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morpeh [17]
2 years ago
11

How do the following factors affect the resistance of a wire?

Physics
1 answer:
Alborosie2 years ago
4 0
The temp would affect the length of the wire 
the length would affect how much support there is as well as the diameter of the wire 
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A current-carrying wire 1.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10 A and there is
LenaWriter [7]

Answer:

0.2

Explanation:

F= BIL sin©

3= B×10×1.5 sin90

B=3/15

B= 0.2

8 0
2 years ago
a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial sp
stiks02 [169]

Answer:

1 / 2 m v^2 = L m g (1 - cos θ)

This is the KE due to the pendulum falling from a 25 deg displacement

v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2

v = 1.92 m/s      this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

1.92 + 1.2 = 3.12 m/s   since it gains 1.92 m/s from its initial displacement

3 0
1 year ago
A cube-shaped piece of copper has sides of 4cm each and it's density is
iris [78.8K]

Answer:

64 cm^3

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

\displaystyle \rho=\frac{m}{V}

We have a cube-shaped piece of copper of 4 cm of side length. The volume of the piece is:

V=(4\ cm)^3=64\ cm^3

Surprisingly, no other magnitude is required, thus the answer is:

\mathbf{64 cm^3}

5 0
2 years ago
PLEASE HELP
jasenka [17]
Don't over think it!

For example, if you do 5 laps in 5 minutes then you do one lap per minute. If you use that example and you apply it to your problem you'll notice that the answer is simply C. 1km/min.

Best wishes!
4 0
3 years ago
Read 2 more answers
A gas is compressed from an initial volume of 5.55 L to a final volume of 1.22 L by an external pressure of 1.00 atm. During the
algol13

Answer:

Explanation:

change in the volume of the gas = 5.55 - 1.22

= 4.33 X 10⁻³ m³

external pressure ( constant ) P = 1 x 10⁵ Pa

work done on the gas

=external pressure x change in volume

= 10⁵ x  4.33 X 10⁻³

=4.33 x 10²

433 J

Using the formula

Q = ΔE + W , Q is heat added , ΔE is change in internal energy , W is work done by the gas

Given

Q = - 124 J ( heat is released so negative )

W = - 433 J . ( work done by gas is negative, because it is done on gas  )

- 124  = ΔE - 433

ΔE = 433  - 124

= 309 J

There is increase of 309 J in the internal energy of the gas.

3 0
3 years ago
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