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a_sh-v [17]
2 years ago
15

A rock is dropped from a height of 3,245 m. If we ignore air resistance, how fast will it be travelling after it falls for 3.4 s

?
Physics
1 answer:
Marta_Voda [28]2 years ago
3 0

Answer:

<u>954.4m/s</u>

Explanation:

For a free falling object,it has constant acceleration and a changing velocity.

By using the velocity-time formula, the velocity can be obtained.

The height the rock travelled is the distance.

From,

Velocity (v) = Distance (d) / Time(t)

v = 3245m/3.4s

v = <u>954.4m/s</u>

That js the answer I got. Hope it's right.

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What is the most commonly used reference point
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Explanation:

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A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
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Answer:

1408.685 KN/C

Explanation:

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k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

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3 years ago
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The y-component of the force F which a person exerts on the handle of the box wrench is known to be 86 lb. Determine the x-compo
emmainna [20.7K]

Answer:

x-component of force is  38.18 lb where as magnitude of Force is 93.16

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magnitude of perpendicular = P =  12

magnitude of base = B = 5

using Pythagoras theorem

                        H= \sqrt{P^{2} + B^{2}}

                 H= \sqrt{12^{2} + 5^{2}}

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y-component of force is given given as:

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