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Artemon [7]
2 years ago
6

An object is dropped from rest into a pit, and accelerates due to gravity at roughly 10 m/s2. It hits the ground in 5 seconds. A

rock is then dropped from rest into a second pit, and hits the ground in 10 seconds. How much deeper is the second pit, compared to the first pit?
Physics
1 answer:
vitfil [10]2 years ago
7 0

Answer:

Second pit is 375 m deeper compared to first pit.

Explanation:

We have equation of motion s = ut + 0.5at²

First object hits the ground after 5 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 5 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 5 + 0.5 x 10 x 5²

                    s = 125 m

           Depth of pit 1 = 125 m

Second object hits the ground after 10 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 10 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 10 + 0.5 x 10 x 10²

                    s = 500 m

           Depth of pit 2 = 500 m

Difference in depths = 500 - 125 = 375 m

Second pit is 375 m deeper compared to first pit.

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Answer:

C = 3.77*10⁻¹⁰ F = 377 pF

Q = 1.13*10⁻⁵ C

Explanation:

Given

D = 8.0 cm = 0.08 m

d = 0.95 cm = 0.95*10⁻² m

k = 80.4  (dielectric constant of the milk)

V = 30000 V

C = ?

Q = ?

We can get the capacitance of the system applying the formula

C = k*ε₀*A / d

where

ε₀ = 8.854*10⁻¹² F/m

and   A = π*D²/4 = π*(0.08 m)²/4

⇒  A = 0.00502655 m²

then

C = (80.4)*(8.854*10⁻¹² F/m)*(0.00502655 m²) / (0.95*10⁻² m)

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Now, we use the following equation in order to obtain the charge on each plate when they are fully charged

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7 0
2 years ago
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Answer:

Part a)

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Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

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t_1 = \frac{\omega_1}{\alpha}

Part b)

angular displacement is given as

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Part c)

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t_1 = \frac{\omega_1}{\alpha}

so we have

t = \frac{t_1}{5}

6 0
2 years ago
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