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Serga [27]
3 years ago
8

What is the relationship between Potential and Kinetic Energy

Physics
1 answer:
Wittaler [7]3 years ago
3 0
Potential energy is to stored with in an object due to position and arrangement. Kinetic energy is to energy with in an object due its to movement with their motion.

It will help you.

-Charlie

Have a great day
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Sound source A is located at x = 0, y = 0, and sound source B is located at x = 0, y = 2.3 m. The two sources radiate coherently
Oliga [24]

Answer:

Solution is given in the attachments

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3 years ago
A rifle fires a bullet at a velocity of 78 m/s, 40 degrees above the horizontal. Determine the height (h) above the starting pos
qwelly [4]

Answer:

H = Vy t - 1/2 g t^2  height of an object with an initial "vertical" velocity

                                 at t sec after firing

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H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m

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bagirrra123 [75]

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5 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
3 years ago
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