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Y_Kistochka [10]
3 years ago
6

A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper nee

ds to be drawn into a wire with a diameter of 9.50 mm. What length of wire, in meters, can be produced?
Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:

120.125 m

Explanation:

Density = Mass/volume

D = m/v .............................. Equation 1.

Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.

Making v the subject of the equation,

v = m/D............................... Equation 2

Given: m = 76.5 kg,

Constant: D = 8960 kg/m .

Substituting into equation 2

v = 76.5/8960

v = 0.0085379 m³

Since the copper sphere is to be drawn into wire,

Volume of the copper sphere = volume of the wire

v = volume of the wire

Volume of wire = πd²L/4

Where d = diameter of the wire, L = length of the wire.

Note: A wire takes the shape of a cylinder.

v = πd²L/4 ........................ equation 3.

making L the subject of the equation,

L = 4v/πd²..................... Equation 4

Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095  and π = 3.14

Substitute into equation 4

L = 4×0.0085379/(3.15×0.0095²)

L = 0.0341516/0.0002843

L = 120.125 m.

L = 120.125 m

Thus the length of the wire produced = 120.125 m

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Explanation:

From the question,

Spring scale reading = W-U............... Equation 1

Where W = weight of the cube, U = upthrust.

W = mg

Where m =  mass of the cube, g = acceleration due to gravity.

Given: m = 11 g = 0.011 kg, g = 9.8 m/s².

W = 0.011(9.8)

W = 0.1078 N.

From Archimedes principle,

Upthrust = weight of water displaced.

U = (Density of water×volume of metal cube)×acceleration due to gravity.

U = (D×V)g

Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²

U = 1000(9.8)(10⁻⁶)

U = 0.0098 N.

Substitute the value of W and U into equation 1

Reading of the spring scale = 0.1078-0.0098

Reading of the spring scale = 0.098 N

7 0
3 years ago
When velocity is graphed with respect to time, what does the area under
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In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that i
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As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here

v_f^2 - v_i^2 = 2 a d

since we released it from rest so we have

v_i = 0

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v_f = \sqrt{2gd}

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v_f' = \sqrt{2g(2d)}

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3 years ago
A tow truck drags a stalled car along a road. The chain makes an angle of 30° with the road and the tension in the chain is 1400
AURORKA [14]

Answer:

The work is done by the truck pulling the car 1 km is 1,212,436 J

Explanation:

Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.

So work is one of the forms of energy transmission between bodies. To perform a job, you must exert a force on a body and it moves.

In the International System of Units, work is measured in Joule. A Joule is the work that a constant force of 1 Newton does on a body that moves 1 meter in the same direction and direction as the force. Then, Joule is equivalent to Newton per meter.

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves:

Work= Force*Distante* cos (θ)

In this case:

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Replacing:

Work= 1,400 N* 1,000 m* cos (30°)

Work= 1,212,435. 565 Joule≅ 1,212,436 J

<u><em> The work is done by the truck pulling the car 1 km is 1,212,436 J</em></u>

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3 years ago
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