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Rom4ik [11]
3 years ago
9

Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and plac

e it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6370 km.) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.
Physics
1 answer:
mylen [45]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

W = U_f - U_i

Where,

W = Work

U_f = Final gravitational Potential energy

U_i = Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

U_i = \frac{-GMm}{R}

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is R_{orbit} \approx R then the energy possessed by the spacecraft is

U_f = \frac{-GMm}{2R}

Work needed to put it in orbit is the difference between the above two

W = U_f - U_i

W = -GMm (\frac{1}{2R}-\frac{1}{R})

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

W = \frac{GMm}{2R}

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Serga [27]

Potential energy does.

A book on the floor has negative potential energy relative to the seat
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8 0
3 years ago
When a moving car is brought to a stop with the brakes, its acceleration vector is 29. (a) in the same direction as its velocity
professor190 [17]

To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is

\vec{a} = \frac{\vec{v_2}-\vec{v_1}}{\Delta t}

If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.

Therefore the correct answer is B.

7 0
3 years ago
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Marrrta [24]
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8 0
3 years ago
A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on
svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}

Put the value into the formula

\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}

\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}

\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}

\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}

\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1

x=0.32\ m

Hence, The position on the x axis is 0.32 m.

5 0
3 years ago
A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The pai
RSB [31]

Explanation:

For equilibrium, \sum M = 0.

So,   8 m \times mg - (10 m) T_{1} = 0

             T_{1} = \frac{8 \times mg}{10}

                        = \frac{8 \times 90 \times 9.8}{10}

                        = 705.6 N

Also, for equilibrium \sum F_{y} = 0

              T_{1} + T_{2} - mg = 0

or,         T_{2} = mg - T_{1}

                        = 90 \times 9.8 - 705.6

                        = 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

8 0
4 years ago
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