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olga55 [171]
3 years ago
5

We move a square loop of wire 2.3 cm on a side into a region of uniform magnetic field of 1.5 T with the plane of the loop perpe

ndicular to the direction of the field. The loop is moving with a constant velocity of 2.1 m/s into the field region starting from a point outside the field. If the loop has a resistance of 3 Ω, find the force on the loop as it just enters the magnetic field region. (Enter the magnitude.)
Physics
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

Force will be F=8.33\times 10^{-4}N

Explanation:

We have given side of wire d = 2.3 cm = 0.023 m

Magnetic field B = 1.5 T

Velocity v = 2.1 m/sec

Resistance R = 3 ohm

We know that current is given by i=\frac{Bvd}{R}=\frac{1.5\times 2.1\times 0.023}{3}=0.02415A

Power is given by P=i^2R=0.02415^2\times 3=0.00174watt

We know that power is also given by P=Fv

So 0.00174=F\times 2.1

F=8.33\times 10^{-4}N

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A) initial volume
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p_i V_i = nRT_i
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V_i is the initial volume of the gas
n=2.3 mol is the number of moles
R=8.31 J/K mol is the gas constant
T_i=240^{\circ}C=513 K is the initial temperature of the gas

By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
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3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
W=p \Delta V= p(V_f -V_i)
by using the data we found at point 1) and 2), we find
W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J
where the negative sign means the work is done by the surrounding on the gas.
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A copper wire has a square cross section 2.1 mm on a side. The wire is 4.3 m long and carries a current of 3.6 A. The density of
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Explanation:

Below is an attachment containing the solution.

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When waves superimpose and make bigger amplitudes what form of interference is that
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Answer:

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Explanation:

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In constructive interference, the crests of one wave fall on the crests of second wave and the amplitudes add up. The amplitude of the resultant wave is equal to sum of the amplitude of the individual waves. Similarly, the trough of first wave falls on the trough of other wave and they superimpose to create the trough of the resultant wave.

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Answer:

4.15 m/s

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ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

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As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

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From (1) and (2) we can solve for vf, as follows:

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