Answer:
A. Zero
Explanation:
The force on a coil of N turns, enclosing an area, A and carrying a current I in the presence of a magnetic field B, is :
F = N * I * A * B * sinθ
Where θ is the angle between the normal of the enclosed area and the magnetic field.
Since the normal of the area is parallel to the magnetic field, θ = 0
Hence:
F = NIABsin0
F = 0 or Zero
Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J
Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = ![X_{L} = j\omega L = 2\pi fL](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%20j%5Comega%20L%20%3D%202%5Cpi%20fL)
where
R = resistance
![X_{L} = Inductive Reactance](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%20Inductive%20Reactance)
f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:
![C = \frac{\epsilon_{o}A}{x}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B%5Cepsilon_%7Bo%7DA%7D%7Bx%7D)
where
x = separation between the parallel plates
Thus
C ∝ ![\frac{1}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D)
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,
![Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28X_%7BL%7D%20-%20X_%7BC%7D%29%5E%7B2%7D%7D)
Also,
Z ∝ I
Therefore,
![\frac{Z}{I} = \frac{Z'}{I'}](https://tex.z-dn.net/?f=%5Cfrac%7BZ%7D%7BI%7D%20%3D%20%5Cfrac%7BZ%27%7D%7BI%27%7D)
![\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28R%20-%20X_%7BC%7D%29%5E%7B2%7D%7D%7D%7B3I%7D%20%3D%20%5Cfrac%7B%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28R%20-%20%5Cfrac%7BX_%7BC%7D%7D%7B3%7D%29%5E%7B2%7D%7D%7D%7BI%7D)
![{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})](https://tex.z-dn.net/?f=%7BR%5E%7B2%7D%20%2B%20%28R%20-%20X_%7BC%7D%29%5E%7B2%7D%7D%20%3D%209%28%7BR%5E%7B2%7D%20%2B%20%28R%20-%20%5Cfrac%7BX_%7BC%7D%7D%7B3%7D%29%5E%7B2%7D%7D%29)
![{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})](https://tex.z-dn.net/?f=%7BR%5E%7B2%7D%20%2B%20R%5E%7B2%7D%20%2B%20X_%7BC%7D%5E%7B2%7D%20-%202RX_%7BC%7D%20%3D%209%28%7BR%5E%7B2%7D%20%2B%20R%5E%7B2%7D%20%2B%20%5Cfrac%7BX_%7BC%7D%5E%7B2%7D%7D%7B9%7D%20-%202RX_%7BC%7D%29)
Solving the above eqn:
![X_{C} = 4R](https://tex.z-dn.net/?f=X_%7BC%7D%20%3D%204R)
Answer:
answer a, 4
Explanation:
when the 4 is before the compound it applies to the whole compound