A 3.00-kg box that is several hundred meters above the earth’s surface is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope and results in a tension in the rope of T(t) = (38.0 N/s)t. The box is at rest at t = 0. The only forces on the box are the tension in the rope and gravity. A) What is the velocity of the box at t = 1.00 s? B) What is the maximum distance that the box descends below its initial position? C) At what value of t does the box return to its initial position? D) What is the velocity of the box at t = 3.00 s?

A) 3.48m/s

B) 3.92m

C) 2.32m

D 23.33m/s

Explanation:

ma(t)=mg-At

a(t)= g - (At/m)

V(t)= integrala(T)dT = gt- (At^2/2m)

Initial x coordinator of the box is zero

X(t)= integralV(t)dt= 1/2gt^2-(At^3/6m)

a) V =( 9.81×1) -(38×1^2/2×3)

V= 9.81-6.33= 3.48m/s

b)-AT^2/2m + gT= 0

T=2mg/A= (2×3×9.81)/38

T= 1.549m

X(T)= (1/2×9.81×1.549^3)- (38×1.549^3/6×3)

X(T)= 11.768- (141.23/18) = 11.768 - 7.85= 3.92m

C) 1/2gT''^2 - AT''^3/6m =0

The only non trivial solution is T''= 3mg/A

T=(3×3×9.81)/38 = 2.32m

D) V = 9.81×3) - (38×3^2/6)

V= 29 - 5.667= 23.33m/s

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

happy to

pls mark as brainliest.

the answer is 3.1 his answer is wrong mines is right