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) 0.1832 A
b) 11.91 Volts
c) 2.18 Watt , 0.0168 Watt
Explanation:
(a)
R = external resistor connected to the terminals of the battery = 65 Ω
E = Emf of the battery = 12.0 Volts
r = internal resistance of the battery = 0.5 Ω
i = current flowing in the circuit
Using ohm's law
E = i (R + r)
12 = i (65 + 0.5)
i = 0.1832 A
(b)
Terminal voltage is given as
= i R
= (0.1832) (65)
= 11.91 Volts
(c)
Power dissipated in the resister R is given as
= i²R
= (0.1832)²(65)
= 2.18 Watt
Power dissipated in the internal resistance is given as
= i²r
= (0.1832)²(0.5)
= 0.0168 Watt
Answer:
B. less
Explanation:
acceleration due to gravity on Earth, g = 9.8 m/s²
acceleration due to gravity on Moon, g = 1.6 m/s²
Given mass of the object as, m = 5 kg
Weight of an object is given as, W = mg
Weight of the object on Earth, W = 5 x 9.8 = 49 N
Weight of the object on Moon, W = 5 x 1.6 = 8 N
Therefore, the object weighs less on the moon compared to its weight on Earth.
The correct option is "B. less"
Well I have 2 examples.
AM/FM radio is a perfect type of technology.
A smart phone is also a great presentation of technology.
