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Mkey [24]
2 years ago
13

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth

rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58×107m(≈22,000miles).What is the period of a satellite in a geosynchronous orbit?
Physics
1 answer:
Gala2k [10]2 years ago
5 0

Answer:

T = approximately 24 hs.

Explanation:

In order to keep the satellite over a fixed point on the equator, as the earth rotates, the satellite must have the same angular velocity that Earth has, which means that it must have a period equal to the time used by Earth to complete an entire rotation on itself, which is almost exactly 24 Hs.

Mathematically, this can be obtained taking into account that the force that keeps the satellite in orbit is the centripetal force, which is actually the gravitational force exerted by Earth, so we can write the following equality:

Fg = Fc ⇒ G*ms*me / (re +rsat)² = ms*ω²*(re +rsat)

By definition, ω =ΔФ / Δt

For a complete revolution, ΔФ = 2*π, and Δt = T (period of the rotation),

so we can replace ω by (2*π/T), solving then for T:

T= 86,313 sec. (24 hs are exactly 86,400 sec, so the value is actually very close to the theorical one).

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A professor's office door is 0.89 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg ; and pivots on frictionless hinges.
taurus [48]
In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.

The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it.  All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.

Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges.  That
distance from the hinges is the width of the door ... 0.89 m.

We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.

Torque is (force) x (distance from the hinge).

                                    4.9 N-m  =  (force) x (0.89 m) 

Divide each side by 0.89m:    Force = (4.9 N-m) / (0.89 m)

                                                             =  5.506 N .
7 0
2 years ago
A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver
Akimi4 [234]

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = \frac{distance}{time}

Solution:

The speed of the ball is given by,

speed = \frac{distance}{time}

Thus,

Time = \frac{distance}{speed}

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

3 0
2 years ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
3 years ago
A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0
2 years ago
In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of 20 MV. What is their ki
myrzilka [38]

Answer:

a) 2 x10^7 eV

b) 2 x10^4 keV

c) 20 MeV

d) 0.02 Gev

e) 3.2 x 10^-12J

Explanation:

The potential difference = 20 x 10^6 V

The charge on the proton = 1.6 x10^-19

The work done to move the proton will be basically the proton will acquire if it accelerates.

Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19

                                                 =3.2 x 10^-12J or 2 x10^7 eV

2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev

5 0
2 years ago
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