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Gemiola [76]
3 years ago
10

In a parallel connection, which of the following is true?

Physics
1 answer:
pashok25 [27]3 years ago
4 0
I think it’s 3 sorry if I’m wrong
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A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
Electricity costs 6 cents per kilowatt hour. In one month one home uses one megawatt hour of electricity. How much will the elec
Luda [366]

Answer:

Cost of 1000  kilowatt hour = 6000 cents

Explanation:

Given that

Electricity cost is 6 cents per kilowatt hour.

And we have to found out the cost for one megawatt hour

We know that

1 kilowatt = 1000 watt

1 megawatt = = 1000000 watt

1 megawatt = 1000 kilowatt

1 megawatt hour = 1000 kilowatt hour  

Given that cost of 1 kilowatt hour = 6 cents

So the cost of 1000  kilowatt hour = 6 x 1000 cents

Cost of 1000  kilowatt hour = 6000 cents

3 0
3 years ago
A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric fiel
AveGali [126]

Answer:

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

j=\frac {I}{A} where I is current and A is area and A=\pi r^{2}

Substituting this into the first equation then E=P\times \frac {I}{\pi r^{2}}

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

E=1.72\times 10^{-8}\times \frac {30}{\pi \times 0.001295^{2}}=9.79\times 10^{-2} N/c=0.0979 N/c

4 0
3 years ago
A school bus takes 0.53 hours to reach the school from your house. If the average speed of the bus is 19km/h, what is the displa
Artist 52 [7]
D = v*t= 19*0,53 = 10.07 km
5 0
3 years ago
Read 2 more answers
A Ground Fault Circuit Interrupter (GFCI) works by comparing the amount of current going to and returning from equipment along t
DanielleElmas [232]
Choice-B is the correct one.
5 0
3 years ago
Read 2 more answers
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