Answer:
W = 28226.88 N
Explanation:
Given,
Mass of the satellite, m = 5832 Kg
Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m
The time period of the orbit, T = 1.9 h
= 6840 s
The radius of the planet, R = 4.38 x 10⁶ m
The time period of the satellite is given by the formula
second
Squaring the terms and solving it for 'g'
g = 4 π² 
m/s²
Substituting the values in the above equation
g = 4 π² 
g = 4.84 m/s²
Therefore, the weight
w = m x g newton
= 5832 Kg x 4.84 m/s²
= 28226.88 N
Hence, the weight of the satellite at the surface, W = 28226.88 N
Answer:
Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?
Using Coulomb's law:
F = 1/(4πE) x Q1 x Q2/ r^2
Where
k= 1/(4πE) = 9 x 10^9Nm2/C2
Therefore,
F = 9x 10^9 x 2.50 x 10^-5 x2.50 x
10^-5/. ( 0.5)^2
F= 5.625/ 0.25
F= 22.5N approximately
F= 23N.
To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.
Hence To = 23N, attractive. C ans.
Thanks.
Answer:
a = - 50 [m/s²]
Explanation:
To solve this problem we simply have to replace the values supplied in the given equation.
Vf = final velocity = 0.5 [m/s]
Vi = initial velocity = 10 [m/s]
s = distance = 100 [m]
a = acceleration [m/s²]
Now replacing we have:
![(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]](https://tex.z-dn.net/?f=%280.5%29%5E%7B2%7D-%2810%29%5E%7B2%7D%20%3D%202%2Aa%2A%28100%29%5C%5C0.25-10000%3D200%2Aa%5C%5C200%2Aa%3D-9999.75%5C%5Ca%20%3D-50%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign of acceleration means that the ship slows down its velocity in order to land.
The answer is 60 km. I hope it helps i dont know if this is right or wrong.
Answer:
option ( a ) is correct .
Explanation:
Escape velocity on the earth = √ ( 2 GM / R )
where G is universal gravitational constant , M is mass of the earth and R is radius .
V₀ = √ ( 2 GM / R )
escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth
Radius of the planet = 4 R
escape velocity of planet = √ ( 2 GM / 4R )
= .5 x √ ( 2 GM / R )
= .5 V₀
option ( a ) is correct .
