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:
option 1
Explanation:
i just used the SOH CAH TOA, and since the given is tan=opposite/adjacent, that should be the answer
Answer:
r = 0.5 m
Explanation:
First we find the angular speed of the ball by using its period:
ω = θ/t
For the time period:
ω = angular speed = ?
θ = angular displacement = 2π rad
t = time period = 0.5 s
Therefore,
ω = 2π rad/0.5 s
ω = 12.56 rad/s
Now, for the radius:
v = rω
r = v/ω
where,
v = linear speed = 6.29 m/s
r = radius = ?
r = (6.29 m/s)/(12.56 rad/s)
<u>r = 0.5 m</u>
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
v = √(kgr)
v = √(0.20·9.8 m/s²·50 m)
= 7√2 m/s ≈ 9.90 m/s
Answer:
The refraction of light at the surface of water makes ponds and swimming pools appear shallower than they really are. A 1m deep pond would only appear to be 0.75 m deep when viewed from directly above. When light emerges from glass or water into air it speeds up again.
Explanation:
