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3 years ago

A marble rolls off a table with a speed of 2.30 m/s. If the table is 1.12 m high, how far from the

Physics

1 answer:

Vlada [557]3 years ago

6 0

Answer:

1.10 m

Explanation:

First, find the time it takes to land.

Given, in the y direction:

Δy = 1.12 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(1.12 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.478 s

Next, find the distance traveled in that time.

Given, in the x direction:

v₀ = 2.30 m/s

a = 0 m/s²

t = 0.478 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.30 m/s) (0.478 s) + ½ (0 m/s²) (0.478 s)²

Δx = 1.10 m

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3 years ago

The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?

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Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

$Initial\ Distance = 2cm$ ---- $r_1$

$New\ Distance = 4cm$ --- $r_2$

Required

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The initial force F1 is:

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$F_1 = \frac{kq_1q_2}{2^2}$

$F_1 = \frac{kq_1q_2}{4}$

The new force (F2) is

$F_2 = \frac{kq_1q_2}{r_2^2}$

Substitute 4 for r2

$F_2 = \frac{kq_1q_2}{4^2}$

$F_2 = \frac{kq_1q_2}{4*4}$

$F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}$

Substitute $F_1 = \frac{kq_1q_2}{4}$

$F_2 = \frac{1}{4}*F_1$

$F_2 = \frac{F_1}{4}$

The new force is 1/4 of the previous force.

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2 years ago

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Answer:

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