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
The slope represents the acceleration
Answer:
The acceleration of the wagon is 3 m/s².
To calculate the acceleration of the wagon, we use the formula below.
Formula:
F = ma............. Equation 1
Where:
F = horizontal Force
m = mass of the wagon
a = acceleration of the wagon.
make a the subject of the equation
a = F/m.............. Equation 2
From the question,
Given:
F = 30 N
m = 10 kg
Substitute these values into equation 2
a = 30/10
a = 3 m/s²
Hence, the acceleration of the wagon is 3 m/s².
