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:
a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.
b) The frequency is 1404.08 Hz
Explanation:
If the police car is a stationary source, the frequency is:
(eq. 1)
fs = frequency of police car = 1200 Hz
fa = frequency of moving car as listener
v = speed of sound of air
vc = speed of moving car
If the police car is a stationary observer, the frequency is:
(eq. 2)
Now,
fL = frequecy police car receives
fs = frequency police car as observer
a) The velocity of car is from eq. 2:
b) Substitute eq. 1 in eq. 2:
The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.
Explanation:
Let at A both net electric field is zero then
At A ,E1=E2
E1=k*Iq1I / (d+x)^2
E2=k*Iq2I /x^2
Equating both
Answer:
The pressure of liquid column is given by p=hpg, where h is depth, p is density and g is acceleration due to gravity.
Therefore, pressure of the liquid column increases with depth. The height of the blood column in a human body is more at feet than at the brain. Therefore, the blood pressure in humans is greater at the feet than the brain.
