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3 years ago

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A 4.0 kilogram object is dropped from a height of 1.0 meter onto a spring with a spring constant of 850 newtons per meter. Appro

ximately what distance will the spring compress when the object lands on it? (1) 3.0 dm
(2) 9.0 cm

(3) 9.0 mm

(4) 2.0 dm

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1 answer:

lisov135 [29]3 years ago

7 0

Answer:

(1) 3.0 dm

Explanation:

Due to the law of conservation of energy, the initial gravitational potential energy of the object will be all converted into elastic potential energy of the spring as the object hits the spring:

$mgh = \frac{1}{2}kx^2$

where

m = 4.0 kg is the mass of the object

g = 9.8 m/s^2 is the acceleration due to gravity

h = 1.0 m is the initial height of the object

k = 850 N/m is the spring constant

x is the compression of the spring

Solving the equation for x, we find:

$x=\sqrt{\frac{2mgh}{k}}=\sqrt{\frac{2(4.0 kg)(9.8 m/s^2)(1.0 m)}{850 N/m}}=0.30 m=3.0 dm$

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The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

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4 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

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Answer:

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3 years ago