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Nady [450]
3 years ago
10

The international space station orbits the earth at a distance of 6,783 km from the center of the earth it travels at 7.66 km/s.

Find the time it takes for the ISS to complete one orbit.
Physics
1 answer:
frosja888 [35]3 years ago
4 0

Answer:

886 seconds

Explanation:

The circumference of the cross section of sphere of radius 6,783 km is the distance the space station has to cover for a complete orbit. The international space station orbits at about 7.66km/s. The time taken is given by distance/speed. Time taken=6783/7.66;

Evaluating gives 885.5 seconds or about 14.8 minutes.

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A 1400-kg car moving with a speed of 32 m/s is approaching a stoplight. The light turns yellow and the car makes an abrupt stop
inna [77]

Answer:

c. 716, 800 J

Explanation:

t = Time taken

u = Initial velocity = 32 m/s

v = Final velocity = 0

s = Displacement = 60 m

a = Acceleration

m = Mass of car = 1400 kg

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-32^2}{2\times 60}

Work done is given by

W=Fscos\theta\\\Rightarrow W=mascos\theta\\\Rightarrow W=1400\times \dfrac{0^2-32^2}{2\times 60}\times 60\times cos0\\\Rightarrow W=-716800\ J

The amount of work done to stop the car is 716800 J

8 0
3 years ago
A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base
-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g=426\times 10^{-3} kg

1 kg=1000 g

Radius,r=\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m

1m=100 cm

Height,h=5m

I=\frac{2}{2}mr^2

a.By law of conservation of energy

\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh

\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh

v=\omega r

gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2

\omega^2=\frac{6}{5r^2}gh

\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s

Where g=9.8m/s^2

b.Rotational kinetic energy=\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J

Rotational kinetic energy=8.35 J

6 0
3 years ago
When is a zero not significant?
maria [59]

Answer:

Not between significant digits.

Explanation:

A zero not significant when it's not between significant digits.

8 0
3 years ago
Read 2 more answers
How much would you have to eat for your appendix to explode ? I ate so much on thanksgiving and i need to know i’m very scared
dimulka [17.4K]

Answer:

I don't think your appendix can explode because you ate too much honestly. It's not even possible to eat so much that your appendix explodes, and if you're feeling any pain it definitely isn't because your appendix is about to explode, believe me. Also you could just type it into the internet, that'd be a much faster solution.

6 0
3 years ago
A spinning turbine can generate electricity only in the form of a/an _______ current.
lukranit [14]

Answer:

A spinning turbine can generate electricity only in the form of an alternating current.

8 0
3 years ago
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