 [50]3 years ago

4 0

Answer:

18,316.2

Explanation:

The formula for GPE is mgh, where

M = Mass of the object

G = Acceleration due to gravity (9.8 m/s^2 on Earth)

H = Height above ground

890 × 9.8 × 2.1 = 18,316.2

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3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

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Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

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For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

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We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

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Answer:

Explanation:

Given

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Required

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Hence the rider's velocity after the acceleration is 14.97m/s

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6 0

3 years ago

A crate of 890 kg gets raised to a height of 2.1m. Calculate the crate's gravitational potential energy.​

A crate of 890 kg gets raised to a height of 2.1m. Calculate the crate's gravitational potential energy.