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lina2011 [118]
3 years ago
13

Two long, parallel wires are separated by a distance of 2.2 cm. The force per unit length that each wire exerts on the other is

3.6 × 10^-5 N/m, and the wires repel each other. The current in one wire is 0.52 A. What is the magnitude of the current in the second wire? (Give your answer in decimal using "A" (Ampere) as unit)
Physics
1 answer:
VikaD [51]3 years ago
5 0

Answer:

i_2 = 7.6 A

Explanation:

As we know that the force per unit length of two parallel current carrying wires is given as

F = \frac{\mu_o i_1 i_2}{2\pi d}

here we know that

F = 3.6 \times 10^{-5} N/m

i_1 = 0.52 A

d = 2.2 cm

now from above equation we have

3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7} (0.52)(i_2)}{2\pi (0.022)}

3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2

i_2 = 7.6 A

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Vedmedyk [2.9K]

Answer:

A star that always remains above your horizon and appears to rotate around the celestial pole.

Explanation:

A) a star that is close to the north celestial pole: a circumpolar star could be close to the north celestial pole, but this answer is omitting the south celestial pole.

B) a star that is close to the south celestial pole: a circumpolar star could be close to the south celestial pole, but this answer is omitting the north celestial pole.

C) a star that always remains above your horizon and appears to rotate around the celestial pole: this is the definition of a circumpolar star.

D) a star that makes a daily circle around the celestial sphere: every star does this.

E) a star that is visible from the Arctic or Antarctic circles : there are many starts visible from there that are not circumpolar.

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4 years ago
A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe
Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

I_R=\frac{1}{12}ML^2

where

M=3.30\cdot 10^{-2} kg is its mass

L = 0.450 m is its length

Substituting,

I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2

The moment of inertia of the two rings at the beginning is

I_r = 2mr^2

where

m = 0.200 kg is the mass of each ring

r=5.20\cdot 10^{-2} m is their distance from the center of the rod

Substituting,

I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2

So the total moment of inertia at the beginning is

I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2

The initial angular velocity of the system is

\omega_1 = 35.0 rev/min

The angular momentum must be conserved, so we can write:

L=I_1 \omega_1 = I_2 \omega_2 (1)

where I_2 is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

r=\frac{0.450 m}{2}=0.225 m

so the moment of inertia of the rings is

I_r=2(0.200)(0.225)^2=0.0203 kg m^2

and the total moment of inertia is

I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2

Substituting into (1), we find the final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

I_2 = I_R = 5.57\cdot 10^{-4}kg m^2

So using again equation of conservation of the angular momentum:

L=I_1 \omega_1 = I_2 \omega_2

We find the new final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min

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Answer:

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Explanation:

The rate of change in position at a given point in time is called velocity.

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