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katrin [286]
3 years ago
7

Problem 13.145 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once an answer is submitte

d, you will be unable to return to this part. A 100-ton tugboat is moving at 6 ft/s with a slack towing cable attached to a 100-ton barge that is at rest. The cable is being unwound from a drum on the tugboat at a constant rate of 5.4 ft/s and that rate is maintained after the cable becomes taut. Problem 13.145.a - Tugboat velocity Neglecting the resistance of the water, determine the velocity of the barge after the cable becomes taut. (You must provide an answer before moving to the next part.) The velocity of the TUGBOAT after the cable becomes taut is ft/s.

Physics
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer:

Please see the attached picture for the complete answer.

Explanation:

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If an object is thrown upward with an initial velocity of 128 ​ft/second, then its height after t seconds is given by the follow
IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

h = 128t-32t^2....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

\dfrac{dh}{dt}=0

From equation (I)

\dfrac{dh}{dt}=128-64t

128-64t=0

t=\dfrac{128}{64}

t=2\ sec

Now, for maximum height

Put the value of t in equation (I)

h =128\times2-32\times4

h=128\ ft

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

128t-32t^2=0

t(128-32t)=0

128=32t

t=4\ sec

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

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A measure of the body's resting energy expenditure based on data that is collected four hours after eating or physical activity
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Electricity. I took something like this hope this helps :)
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Which type of radiation travels at the speed of light and penetrates matter easily?
goblinko [34]
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Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on
lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

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What problim in flywheel
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There are several tell-tale signs that there is a problem with either the starter or flywheel that is preventing your car from starting or causing starting issues. Free spinning is when, for some reason, the starter gear is just spinning without contacting anything. The sound is similar to that of a small drill spinning.

please can u mark me as branliest

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