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3 years ago

What force is needed to give a 0.25-kg arrow an acceleration of 196 m/s/s

2 answers:

8 0

Force (f) = ?

Acceleration (a) = 196 m/s^2

Mass (m) = 0.25 kg

F = (m) • (a)

F = (0.25) • (196)

F = 49 N

Answer : 49 N

I hope that helps you!! Any more questions??

Nataly [62]3 years ago

6 0

the answer is 49

hope this helps

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A battery of voltage V delivers power P to a resistor of resistance R connected to it. By what factor will the power delivered t

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Answer:

Explanation:

Power P = V² / R

a ) The resistance is changed to 2.90R

Power will become 1 / 2.9 times .

b )The voltage of the battery is now 2.90V, but the resistance is R

P = (2.9V)² / R

= 8.41 x V² / R

So power becomes 8.41 times

c )The resistance is 2.90R and voltage is 2.90V

Power P = (2.9V)² / 2.9 R

= 2.9 V²/R

So power becomes 2.9 times

d ) The resistance is 2.90R and the voltage is V/2.90

Power P = ( V/2.90)² x 1 / 2.90R

1 / ( 2.9 )³ x V² / R

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3 years ago

Why are you able to observe the Doppler effect on earth with sound waves but not with light waves?

VladimirAG [237]

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Explanation:

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3 years ago

The points plotted on a graph represent the actual data recorded by an experiment.

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3 years ago

Read 2 more answers

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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