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3 years ago

14

I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second

s. The acceleration is a constant. How much is the acceleration?

2 answers:

kozerog [31]3 years ago

8 0

Answer:

-1.4 m/s²

Explanation:

The initial velocity is 7 m/s. When the ball rolls back, its velocity is -7 m/s.

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (-7 m/s − 7 m/s) / 10 s

a = -1.4 m/s²

Degger [83]3 years ago

3 0

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

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1 year ago

Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

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With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

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Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

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$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

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For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

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$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

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∅ = 65.7 degrees

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3 years ago

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