your answer is the letter (b)
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
800 J Got it right on edgenuity
Answer:
Explanation:
Work done = ∫Fdx
= ∫(cx-3.00x²) dx
[ c x² / 2 - 3 x³ / 3 ]₀²
= change in kinetic energy
= 11-20
= - 9 J
[ c x² / 2 - x³ ]₀² = - 9
c x 2² / 2 - 2³ = -9
2c - 8 = -9
2c = -1
c = - 1/2
