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2 years ago

7

A satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther a

way from the same planet. How do the speeds of the two satellites compare

1 answer:

irinina [24]2 years ago

3 0

The speed of the second satellite is less than the speed of the first satellite.

<h3>What is speed?</h3>

The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.

Given is a satellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet.

When the distance from the center of the orbit increases, the time to complete the orbit will be greater.

Thus, the speed of the second satellite is less than the speed of the first satellite.

Learn more about speed.

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. At what velocity will the box in Problem 6 be traveling when it hits the ground? Use formula: Va = Vf - Vi / 2

VladimirAG [237]

Answer:

How much time does his victim on the ground below have to move out of harm's way? At what velocity will the safe hit the ground? sownt d-200m. n a = 100/2. Vi-o.

Explanation:

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2 years ago

What is the coldest place on earth

exis [7]

The continent of Antartica is located at the bottom of the world. the South Pole is at its center. Antarctica is the coldest and windiest place on earth. It is covered with ice up to 3 miles thick. Very few plants and animals can survive here, but penguins, fish, and seals live on the coast and in the seas. No people live on Antarctica permanently, but scientists and tourists visit.

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3 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

2 years ago

150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit

zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

$I=\frac{150}{30}=5 A$

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6 0

3 years ago

A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats

ololo11 [35]

Answer:

Part a)

$f_B = 290 Hz$

Part B)

percentage increase is

$percentage = 1.38$ %

Explanation:

Part a)

As we know that the beat frequency is

$f_A - f_B = 3$

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

$293 - f_B = 3$

$f_B = 290 Hz$

Part B)

percentage increase in the tension of the string will be given as

$f_A - f_B' = 1$

$f_B' = 292 Hz$

now we have

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

so we have

$T_1 = C (290)^2$

$T_2 = C(292)^2$

so we have

$\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}$

percentage increase is

$percentage = 1.38$

4 0

3 years ago