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alina1380 [7]
3 years ago
6

A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s

acceleration compared to the gravitational field at the location of the satellite?
Physics
2 answers:
bulgar [2K]3 years ago
4 0

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

E = \frac{F}{m}

now in order to find the acceleration of the satellite we know by Newton's II law

F = ma

so we will have

a = \frac{F}{m}

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

PSYCHO15rus [73]3 years ago
4 0

Answer:

The gravitational field and the acceleration point in the same direction.

The magnitudes of the acceleration and the gravitational field strength are equal.

These are the two answers

Explanation:

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3 years ago
How is a net electric charge produced?​
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Answer:

Net charge produce due to presence of different chargers in the vicinity or in the considered region.

Explanation:

Net electric charge is just a concept introduce to identify the result of the presence of number of charges which interact with each other. It can be found by treating electric chargers in algebraic form like numbers, where +charges get added with +charges

-charges get added with -charges

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7 0
2 years ago
During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima
vova2212 [387]

Answer:

The value  F  =  0.1396 \ N

Explanation:

From the question we are told that

   The volume blood  ejected is  V  =  65 \ cm^3  = 65*10^{-6} \  m^3

    The velocity of the blood ejected is  v  = 103 \  cm/s  = \frac{103}{100} = 1.03 \ m/s

    The density of blood is  \rho = 1060 \  kg/m^3

     The heart beat is R = 59 \  bpm(beats \  per \  minute) = \frac{59}{60}= 0.9833\ bps

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      F  =  2 * \rho  *  V  *  R *  v

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=>    F  =  0.1396 \ N

8 0
3 years ago
The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob
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Answer:

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Explanation:

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\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

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2 years ago
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Answer:

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