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JulsSmile [24]
3 years ago
9

What is the relationship between elevation and energy

Physics
1 answer:
jasenka [17]3 years ago
7 0

The Energy related to object motion.Potential energy most commonly refers to gravitational potential energy, which is energy associated with an object's elevation. For an object of mass m and elevation z, its potential energy is mgz where g is the gravitational constant .............. HOPE IT HELPS YOU :)

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(please help ASAP!)
makkiz [27]

Answer:

It's C. Length

Explanation:

Just think of a ruler. Hope I helped! :)

8 0
3 years ago
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Bicycle in its writers have combined a mass of 80 kg in a speed of 6.0 meters per Second what is the magnitude of the average fo
erik [133]

Answer:

F = 120 N

Explanation:

Force x distance = energy

The bike has energy 1/2 . 80 . 6^2   = 1440 J

You are looking at an example of not reading the question properly.

Impulse = Force . time  = change in momentum

F . 4  = 80 .6

F = 120 N

7 0
3 years ago
What process occurs when all of the energy from light waves is transferred to a medium?
Wittaler [7]
Absorption happens when <span>all of the energy from light waves is transferred to a medium.</span>
8 0
3 years ago
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A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

P = VI

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

V = IR

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

3 0
3 years ago
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Determine la fuerza aplicada sobre un émbolo en una prensa hidráulica si se quiere cargar una lanza de 2970kg; toma en cuenta qu
Artist 52 [7]
I’m not sure sorry I really wish I could help
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2 years ago
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