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kolbaska11 [484]
3 years ago
8

Calculate the force of Earth's gravity on 1350kg spacecraft that is 1.28 * 10 ^ 6 * m above Earth surface ( Earth's mass is 6*10

^ 24 kgg) .
Physics
1 answer:
Darya [45]3 years ago
8 0

Answer:

The force of Earth's gravity is 327954 N

Explanation:

Given:

Mass of the space craft = 1350 kg

Mass of the earth = 6*10^ 24

Distance = 1.28 * 10 ^ 6

To Find:

The  force of Earth's gravity = ?

Solution:

The force of attraction between a planet and an object kept in space is given by the expression

F =\frac{GMm}{R^2}

where

M is the mass of the earth

m is the mass of the space craft

R is the distance between the earth and the space craft and

G is the gravitational constant

On substituting the values

F =\frac{(6.67 \times 10^{-11})(6\times10^ {24})(1350)}{(1.28\times 10^6)^2}

F =\frac{(54027\times 10^{13})}{(1.28\times 10^6)^2}

F =\frac{(54027\times 10^{13})}{(1.6384\times 10^{12})}

F = 327954 N

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We want to explain why two different observes may measure different frequencies for the same vibrating object.

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Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

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For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

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Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120
igomit [66]

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3 years ago
A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant
atroni [7]

Answer:

The  time constant is  \tau  = 1.265 s

Explanation:

From the question we are told that

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The mathematically representation for voltage potential of a capacitor at different time is

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Where  \tau  is the time constant  

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     So  the capacitor potential will be  100%  when it is full thus  V_o  =100%  =  1  

and from the  question we are told that the  at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

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Hence

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       - {\frac{2.4}{\tau } }  =  ln0.15

        \tau  = 1.265 s

     

7 0
3 years ago
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