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Leviafan [203]
3 years ago
7

When the Moon is at a distance of 0.25 D from the Earth, the gravitational force on the Moon from the Earth is 16.00 F. At what

distance will the force on the Moon be 4.00 F?
Physics
1 answer:
puteri [66]3 years ago
6 0

Answer:

On the way to the Moon the Apollo astronauts reached a point where the. Moon's gravitational pull became stronger, than the Earth's. (a) Determine the distance of ...

Explanation:

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Oscillation whose amplitude reduce with time are called damped oscillation. This happen because of the friction. In oscillation if its amplitude doesn't change with time then they are called Undamped oscillation

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The best way to prevent heat exhaustion while exercising is to
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Take breaks and drink plenty of water not to much because it will make you sick sorry apparently pu ke is a bad word.
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A 75 kV x-ray machine gives an exposure of 200 mR for a 0.2-sec chest x-ray exam.
pashok25 [27]

Explanation:

Given that,

Potential = 75 kV

Exposure = 200 mR

Time = 0.2 sec

We need to calculate the x-ray fluence during this chest x-ray exam

Using formula of fluence

x-ray\ fluence=exposure\times time

Put the value into the formula

x-ray\ fluence=200\times0.2

x-ray\ fluence=40\ mRs

We need to calculate the energy fluence

Using formula of energy fluence

E_{f}=40\times10^{-3}\times75\times10^{3}

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We need to calculate dose -equivalent delivered to the bone, muscle, and fat

Using formula of dose

D=\dfrac{E}{t}

Where, D = dose

E = energy

t = time

Put the value into the formula

D=\dfrac{3000}{0.2}

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Hence, This is the required solution.

7 0
4 years ago
A crane lifts a car with 9023 J of kinetic energy. The crane lifts the car with a speed
lapo4ka [179]

Answer:

m = 2301.8 [kg]

Explanation:

This problem can be solved in an easy way using the definition of kinetic energy, in this way you can clear the mass, depending on the known variables.

E_{k}=0.5*m*v^2\\replacing\\\\9023 = 0.5 *m*(2.8)^2\\m = 2301.8[kg]

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3 years ago
A particular electric car is supplied with 300 kJ of chemical energy by the battery. Of this, a total of 70.5 kJ of energy is wa
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Supplied energy=300kJ

  • Wasted energy=70.5J

Used energy:-

\\ \sf\longmapsto 300-70.5=229.5kJ

We know

\boxed{\sf Efficiency=\dfrac{Used\:Energy}{Supplied\:Energy}\times 100}

\\ \sf\longmapsto Efficiency=\dfrac{229.5}{300}\times 100

\\ \sf\longmapsto Efficiency=\dfrac{229.5}{3}

\\ \sf\longmapsto Efficiency=76.5\%

3 0
3 years ago
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