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Free_Kalibri [48]
3 years ago
12

If a 5 kg bowling ball is lifted to a height of 2 meters, how much potential

Physics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 98.1 \ Joules}}

Explanation:

The formula for potential energy is:

E_p=mgh

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass is 5 kilograms and the height is 2 meters. Assuming this ball is on Earth, then the gravitational acceleration is 9.81 meters per square second.

m= 5 \ kg \\g= 9.81 \ m/s^2 \\\h= 2 \ m

Substitute the values into the formula.

E_p= ( 5 \ kg )(9.81 \ m/s^2)( 2 \ m)

Multiply the first two numbers.

E_p=49.05 \ kg*m/s^2 (2 \ m )

Multiply again.

E_p= 98.1 \ kg*m^2/s^2

  • 1 kilogram square meter per square second is equal to 1 Joule.
  • Our answer is equal to 98.1 Joules

E_p= 98.1 \ J

The ball has 98.1 Joules of potential energy.

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Answer

Nature of the surfaces.


Explanation

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Friction is affected by a number of factors.

One is the weight of the object. The more the weight of the object the higher the friction between it and the surface.

The other factor is the nature of the surfaces. Rough surfaces contribute to high friction while smooth surfaces reduces the friction between surfaces.  

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3 years ago
Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema - a sunburn. The amount of exposure ne
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Answer:

Energy = 7.83 x 10⁻¹⁹ J

Energy = 6.63 x 10⁻¹⁹ J

Explanation:

The energy of a photon in terms of wavelength can be calculated by the following formula:

Energy = \frac{hc}{\lambda}\\

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λ = wavelength of light

Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:

Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\

<u>Energy = 7.83 x 10⁻¹⁹ J</u>

<u></u>

Now, for λ = 300 nm = 3 x 10⁻⁷ m:

Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\

<u>Energy = 6.63 x 10⁻¹⁹ J</u>

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3 years ago
Which of the following properties must be known in order to calculate the amount of heat needed to melt 1.0 kg of ice at 0ºC?
meriva
If you're only melting ice ... turning solid ice at 0°C into liquid water at 0°C ...
then you only need to know t<span>he latent heat of fusion for water</span>. 

That's exactly what it means ... the amount of energy involved when a gram
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Well what are you comparing or try to pair together?

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