Latitude, elevation, ocean currents, topography, and prevailing winds. There's probably a few others but these are the most important.
Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.
Answer:
1.1299 x 10^-8 second
Explanation:
Period = 1 / f = 1 / (8.85 * 10^7) = 1.1299 x 01^-8 sec
Answer:
The Resultant Induced Emf in coil is 4∈.
Explanation:
Given that,
A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.
To find :-
find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).
So,
Emf induced in the coil represented by formula
∈ =
...................(1)
Where:
.
{ B is magnetic field }
{A is cross-sectional area}
.
No. of turns in coil.
.
Rate change of induced Emf.
Here,
Considering the case :-
&
Putting these value in the equation (1) and finding the new emf induced (∈1)
∈1 =
∈1 =
∈1 =![4 [-N\times\frac{d\phi}{dt}]](https://tex.z-dn.net/?f=4%20%5B-N%5Ctimes%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D%5D)
∈1 = 4∈ ...............{from Equation (1)}
Hence,
The Resultant Induced Emf in coil is 4∈.
The change in potential energy when the block falls to ground is -480J.
The maximum change in kinetic energy of the ball is 480 J.
The initial kinetic energy of the ball is 0 J.
The final kinetic energy of the ball is 0.148J.
The initial potential energy of the ball is 0.187 J.
The final potential energy of the ball is 0 J.
The work done by the air resistance is 0.039 J.
<h3>Change in potential energy when the block falls to ground</h3>
ΔP.E = -mgh
ΔP.E = -Wh
ΔP.E = - 40 x 12
ΔP.E = -480 J
<h3>Maximum change in kinetic energy of the ball</h3>
ΔK.E = - ΔP.E
ΔK.E = - (-480 J)
ΔK.E = 480 J
<h3>Initial kinetic energy of the ball</h3>
K.Ei = 0.5mv²
where;
- v is zero since it is initially at rest
K.Ei = 0.5m(0) = 0
<h3>Final kinetic energy</h3>
K.Ef = 0.5mv²
K.Ef = 0.5(0.0091)(5.7)²
K.Ef = 0.148 J
<h3>Initial potential energy of the ball</h3>
P.Ei = mghi
P.Ei = 0.0091 x 9.8 x 2.1
P.Ei = 0.187 J
<h3>Final potential energy</h3>
P.Ef = mghf
P.Ef = 0.0091 x 9.8 x 0
P.Ef = 0
<h3>Work done by the air resistance</h3>
W = ΔE
W = P.E - K.E
W = 0.187 J - 0.148 J
W = 0.039 J
Learn more about potential energy here: brainly.com/question/1242059
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