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3 years ago

A space shuttle orbits Earth at a speed of 21,000 km/hr. How far will it travel in 5 hours?

Physics

1 answer:

lianna [129]3 years ago

6 0

Answer:

105000km

Explanation:

Given parameters:

Speed of shuttle = 21000km/hr

Time of travel = 5hrs

Unknown:

The distance covered by the space shuttle;

Solution:

Distance is the length of the path traveled.

Distance = speed x time

Input the parameters and solve;

Distance = 21000km/hr x 5hrs

Distance = 105000km

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a ultrasonic wave at 8x10^4 Hz is emitted into a vien where the speed of sound in blood is 1570 m/s. the wave reflects off the r

Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

$f'=\frac{V+V_{o}}{V-V_{s}} f$

Where:

$f=8(10)^{4} Hz$ is the actual frequency of the sound wave

$f'=8.002(10)^{4} Hz$ is the "observed" frequency

$V=1570 m/s$ is the speed of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which are the red blood cells

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$

$V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}$

Finally:

$V_{s}=0.392 m/s$

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3 years ago

I NEED THIS A SOON AS POSSIBLE

vivado [14]

Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

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3 years ago

Read 2 more answers

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:

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

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

$t'\approx463.733\ s$

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10 kg box accelerates at 2 meters per second as it slides down a ramp at an angle of 25 degrees. What is the coefficient of fric

maria [59]

Answer:

0.241

Explanation:

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Anastasy [175]

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. Y = (the number) .

The correct choice is ' A ' .

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