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Kruka [31]
3 years ago
8

When a rubberband is stretched all the way back, it is an example of which type of energy?

Physics
2 answers:
Murljashka [212]3 years ago
8 0
Elastic potential energy. because elastic is were something is pulled and when let go it pulls its self back to the original position so when you pull the rubberband it builds up energy to pull its self back together
 
avanturin [10]3 years ago
5 0
The answer is C. elastic potential energy
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What is the unit of the quotient of inductance and resistance?
Scorpion4ik [409]

The unit of the quotient of inductance and resistance will be Henry and ohm

Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. The flow of electric current creates a magnetic field around the conductor. The field strength depends on the magnitude of the current, and follows any changes in current.

Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.

unit of quotient of inductance = henry (H)

unit of resistance = ohm

To learn more about resistance here

brainly.com/question/14547003

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5 0
3 years ago
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A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this
krok68 [10]

Answer:

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Explanation:

8 0
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7 0
3 years ago
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Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uni
nexus9112 [7]

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

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3 years ago
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