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3 years ago

LIULIS. What would happen to speed if distance increased in the same amount of time?

Physics

1 answer:

Diano4ka-milaya [45]3 years ago

6 0

Answer:

The speed will be increased.

Explanation:

We can better understand this analysis, if we use numerical values and remember the expression or formula for speed. We know that speed is defined as the distance travelled at a given time.

$v=\frac{x}{t} \\where:\\v = velocity [m/s]\\x = distance [m]\\t = time [s]\\$

So if we have the distance x = 4 [m] and the time 2 [s], the speed will be:

v = (4/2) = 2 [m/s]

And if we increase the distance to x = 10 [m] at the same time 2[s], the speed will be:

v = (10/2) = 5 [m/s]

So we can see that the speed is increased.

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An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int

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3 years ago

Please help me , thank you

mash [69]

Answer:

Since the reading wasn't specified, it would be most likely A

Explanation:

A is the most similar to a protoplanetary disk, so it'd be A most likely

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3 years ago

Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electr

Eduardwww [97]

Answer:

$D=1693742.7m$

Explanation:

For sound waves we have v=d/t where v is the speed of sound and d the distance between the astronauts, while for electromagnetic waves we have c=D/t where c is the speed of light and D the distance between the spaceship and Earth. <em>We have written both times as the same</em> because is what is imposed by the problem, so we have t=d/v=D/c, which means:

$D=\frac{dc}{v}$

And for our values:

$D=\frac{(2m)(299792458m/s)}{354m/s}=1693742.7m$

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3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :