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Drupady [299]
2 years ago
12

A geostationary satellite orbits the Earth in 24 hours along an orbital path that is parallel to an imaginary plane drawn throug

h the Earth's equator. Such a satellite appears permanently fixed above the same location on the Earth. How high above the Earth's surface must it be located?
Physics
1 answer:
PIT_PIT [208]2 years ago
7 0

Answer:

approximately 35,786 km

Explanation:

A satellite in such an orbit is at an altitude of approximately 35,786 km (22,236 mi) above mean sea level. It maintains the same position relative to the Earth's surface.

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Change of 5000 c flows through a circuit in one hour. what is the current intensity?​
AleksandrR [38]

Answer:

I = 1.38 A

Explanation:

Given that,

Charge, q = 5000 C

Time, t = 1 hour = 3600 s

We need to find the current intensity. The current intensity is equal to the electric charge per unit time. It can be given by :

I=\dfrac{q}{t}

Substitute all the values in the above formula

I=\dfrac{5000\ C}{3600\ s}\\\\=1.38\ A

So, the current intensity is 1.38 A.

8 0
3 years ago
A net force is applied on a 100 kg rocket which causes the rocket to acceleration at 10 m/s2. The same net force is applied on a
ELEN [110]

Answer:

i dont know lo

Explanation:

8 0
2 years ago
Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days
Vinil7 [7]

Answer:

The distance is r = 55430496 \  m  

Explanation:

From the question we are told that

   The period of the moon T =  1.2 days = 1.2 * 24 * 3600 = 103680 \  s

    The mass of the planet is  m_p =  9.38*10^{24} kg

Generally the period of the moon is mathematically represented as

       T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

Here G is the gravitational constant with value

        G  =  6.67 *10^{-11} \  N \cdot m^2/kg^2

=>   T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

=>   103680   =  2 *  3.142  * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }

=>    272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}

=>    r = \sqrt[3]{ 1.7031241*10^{23}}j

=>   r = 55430496 \  m        

8 0
3 years ago
If the mass of a material is 120 grams and the volume of the material is 23 cm3, what would the density of the material be?
oee [108]
The density is 5.22 g/cm³.

Density=\frac{mass}{volume}
Density=\frac{120 g}{23 cm {3} }
Density= 5.22 g/cm³
4 0
3 years ago
A plastic rod is charged up by rubbing a wool cloth, and brought to an initially neutral metallic sphere that is insulated from
andrew-mc [135]

Answer:

Yes option A is right.

Explanation:

As we know that the "Opposite charges attract and like charges repel eachother". So based upon that fact we find out the sphere will be repelled or attract by the rod. As in this case metallic sphere was neutral initially but then we touched the rod with it. Although it was for few seconds but the charge is transferred to the sphere. Now both sphere and the rod have charge. After the seperation we look towards their respond If both have the opposite charge they will attract eachother. But here in this case they repel because they have the same charge, as we have charged the neutral sphere with the rod so we already know that they have the same charges that is why they are repelling eachother.

       Insulation from the ground means that blocking the way of charges or free electrons from earth to metallic sphere and vice versa. As there exists free electrons and charges in earth they would flow into the metallic objects. So for more precise and accurate experiments we insulate the metals or prevent the metals from touching the earth surface to avoid the flow of charges through  them. I hope it will help you.

5 0
3 years ago
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