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3 years ago

Rest and Motion are the relative term. why? explain with example(please help me (╥﹏╥))

Physics

1 answer:

klemol [59]3 years ago

3 0

Explanation:

it depens on the subject and object. Let's example

you are driving a tesla car with your dog sitting on your side. You will say that the dog is at REST

but your friend, standing in sidewalk, seeing the same dog, will say that your dog is moving because it has MOTION from your car

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Pick the right answer

Alex777 [14]

I agree with Todd. One time when we went to the beach this happened and the waves brought back the ball so…

:) (can you mark me brainliest?)

8 0

3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

Now the resultant of the two velocities perpendicular to each other:

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

CALCULATE THE DECELERATION OF A RUNNER

AlexFokin [52]

Answer:

$a = \frac{ {v}^{2} - {u}^{2} }{2s} \\ or \\ a = \frac{v - u}{t} \\ say \: a = - 10m. {s}^{ - 2} \\ \therefore \: deceleration = - 10m. {s}^{ - 2}$

5 0

3 years ago

You are traveling on the interstate highway at a speed of 65 mph. What is your speed in km/h? The conversion factor is: 1.0 mph=

chubhunter [2.5K]

You have to cross multiply. So set up the equation like 1.6/x = 1/65 then multiply x by 1 and 1.6 by 65. The answer is 104 km/h.

6 0

4 years ago

Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If

Sphinxa [80]

Answer:

Distance between them is 4,200 meters.

Explanation:

Consinder car A:

${ \bf{distance = speed \times time }}$

substitute:

$distance = 20 \times (2 \times 60) \\ = 2400 \: m$

Consider car B:

$distance = 15 \times (2 \times 60) \\ = 1800 \: m$

since these cars move in opposite directions, distance between them is their summation:

$distance \: between = { \sum(distance \: of \: each \: car)} \\ = 2400 + 1800 \\ = 4200 \: m$

3 0

3 years ago

Rest and Motion are the relative term. why? explain with example(please help me (╥﹏╥))​

Rest and Motion are the relative term. why? explain with example(please help me (╥﹏╥))