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natima [27]
2 years ago
8

When a car is coasting downhill, how are its potential and kinetic energies changing?

Physics
1 answer:
olasank [31]2 years ago
3 0

When a car is coasting downhill, the kinetic and potential energies are increasing and decreasing respectively.

<h3>What are kinetic and potential energy?</h3>

Kinetic energy is the energy possessed by an object because of its motion, equal to one half the mass of the body times the square of its speed.

Potential energy, on the other hand, is the energy possessed by an object because of its position (in a gravitational or electric field), or its condition (as a stretched or compressed spring, as a chemical reactant, or by having rest mass).

According to this question, a car going downhill will begin to speed because there is lesser friction. This suggests that the kinetic energy increases while the potential energy decreases.

Learn more about potential energy at: brainly.com/question/24284560

#SPJ1

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If the car moves along the distance it will be 16 of the line graph where is independent of the graph
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An object travels a distance of 56 meters to the right in 7 seconds. What is the object's velocity?
Yakvenalex [24]

Explanation:

Distance travelled (d) = 56 metres

Time taken (t) = 7 seconds

velocity of the object (V)

= d / t

= 56 / 7

= 8 m/s

The velocity of the object is 8 m/s.

Hope it will help :)

8 0
3 years ago
How will physical and/or chemical weathering contribute to create rounded shapes during spheroidal weathering?
Vilka [71]
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6 0
4 years ago
How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 45 j of work?
bija089 [108]

Here is your answer

5 coulomb

REASON :

We know that

Potential difference, V= W/q

where, W is work done

and, q is magnitude of charge

Given,

V= 9.0 v and W= 45 J

So,

using above relation, we get

9= 45/q

q= 45/9

q= 5 coulomb

HOPE IT IS USEFUL

6 0
3 years ago
Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea
Semmy [17]

Answer:

{\rm K} = 2.4\times 10^{-19}~J

Explanation:

The electric field inside a parallel plate capacitor is

E = \frac{Q}{2\epsilon_0 A}

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

F = qE = \frac{qQ}{2\epsilon_0 A}

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60

Plugging this identity into the force equation above gives

F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is 1.6 \times 10^{-19}

Therefore, the kinetic energy is 2.4\times 10^{-19}

8 0
3 years ago
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