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3 years ago

A cat runs in a straight line. Which of the following statements about the cat's motion must be true?

Physics

1 answer:

alexira [117]3 years ago

5 0

Answer:

I would think it's A cat runs in a straight line.

Explanation:

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Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ

bija089 [108]

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

we can measure it using the following relation

∅= s/r

where

s = the distance travelled by the body, and

r = radius of the circle along which it is moving.

given that

circumference c, s= 400 m

r= ?

we have to solve for the radius

we know that circumference

$c= 2\pi r$

400= 2*3.142*r

400= 6.282*r

divide both sides by 6.284 we have

400/6.284

r= 63.63 m

Angular displcament

∅= 400/63.63

∅= 6.28 radians

8 0

3 years ago

What happens to the magnitude of the force of gravitation between two objects if: 1. distance between the objects is tripled? 2.

Ludmilka [50]

<span>2. mass of both objects doubled?
Hope it helped!</span>

8 0

3 years ago

Read 2 more answers

In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a

pickupchik [31]

Answer:

time of collision is

t = 0.395 s

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

$v_1 = 0$

similarly initial speed of the object which is projected by spring is given as

$v_2 = 16.2 m/s$

now relative velocity of object with respect to ball

$v_r = 16.2 m/s$

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

$d = v_r t$

$6.4 = 16.2 t$

$t = 0.395 m$

Now the height attained by the object in the same time is given as

$h = v_2 t - \frac{1}{2}gt^2$

$h = 16.2(0.395) - \frac{1}{2}(9.81).395^2$

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

4 0

3 years ago

A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

zvonat [6]

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh we need to know h

h = 50 sin 10 = 8.68 meters

then: PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j

KE = 1/2 m v^2 = <u>1703 .49</u>

v = 13 m/s

7 0

2 years ago

Frequencies of sound waves higher than those we can hear is called?

ludmilkaskok [199]

Answer:

utrasonic

Explanation:

these are sounds beyond our hearing capacity range of 20-20kHz

8 0

3 years ago

Read 2 more answers