Answer:
R=4.22*10⁴km
Explanation:
The tangential speed 
of the geosynchronous satellite is given by:
Because 
is the circumference length (the distance traveled) and T is the period (the interval of time).
Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:
If we substitute the expression for in this formula, we get:
Since the centripetal force is the gravitational force 
between the satellite and the Earth, we know that:
![F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }](https://tex.z-dn.net/?f=F_g%3D%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%5C%5C%5C%5C%5Cimplies%20%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%3D%5Cfrac%7B4m%5Cpi%20%5E%7B2%7DR%7D%7BT%5E%7B2%7D%7D%5C%5C%5C%5CR%5E%7B3%7D%3D%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%5C%5C%5C%5CR%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%7D)
Where G is the gravitational constant (
) and M is the mass of the Earth (
). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:
![R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%2A10%5E%7B-11%7DNm%5E%7B2%7D%2Fkg%5E%7B2%7D%29%285.97%2A10%5E%7B24%7Dkg%29%2886400s%29%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%7D%5C%5C%5C%5CR%3D4.22%2A10%5E%7B7%7Dm%3D4.22%2A10%5E%7B4%7Dkm)
This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.
Answer:
The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.
Explanation:
We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.
We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.
Using ohm's law we have
V = I*R
V= voltage
I= current
R= resistance
10 = 2.5*R
R = 10/2.5 = 4ohm
that means we need a total of 4ohm resistance from these two resistors.
since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.
R(net) =4ohm
Now the orientation of the resistors are in parallel so the current will be divided.
we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.
We know that the voltage in parallel remains same
In 20 ohm resistance
again using ohms law
V = i1*R1
10 = i1*20
i1 = 0.5A
in 5ohm resistor
V=i2*R2
10 = I2*5
i2 =2A
and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.
Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.
Answer:
The electromagnetic spectrum comprise a lot of waves length. Usually, different waves length are called as different lights, and a light source can emit in more than a different wave length, as the sun does, for example. The sun emit the visible light, UV light, infrared, etc.
Answer:
a)= 98kJ
b)=108kJ
c) = 10kJ
Explanation:
a. The work that is done by gravity on the elevator is:
Work = force * distance
= mass * gravity * distance
= 1000 * 9.81 * 10
= 98,000 J
= 98kJ
b)The net force equation in the cable
T - mg = ma
T = m(g+a)
T = 1000(9.8 + 10)
T = 10800N
The work done by the cable is
W = T × d
= 10800N × 10
= 108000
=108kJ
c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J
Work done by cable = PE +KE
108,100 J = KE + 98,100 J
KE = 10,000 J
= 10kJ
=
Answer:
<em>the phase relationship between two waves.</em>
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Explanation:
Coherence describes all properties of the correlation between physical quantities between waves. It is an ideal property of waves that determines their interference. In a situation in which there is a correlation or phase relationship between two waves. If the properties of one of the waves can be measure directly, then, some of the properties of the other wave can be calculated.
