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3 years ago

A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.

Physics

1 answer:

AlexFokin [52]3 years ago

4 0

Initial velocity (u) = 0 m/s
Time taken (t) = 2.9s
Acceleration due to gravity (g) = + 10 m/s² [Down]

<h2><u>To calculate</u>,</h2>

Final velocity (v)
Height (h)

<h2><u>Solution</u><u>,</u></h2>

→ v = u + gt

→ v = 0 + 10(2.9)

→ v = 29 m/s $\qquad$ … ( Ans )

And,

→ h = ut + ½gt²

→ h = 0(2.9) + ½ × 10 × (2.9)²

→ h = 5 × 8.41

→ h = 42.05 m $\qquad$ … ( Ans )

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Answer:

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Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

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$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
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$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

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