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4vir4ik [10]
3 years ago
13

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 neg

ligible 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?
Physics
1 answer:
marshall27 [118]3 years ago
6 0

Answer:

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

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Answer:

d

Explanation:

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           V = Velocity of money exchange

           P = The price level

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3 years ago
Can the big bang's sound still be heard?
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There is no sound in space because there is no medium to move it. However i think the big bang can still been seen in the form of background radiation but i am unsure. 
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3 years ago
The figure below shows a man in a boat on a lake. The man's mass is 74 kg, and the boat's is 135 kg. The man and boat are initia
vazorg [7]

The velocity of the boat after the package is thrown is 0.36 m/s.

<h3>Final velocity of the boat</h3>

Apply the principle of conservation of linear momentum;

Pi = Pf

where;

  • Pi is initial momentum
  • Pf is final momentum

v(74 + 135) = 15 x 5

v(209) = 75

v = 75/209

v = 0.36 m/s

Thus, the velocity of the boat after the package is thrown is 0.36 m/s.

Learn more about velocity here: brainly.com/question/6504879

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3 years ago
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A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum  

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is  

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0
3 years ago
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