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3 years ago

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What is an example of kinetic energy due to gravity?

1 answer:

faust18 [17]3 years ago

4 0

Answer:

An example of kinetic energy due to gravity can be a ball held in the air (Kinetic energy is energy an object has because of it's motion.).

Explanation:

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If energy cannot be created or destroyed, where does it go?

blondinia [14]

It transfers and changes into different types of energy, this is why the ground feels hot when something moves fast over it.

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3 years ago

Read 2 more answers

What would be the first thing you would do if your clothes caught fire while working in a laboratory? select one of the options

Ksivusya [100]

<span>c. run towards a source of water to extinguish the fire </span>

3 0

3 years ago

a net force of 219 N is exterted on a rock. the rock has an acceleration of 3m/s^2 due to this force. what is the mass of the ro

Sonbull [250]

Answer:

<h2>73 kg</h2>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{219}{3} \\$

We have the final answer as

<h3>73 kg</h3>

Hope this helps you

6 0

2 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

There are 3.3V passing through an orange power supply cable, and there are 0.025 ohms of resistance in the orange wire. How much

PIT_PIT [208]

P = V^2 / R.

So, 3.3^2 / 0.025 = 435.6W.

Note, you can get the power equation from:
P = V*I. Also, I = V/R.
Substituting V/R in for I in the 1st equation, you get P = V^2 / R.

5 0

3 years ago