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aleksley [76]
3 years ago
5

Which statement best compares potential and kinetic energy?

Physics
1 answer:
xenn [34]3 years ago
8 0

Answer:

D

Explanation:

Because kinetic energy do have more than potential energy: kinetic energy is when a object is moving. Potential energy is when something is at rest and has no movement what so ever

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A bowling ball has the most gravitational force.
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Find the velocity of a dropped brick after 5.0 s using v =a ∆ t and a = - 9.8 m/s^2
Verizon [17]

Answer:

49

Explanation:

9.8 x 5 = 49

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3 years ago
Which is not a way to conserve existing energy resources?
zimovet [89]
For the answer to the question above, the answer is simple.<span> among</span> the choices given the only way on not to conserve energy is by using the available fossil fuel.
I hope my answer helped you. feel free to ask more questions. Have a nice day!
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3 years ago
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What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

6 0
4 years ago
If a wave is traveling at 200 m/s and it’s wavelength is 0.5 meters , what is the frequency of the wave
Mrrafil [7]

Answer:

400 Hz

Explanation:

The frequency of a sound wave is found by the velocity/wavelength.  Therefore, f = v/wavelength = (200 m/s)/(0.5 m) = 400 Hz

I hope this helps! :)

5 0
3 years ago
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