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3 years ago

13

Geosynchronous satellites orbit the Earth at a distance of 42 000 km from the Earth's center. Their angular speed at this height

is the same as the rotation rate of the Earth, so they appear stationary at certain locations in the sky. What is the force acting on a 1 500-kg satellite at this height?

1 answer:

frozen [14]3 years ago

7 0

Answer:

Force will be 7.97 N

Explanation:

We have given that earth is at s distance of 42000 km from the earth center

So r = 42000 km = 42000000 m

Mass m = 1500 kg

Angular velocity of earth $\omega =7.29\times 10^{-5}rad/sec$

We have to find the force

We know that force is given by

$F=m\omega ^2r=1500\times (7.29\times 10^{-5})^2\times 42000000=797.1615\times 10^{-2}=7.97N$

You might be interested in

By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl

drek231 [11]

The charge present determines a force to be attractive or repulsive.

The charges acquired by two bodies determines the Force as Attractive Or Repulsive.

Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.

Examples of the experiments and observations:

On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.

This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.

In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.

The more the distance between the charges, the less is the Electric Force.

The lesser the distance between the charges, the more is the Electric Force.

If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.

Hence, the charge present determines a force to be attractive or repulsive.

Learn more about Coulomb Force here, brainly.com/question/15451944

#SPJ4

6 0

1 year ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the c

Semmy [17]

First establish the summation of the forces acting int the ladder

Forces in the x direction Fx = 0 = force of friction (Ff) – normal force in the wall(n2)

Forces in the y direction Fy =0 = normal force in floor (n1) – (12*9.81) –( 60*9.81)

So n1 = 706.32 N

Since Ff = un1 = 0.28*706.32 = 197,77 N = n2

Torque balance along the bottom of the ladder = 0 = n2(4 m) – (12*9.81*2.5 m) – (60*9.81 *x m)

X = 0.844 m

5/ 3 = h/ 0.844

H = 1.4 m can the 60 kg person climb berfore the ladder will slip

7 0

3 years ago

Read 2 more answers

Please help!?!?!?!?!?!?!

adoni [48]

Answer:

D

Explanation:i think but dont get mad if im wrong

5 0

2 years ago

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. if the ball hits the ground 4.0 second

Viktor [21]

Answer:

78.4 m

Explanation:

To obtain the height of the cliff;

We can use the Relation to obtain the final velocity, v

v = u + at

a = acceleration due to gravity = 9.8m/s²

v = 0 + (9.8*4)

v = 0 + 39.2

v = 39.2 m/s

To obtain the Height, S

v² = u² + 2aS

39.2^2 = 0 + 2(9.8)S

39.2^2 = 0 + 19.6S

1536.64 = 19.6S

S = 1536.64 / 19.6

S = 78.4 m

8 0

2 years ago