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3 years ago

8

A tennis player swings her 1000 g racket with a speed of 12 m/s. she hits a 60 g tennis ball that was approaching her at a speed

of 16 m/s. The ball rebounds at 42 m/s.
(a) How fast is her racket moving immediatley after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.

__________ m/s

(b) If the tennis ball and racket are in contatc for 10 ms, what is the average force that the racket exerts on the ball?

________ N

How does this compare to the gravitational force on the ball?

F avg / W ball=________

1 answer:

tensa zangetsu [6.8K]3 years ago

6 0

Answer:

should be the answer B

Explanation:

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3. The expression 0.62 x10^3 is equivalent to...

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$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

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Ksju [112]

Because of the hint we can conclude what equation we need to solve this problem. We have power and duration that means that we need to express energy:

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3 years ago

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

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Part a)

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Part b)

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Explanation:

Part a)

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Part b)

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3 years ago

A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and

Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

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∑ <em>F</em> = <em>N</em> - <em>W</em> = 0

• net horizontal force:

∑ <em>F</em> = <em>Fs</em> = <em>m a</em>

where

<em>N</em> = magnitude of normal force

<em>W</em> = car's weight

<em>Fs</em> = mag. of static friction

<em>m</em> = car's mass

<em>a</em> = <em>v</em> ²/<em>R</em> = mag. of the centripetal acceleration

<em>v</em> = car's speed

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Now,

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<em>W</em> = <em>m g</em> = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

<em>N</em> = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

<em>Fs</em> = 0.5 <em>N</em> = 7350 N

• solve for the (maximum) acceleration:

7350 <em>N</em> = (1500 kg) <em>a</em> → <em>a</em> = 4.9 m/s²

• solve for the (maximum) speed:

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3 years ago

g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9

Elza [17]

Answer:

The answer is " $\bold{dosage = 0.031 rem}$ "

Explanation:

please find the complete question in the attached file.

Given value:

$m = 79\ kg \\\\n = 3.4 \times 10^9 \\\\E = 5.5 \times 10^{-13} \\\\ RBE = 15$

$\to E = n E\\$

$= 3.4 \times 10^9 \times 5.5 \times 10^{-13} \\\\ = 1.87 \times 10^{-3}$

$\to E(absorbed) = 1.87 \times 10^{-3} \times 0.87 = 1.63 \times 10^{-3}$

calculating the radiation absorbed per kg:

$= \frac{1.63 \times 10^{-3}}{79} \\\\ = 2.06 \times 10^{-5} \\\\ = 0.00206 \ rad$

$\to Dosage = 0.00206 \times 15 \\$

$= 0.031 \ \ rem$

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3 years ago