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enyata [817]
3 years ago
15

A satellite is placed in a geosynchronous orbit. In this equatorial orbit with a period of 24 hours, the satellite hovers over o

ne point on the equator. Which statement is true for a satellite in such an orbit?a. There is a tangential force that helps the satellite keep up with the rotation of the Earth.b. There is no acceleration toward the center of the Earth.c. The force toward the center of the Earth is balanced by a force away from the center of the Earth.d. The satellite is in a state of free fall toward the Earth.e. There is no gravitational force on the satellite.
Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

see that the correct one is D

Explanation:

In this problem the satellite is subjected to the universal force of attraction

    F = G M m / r²

using Newton's second law with centripetal acceleration you can find the angular velocity of the satellite

  F = m a

  F = m w² r

  G M / r³ = w²

and   w = 2π / T

In this case, the satellite is in a two-dimensional movement, where the free fall of the satellite is compensated by the horizontal displacement, for which it is always at a distance from the earth, remaining at the same point.

When reviewing the different statements we see that the correct one isD

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3 years ago
Air expands isentropically from 2.2 MPa and 77°C to 0.4 MPa. Calculate the ratio of the initial to the final speed of sound.
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The ratio of initial to final speed of sound is given as 1.28.

Explanation:

As per the thermodynamic relation of isentropic expansion

\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}

Here

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  • T_1 is the temperature at point 1 which is given as 77 °C  or 273+77=350K
  • P_2 is the pressure at point 1 which is given as 0.4 MPa
  • T_2 is the temperature at point 2 which is to be calculated
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Substituting values in the equation

                                      \frac{T_2}{350}=(\frac{0.4}{2.2})^{\frac{1.4-1}{1.4}}\\\frac{T_2}{350}=(0.18)^{0.2857}\\T_2=(0.18)^{0.2857} \times 350 \\T_2=0.61266 \times 350\\T_2=214.43 K

As speed of sound c is given as

c=\sqrt{kRT}

for initial to final values it is given as

\frac{c_i}{c_f}=\frac{\sqrt{k_1R_1T_1}}{\sqrt{k_2R_2T_2}}

As values of k and R is constant so the ratio is given as

\frac{c_i}{c_f}=\sqrt{\frac{T_1}{T_2}}

Substituting values give

\frac{c_i}{c_f}=\sqrt{\frac{350}{214.43}}\\\frac{c_i}{c_f}=\sqrt{1.63}}\\\frac{c_i}{c_f}=1.277  \approx 1.28

So the ratio of initial to final speed of sound is 1.28.

5 0
3 years ago
What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use
marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

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Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

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We need to calculate the torque due to force F₂

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\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

3 0
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