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Minchanka [31]
3 years ago
13

In a shipping yard, a crane operator attaches a cable to a 1240-kg shipping container and then uses the crane to lift the contai

ner vertically at a constant velocity for a distance of 37 m. Determine the amount of work done by each of the following.
(a) the tension in the cable? (J)
(b) the force of gravity? (J)
Physics
2 answers:
lapo4ka [179]3 years ago
7 0

Answer:

(a) W = 449624 J or 449.624 kJ

(b) W = 449624 J or 449.624 kJ

Explanation:

Parameters given:

mass of container = 1240 kg

vertical distance moved = 37 m

(a) The amount of work done by Tension is given as:

W = T * d

where T = Tension in the cable;

d = distance moved by container.

The Tension in the cable is given as:

T = mg - ma

where w = weight of the container

g = acceleration due to gravity

a = normal acceleration

But a = 0, since the crane is accelerating due to gravity, being lifted vertically upwards. Hence, Tension is:

=> T = 1240 * 9.8

T = 12152 N

Hence, work done  by Tension:

W = T * d

W = 12152 * 37

W = 449624 J or 449.624 kJ

(b) The crane operator is lifting the container vertically upward. In this type of case, the Tension is equal to the weight or force of gravity.

This means that the work done by the Tension will be the same as the work done by the force of gravity.

Hence, work done by force of gravity = 449624 J or 449.624 kJ

Firdavs [7]3 years ago
4 0

Answer:

Explanation:

Given

mass of ship =1240 kg

height=37 m

(a)Work done by Tension in cable+work done by gravity =0

Work done by gravity =change in potential energy of mass=-1240\times 9.8\times 37=-449.624 kJ

Thus work done by tension=449.624 kJ

(b)work done by force of gravity=-mg\times h=-1240\times 9.8\times 37=-449.624 kJ

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The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

\;\;\;=42.9°

6 0
2 years ago
What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff
sergiy2304 [10]

The net force on the object as described is; 58.84N

Two forces acting on the object are;

  • The <em>applied force and the frictional force.</em>

In essence; the frictional force can be evaluated as;

  • Frictional force; = coefficient × Weight of object.

  • Frictional force = 0.21 × 20 × 9.8.

  • Frictional force = 41.16N

  • The Net force = Applied force - frictional force

  • Net force = 100 - 41.16N

Net Force = 58.84 N.

Read more:

brainly.com/question/94428

5 0
2 years ago
What is a magnetic field’s shape?
Agata [3.3K]

Answer:

Magnets come in a variety of shapes and one of the more common is the horseshoe (U) magnet. The horseshoe magnet has north and south poles just like a bar magnet but the magnet is curved so the poles lie in the same plane. The magnetic lines of force flow from pole to pole just like in the bar magnet.

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3 years ago
A car travels 50 km in 25 km /h and then travels 60km with a velocity 20 km/h and then travels 60km with a velocity 20 km/h in t
Butoxors [25]

Answer:

v = 21.25 km/h

The average velocity is 21.25km/h

Explanation:

Average velocity = total displacement/time taken

v = d/t

Given;

A car travels 50 km in 25 km /h

d1 = 50km

v1 = 25km/h

time taken = distance/velocity

t1 = d1/v1

t1 = 50/25 = 2 hours

and then travels 60km with a velocity 20 km/h

d2 = 60km

v2 = 20km/h

t2 = d2/v2 = 60/20

t2 = 3 hours

and then travels 60km with a velocity 20 km/h in the same direction

d3 = 60km

v3 = 20km/h

t3 = d3/v3 = 60/20

t3 = 3 hours

Average velocity = total displacement/total time taken

v = (d1+d2+d3)/(t1+t2+t3)

v = (50+60+60)/(2+3+3)

v = 170/8

v = 21.25 km/h

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3 0
3 years ago
A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l
mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be \who_w = 1000 kg/m^3

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

F_s + F_b = W

Where F_s = kxN is the spring force, F_b = W_w = m_wg = \rho_w V_s g is the buoyancy force, which equals to the weight W_w of the water displaced by the submerged portion of the cylinder, which is the product of water density \rho_w, submerged volume V_s and gravitational constant g. W = mg is the weight of the metal cylinder.

kx + \rho_w V_s g = mg

The submerged volume would be the product of cross-section area and the submerged length x

V_s = Ax = \pi(d/2)^2x

Plug that into our force equation and we have

kx + \rho_w \pi(d/2)^2x g = mg

x(k + \rho_w g \pi d^2/4) = mg

x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m

6 0
3 years ago
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