Answer:
E) 6.5 A
Explanation:
Given that
L = 40 m H
C= 1.2 m F
Maximum charge on capacitor ,Q= 45 m C
The maximum current I given as
I = Q.ω
ω =angular frequency

By putting the values


ω = 144.33 rad⁻¹
Maximum current
I = 45 x 10⁻³ x 144.33 A
I= 6.49 A
I = 6.5 A
E) 6.5 A
Answer:
The object would weight 63 N on the Earth surface
Explanation:
We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

Now, if the body is on the surface of the Earth, its weight (w) would be:

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

<u>Answer:</u> Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
<u>Explanation:</u>
To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.
This wave base is equal to half of the wavelength. The equation becomes:
Wave base = 
We are given:
Wavelength = 24 m
Putting values in above equation, we get:
Wave base = 
Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
Answer:
Net force = 800 – 650
= 150 N
150 = (800 ÷ 9.8) a
a = 1470 ÷ 800
= 1.8375 m/s^2, downwards