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sammy [17]
3 years ago
11

Where is the centre of mass of a uniform cube located?​

Physics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

NICE QUESTION

Explanation:

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Is beacuse of the air within our bodys is exerting the same pressure out wards so tjere is no pressure difference

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What's a good way to determine the net force of something
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By adding up all the individual forces of the object
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Eva makes a graphic organizer to compare electromagnets with solenoids. A venn diagram with 2 intersecting circles. The circle o
Eva8 [605]

The metalcore is the central component of any metal that is covered by wires. Metalcore is found in the X denoted area.

<h3>What is a solenoid?</h3>

A solenoid is a coil of wire that conducts an electric current. A solenoid is an electromagnet made out of a wire or helical coil.

When an electric current is sent through the coil, it produces a magnetic field. A solenoid is a coil that generates a magnetic field when an electric current passes through it.

When a conductive wire is used to build a loop, a solenoid is formed.

Hence the metalcore is found in the given X area.

To learn more about the solenoid refer to the link ;

brainly.com/question/16015159

5 0
3 years ago
A submarine is at an ocean depth of 3.00 km.
Alexxx [7]

Answer:

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Explanation:

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7 0
3 years ago
How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?
jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

  • diameter of the steel, d = 4 mm
  • the radius of the wire, r = 2mm = 0.002 m
  • original length of the wire, L₁
  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
  • extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
  • the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}

F = \frac{200 \times 10^9\  \times\  1.257\times 10^{-5}\  \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0
2 years ago
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