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stiks02 [169]
3 years ago
8

Communications satellites are placed in orbits so that they always remain above the same point of the earth's surface.A Part com

pleteWhat must be the period of such a satellite?Express your answer in hours to the nearest integer.TT = 24 h Previous Answers CorrectB What is its angular velocity?
Physics
1 answer:
masya89 [10]3 years ago
3 0

Answer:

24 hours

7\times 10^{-5}\ rad/s

Explanation:

If a satellite is in sync with Earth then the period of each satellite is 24 hours.

T=24\times 60\times 60\ s

Angular velocity is given by

\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{24\times 60\times 60}\\\Rightarrow \omega=7\times 10^{-5}\ rad/s

The angular velocity of the satellite is 7\times 10^{-5}\ rad/s

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The sprinter ran 110 m in 11 seconds. What was her average speed in m/s?
Tanya [424]

Answer:

10 m/s

Explanation:

110/11=10

5 0
3 years ago
Read 2 more answers
I have my exam monday I need some help!
Lana71 [14]

Answer:

Explanation:

If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.

The distance between the 2 crests (or troughs) is the wavelength, λ.

We want the velocity of this wave. The equation that relates these 3 things is

f=\frac{v}{\lambda} and filling in:

2.5=\frac{v}{2.0} so

v = 2.5(2.0) and

v = 5.0 m/s

7 0
3 years ago
As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself.There is no one ar
Ira Lisetskai [31]

Answer: magnitude of applied force is FA = mg + F

Where F is the resultant force downward that the rope moves with

Explanation:

Force downwards F is,

F = FA - T

T is the upwards tension force on the rope

FA is the actual applied force in pulling the rope down.

Therefore, T = FA - F .....equ. (1)

For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)

Therefore, ma = T - mg

T = ma + mg ..... equ. (2)

Equating equ. 1 and 2

T = FA - F = ma + mg

Therefore FA = ma + mg + F

But at constant velocity a = 0

Magnitude of applied force becomes

FA = mg + F

See image below

5 0
3 years ago
What is the role of equations in this course?
Karolina [17]
That will depend on which course you're talking about. It will be a minor role in, say, Maritime Law or Comparitive Religion, but a major one in, say, Particle Physics or Linear Algebra.
4 0
3 years ago
A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia
Bingel [31]

Answer

given,

diameter of the pipe is  =  (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the  normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

F = \dfrac{mv^2}{r}

equating both the force equation

mg = \dfrac{mv^2}{r}

v = \sqrt{gr}

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s²   or g = 9.8 m/s²

v = \sqrt{32 \times 7}

v = 14.96 ft/s

or

v = \sqrt{9.8 \times 2.135}

v = 4.57 m/s

5 0
4 years ago
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