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Levart [38]
3 years ago
6

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

, the tension in the cable was 7000 N {\rm N}. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N {\rm N} drag force.
Part A

What was the tension in the cable when the craft was being lowered to the seafloor?

Express your answer to two significant figures and include the appropriate units.

Part B

What was the tension in the cable when the craft was being raised from the seafloor?

Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Talja [164]3 years ago
4 0

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

F_{g} - F_{d} - T = 0

7000 - 1800 - T = 0

T = 5200 N

T = 5.2 x 10³ N

Part B)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

T  - F_{g} - F_{d} = 0

T - 7000 - 1800  = 0

T = 8800 N

T = 8.8 x 10³ N

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A 425 g block is released from rest at height h0 above a vertical spring with spring constant k = 460 N/m and negligible mass. T
Morgarella [4.7K]

Answer:

(a) = +5.38m (b) = -5.38m (c) = 1.246m (d) = +0.3771m.

Explanation:

Initially the spring is at equilibrium,

Work done by all forces = change in kinetic energy

Work = ∇K.E

Work = Kf -Ki =0

Since the work done = 0 since the body is at rest.

W(spring) + W(gravity) = 0

W(spring) + W(gravity) = 0

W(spring) = -W(gravity)

Work done by the block on the spring = W(block/spring)

W(block/spring) + W(spring) = 0

W(spring) = -∫kx.dx

W(spring) = ½k(X²i - X²f) ; Xi =0, Xf = 15.3cm = 0.153m

W(spring) = -½* 460 * (0.153)²

W(spring) = - 5.38NM

Work done by block on spring = + 5.38NM

(b). Workdone by spring on the block = -5.38NM.

Note: This is so because the displacement of the force is in the opposite direction to the previous one since they counter each other to maintain equilibrium.

(C). W(spring) +W(gravity) = 0

½kx² + mg(h + x) = 0

-5.83 + mg(h + 0.153) =0

5.83 = 0.425*9.8 (h + 0.153)

5.83 = 4.165(h + 0.153)

H = 1.399 - 0.153

H = 1.246m

(D).

If the release height was 6ho

H = 6* 1.246m = 7.476m

W(spring) = W(gravity)

½kx² = mg(7.476 + x)

Note: At maximum compression, the blocks would be at rest.

½Kx² = mg(h + x)

½ * 460 * x² = 0.425 * 9.8 * (7.476 + x)

230x² = 4.165 (7.476 + x)

230x² = 31.137 + 4.165x

230x² - 4.165x - 31.137 = 0

Solving the quadratic equation ( i would suggest you use formula method for easy navigation of the variables)

X = + 0.3771m or -0.3589m

But we can't have a negative compression value,

X = + 0.3771m

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3 years ago
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I would say C: Dam

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3 years ago
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Why are solar winds potentially fatal for organisms living on earth?
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7 0
3 years ago
Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,
insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction \mu_{k} = 0.250

Zak's speed v = 3 \frac{m}{s}

Gravitational acceleration g = 9.8 \frac{m}{s^{2} }

Work done by frictional force is given by,

  W = \Delta K

 \mu _{k} mg d = \frac{1}{2} m v^{2}

  d = \frac{v^{2}  }{2 g \mu _{k} }

  d = \frac{9}{2 \times 9.8 \times 0.250}

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Therefore, the distance travel before stopping is 1.84 m

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3 years ago
For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex
rjkz [21]
Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet

convert feet to inches, for uniformity.
1 foot = 12 inches
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1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f

The focal length of the project lens is 3.84 inches 

4 0
3 years ago
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