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3 years ago

An object is dropped from a height h. during the final second of its fall, it traverses a distance of 41.2 m. what was h?

1 answer:

3 0

41.2 = h-1/2g(t-1)^2
<span> {-h = -1/2gt^2-1/2g+g*t-41.2
</span><span> {h = 1/2gt^2
</span><span> summing them up
</span><span> 0 = -1/2g+g*t-41.2
</span><span> 41.2 +4.9 = g*t
</span><span> t = 46.1/9.8 = 4.70 sec
</span><span> h = 1/2gt^2 =4.9*(4.70^2) = 108.241 m </span>

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Final velocity of the car ( $v_{f}$ ) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)

Deceleration = (( $v_{f}$ ) - ( $v_{i}$ )) ÷ time (m/s²)

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Deceleration = (- 20) ÷ 3 (m/s²)

Deceleration = - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is

Distance traveled by the car is equals to the product of the speed and time

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(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)

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0.5 = e-^(t/RC)

1/2 = e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

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

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

U = ¹/₂Q²/C

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t = 1.59 x 10⁻³ s

t = 1.59 ms

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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