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Daniel [21]
2 years ago
10

Please solve the Problem.

Physics
1 answer:
aalyn [17]2 years ago
5 0

(a) The electrical resistance of the wire is determined as 9.27 x 10⁻⁴ ohms.

(b) The current flowing through the wire is 1.1 mA.

<h3>Electrical resistance of the wire</h3>

The electrical resistance of the wire is determined as follows;

ρ = RA/l

RA = ρL

R = ρL/A

Where;

  • A is area of the iron bar
  • ρ is electrical resistivity = 1/conductivity = 1/1.0299 x 10⁷ = 9.7 x 10⁻⁸

A = πd²/4

A =  π x (0.02)²/4

A = 3.14 x 10⁻⁴ m²

R = (9.7 x 10⁻⁸ x 3)/(3.14 x 10⁻⁴)

R = 9.27 x 10⁻⁴ ohms

<h3>From the chart, voltage after 120 minutes ≈ 1.1 V</h3>

Current, I = V/R

where;

  • R = 1 k-ohms = 1000 ohms

I = 1.1/(1000)

I = 1.1 mA

Learn more about resistance here: brainly.com/question/17563681

#SPJ1

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v_e=\sqrt{\frac{2GM}{R}}

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v=0.421 v_e

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K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}

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At the point of maximum altitude, all this energy is converted into gravitational potential energy:

K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:0.177 \frac{GMm}{R}=\frac{GMm}{nR}

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b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

This time, the projectile has 0.421 times this energy:

K=0.421 \frac{GMm}{R}

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

0.421 \frac{GMm}{R}=\frac{GMm}{nR}

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The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

E=U=\frac{GMm}{R}

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K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

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