A) 1.05 N
The power dissipated in the circuit can be written as the product between the pulling force and the speed of the wire:
![P=Fv](https://tex.z-dn.net/?f=P%3DFv)
where
P = 4.20 W is the power
F is the magnitude of the pulling force
v = 4.0 m/s is the speed of the wire
Solving the equation for F, we find
![F=\frac{P}{v}=\frac{4.20 W}{4.0 m/s}=1.05 N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BP%7D%7Bv%7D%3D%5Cfrac%7B4.20%20W%7D%7B4.0%20m%2Fs%7D%3D1.05%20N)
B) 3.03 T
The electromotive force induced in the circuit is:
(1)
where
B is the strength of the magnetic field
v = 4.0 m/s is the speed of the wire
L = 10.0 cm = 0.10 m is the length of the wire
We also know that the power dissipated is
(2)
where
is the resistance of the wire
Subsituting (1) into (2), we get
![P=\frac{B^2 v^2 L^2}{R}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BB%5E2%20v%5E2%20L%5E2%7D%7BR%7D)
And solving it for B, we find the strength of the magnetic field:
![B=\frac{\sqrt{PR}}{vL}=\frac{\sqrt{(4.20 W)(0.350 \Omega)}}{(4.0 m/s)(0.10 m)}=3.03 T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Csqrt%7BPR%7D%7D%7BvL%7D%3D%5Cfrac%7B%5Csqrt%7B%284.20%20W%29%280.350%20%5COmega%29%7D%7D%7B%284.0%20m%2Fs%29%280.10%20m%29%7D%3D3.03%20T)
S = d/t
s = 1396/4 = 349m/s
Answer:
Ф6 = 3 Wb
Explanation:
the flux is positive when N is odd and positive when N is even so we can write the flux through the all side as
∑ N=1, to5
∑ -1 ^N Ф N * Ф6 =0
∑ -1 ^ N * Ф6 =0
Where :
ФN=N
Replacing and operated so
-1 Wb +2 Wb - 3 Wb + 4Wb - 5Wb + Ф6=0
Ф6 - 3Wb = 0
Ф6 = 3 Wb
Answer:
3
Explanation:
1-3 age is hard to remember
Work = Force x Distance
Work = 45N x 1.2m
Work = 54 N x m
Work = 54J
The amount of work done was 54 joules.