Answer:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
Answer:
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
Explanation:
Given that
Yield strength ,Sy= 240 MPa
Tensile strength = 310 MPa
Elastic modulus ,E= 110 GPa
L=380 mm
ΔL = 1.9 mm
Lets find strain:
Case 1 :
Strain due to elongation (testing)
ε = ΔL/L
ε = 1.9/380
ε = 0.005
Case 2 :
Strain due to yielding


ε '=0.0021
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
For computation of load strain due to testing should be less than the strain due to yielding.
Answer:
opposite the sun. between the Earth and the sun. rising perpendicular to the sun.
Explanation:
Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N