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3 years ago

9. A 0.500 kg ball with a speed of 4.0 m/s strikes a stationary 1.0 kg target. The ball and target

Physics

1 answer:

Virty [35]3 years ago

4 0

Answer:

D. 2.0 kg m/s

Explanation:

Momentum after = momentum before

p = (0.500 kg) (4.0 m/s) + (1.0 kg) (0 m/s)

p = 2.0 kg m/s

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3 years ago

Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour

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Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

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t₁= 16.7 years

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$\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }$

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2 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

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Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

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$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

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3 years ago