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Anuta_ua [19.1K]
3 years ago
12

Two satellites are in circular orbits around Earth. Satellite A has speed vA. Satellite B has an orbital radius nine times that

of satellite A. What is the speed of Satellite B?
A. vA.9
B. vA/3
C. 3vA
D. 9vA
Physics
1 answer:
zaharov [31]3 years ago
5 0

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

v_a =\sqrt{\dfrac{GM}{R}}........(1)

now, orbital velocity of satellite B

v_b=\sqrt{\dfrac{GM}{R'}}

v_b=\sqrt{\dfrac{GM}{9R}}

v_b=\dfrac{1}{3}\sqrt{\dfrac{GM}{R}}

from equation 1

v_b=\dfrac{v_a}{3}

hence, the correct answer is option B

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2 Points
yulyashka [42]

The  force applied to lift the crate is 171 N

Explanation:

The lever works on the principle of equilibrium of moments, so we can write:

F_i d_i = F_o d_o

where

F_i is the  force in input

d_i is the arm of the input force

F_o is the output force

d_o is the arm of the output force

For the lever in this problem, we have:

d_i = 0.25 m

d_o = 0.19 m

F_i = 130 N (force applied)

Solving the equation for F_o, we find the force applied to lift the crate:

F_o = \frac{F_i d_i}{d_o}=\frac{(130)(0.25)}{0.19}=171 N

Learn more about levers:

brainly.com/question/5352966

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6 0
3 years ago
The equation for the chemical reaction shown is not balanced. What number should replace the question mark to balance this equat
Fudgin [204]
The equation is balanced as it is.
7 0
3 years ago
When is the force on a current-carrying wire in a magnetic field at its strongest?
Hitman42 [59]

The forces on a current-carrying wire in a magnetic field are at their strongest when the current is at a 90-degree angle to the field. Option D is correct.

<h3>What is a magnetic field?</h3>

It is the type of field where the magnetic force is obtained. The magnetic force is obtained by the field felt around a moving electric charge.

The complete question is;

"When is the force on a current-carrying wire in a magnetic field at its strongest?

-when the current is at a 0-degree angle to the field

-when the current is at a 30-degree angle to the field

-when the current is at a 45-degree angle to the field

-when the current is at a 90-degree angle to the field"

The magnetic force is found as;

F=BILSINΘ

Where,

Magnetic Field, B

Length of the wire, L

The angle between field and current, Θ

When Θ=90°

The value of the magnetic force is;

F=BIL

When the current is flowing at a 90-degree angle to the magnetic field, the forces acting on a wire carrying a current are the strongest.

Hence, option D is correct.

To learn more about the magnetic field, refer to the link;

brainly.com/question/19542022

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4 0
2 years ago
Kinematics
leonid [27]

Answer:

a)

a = 2 [m/s^2]

b)

a = 1.6 [m/s^2]

c)

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

a)

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

v_{f} = v_{i}+(a*t)

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

40 = 0 + (a*20)

a = 2 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2} =  v_{i} ^{2}+(2*a*x)

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

where:

x2 = distance [m]

t2 = 30 [s]

x2 = 40*30

x2 = 1200 [m]

b)

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

v_{f} = v_{i}-a*t

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2}  = v_{i} ^{2} -2*a*x3\\

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

c)

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

8 0
3 years ago
PLEASE HELP!!!!!!!! NEED DONE SOON!!!
zhannawk [14.2K]

Answer:

the object is decelerating

4 0
3 years ago
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