Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
Answer:
3 m/s^2
Explanation:
acceleration= Change in velocity/time
= 30-0 / 10
= 30/10
=3 m/s^2
Answer:
857.5 m
2.8583×10⁻⁶ seconds
Explanation:
Time taken by the sound of the thunder to reach the student = 2.5 s
Speed of sound in air is 343 m/s
Speed of light is 3×10⁸ m/s
Distance travelled by the sound = Time taken by the sound × Speed of sound in air
⇒Distance travelled by the sound = 2.5×343 = 857.5 m
⇒Distance travelled by the sound = 857.5 m
Time taken by light = Distance the light travelled / Speed of light
Time taken by light = 2.8583×10⁻⁶ seconds
C the third one i think good luck
Answer:
33.6 m
Explanation:
Given:
v₀ = 0 m/s
a = 47.41 m/s²
t = 1.19 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (1.19 s) + ½ (47.41 m/s²) (1.19 s)²
Δx = 33.6 m
