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irakobra [83]
3 years ago
13

Geosynchronous satellites orbit the Earth at a distance of 42 000 km from the Earth's center. Their angular speed at this height

is the same as the rotation rate of the Earth, so they appear stationary at certain locations in the sky. What is the force acting on a 1 500-kg satellite at this height?
Physics
1 answer:
frozen [14]3 years ago
7 0

Answer:

Force will be 7.97 N

Explanation:

We have given that earth is at s distance of 42000 km from the earth center

So r = 42000 km = 42000000 m

Mass m = 1500 kg

Angular velocity of earth \omega =7.29\times 10^{-5}rad/sec

We have to find the force

We know that force is given by

F=m\omega ^2r=1500\times (7.29\times 10^{-5})^2\times 42000000=797.1615\times 10^{-2}=7.97N

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2 resistors of resistance 1000 ohm and 2000 ohm are joined in series with a 100V supply. A voltmeter of internal resistance 4000
Vadim26 [7]
<h2>The voltmeter reading will be 35.7 volt </h2>

Explanation:

The resistor 1000 ohm and 4000 ohm are connected in parallel .

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or R₁ = 800 ohm

Therefore the total resistance in circuit is = 2000 + 800 = 2800 ohm

Because these are in series .

We can find  current flowing through the circuit  I = \frac{V}{R} = \frac{100}{2800} = \frac{1}{28}

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The potential difference across 1000 ohm = \frac{1}{28} x 1000 = 35.7 volt

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3 years ago
The barometric pressure in breckenridge, colorado (elevation 9600 feet) is 580 mm hg. how many atmospheres is this?
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7 0
2 years ago
A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma
goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

F=ma

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m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

F=ma

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

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where a_g = 0.098 m/s^2 is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

x'(t)=15-\frac{1}{2}a_s t^2

where a_s=0.8 m/s^2 is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s

4 0
3 years ago
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