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Yuki888 [10]
3 years ago
13

If a ball is dropped from rest, what is it’s velocity after 4 seconds

Physics
1 answer:
Ghella [55]3 years ago
4 0

Answer: -39.2 m/s directed downwards

Explanation:

We can solve this problem with the following equation:

V=V_{o}+gt

Where:

V is the ball's velocity at 4 s

V_{o}=0m/s is the ball's intial velocity, since it was dropped from rest

g=-9.8 m/s^{2} is the acceleration due gravity, aways directed downwards

t=4 s is the time

Solving:

V=0m/s-9.8 m/s^{2}(4 s)

Finally:

V=-39.2 m/s <u>The negative sign indicates the velocity is downwards</u>

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If the angle of incidence is 30°, degrees what is the angle of reflection? ​
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Answer:

Angle of reflection = 30°

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A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

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<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

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Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

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