All categories

2 years ago

Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .

Physics

1 answer:

svet-max [94.6K]2 years ago

5 0

The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

<h3>Speed of the satellite</h3>

v = √GM/r

where;

M is mass of Earth
G is universal gravitation constant
r is distance from center of Earth = Radius of earth + 4930 km

v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]

v = 5,916.36 m/s

Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

Learn more about speed here: brainly.com/question/6504879

#SPJ1

You might be interested in

40 Points!!!!!!!

marshall27 [118]

D. Ted associated being asked a question with embarrassment.

7 0

3 years ago

Read 2 more answers

A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l

hram777 [196]

Answer:

$t = 37.6 s$

Explanation:

As we know that train is initially moving with the speed

$v_i = 75.1 km/h$

now we know that

$v_i = 20.86 m/s$

now the final speed of the train when it crossed the crossing

$v_f = 15 km/h$

$v_f = 4.17 m/s$

now we can use kinematics here

$v_f^2 - v_i^2 = 2 a d$

$4.17^2 - 20.86^2 = 2 a(471)$

$a = -0.44 m/s^2$

Now the time to cross that junction is given as

$v_f - v_i = at$

$4.17 - 20.86 = a(-0.44)$

$t = 37.6 s$

4 0

2 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n

satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

A student coils a copper wire around a bar magnet. What action will cause the device to generate electricity?

kolbaska11 [484]

I hope the wire is not wound too tightly around the bar magnet.
The device will generate electrical energy when the bar magnet
is moving in or out of the coil of wire.

6 0

3 years ago