Answer:
The magnitude of the emf induced in the coil is 60 mV.
Explanation:
We have,
Side of the square coil, a = 24 cm = 0.24 m
Number of turns in the coil, N = 2
It is placed in a uniform magnetic field that makes an angle of 60 degrees with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, we need to find the magnitude of the emf induced in the coil.
We know that the induced emf is given by the rate of change of magnetic flux throughout the coil. So,

is the angle between magnetic field and the normal to area vector.
But in this case, a uniform magnetic field that makes an angle of 60 degrees with the plane of the coil. So, the angle between magnetic field and the normal to area vector is 90-60=30 degrees.
Now, induced emf becomes :

or

So, the magnitude of the emf induced in the coil is 60 mV. Hence, this is the required solution.
Since the rocket’s acceleration is 3.00 m/s^3 * t, its acceleration is increasing at the rate of 3 m/s^3 each second. The equation for its velocity at a specific time is the integral of the acceleration equation.
<span>vf = vi + 1.5 * t^2, vi = 0 </span>
<span>vf = 1.5 * 10^2 = 150 m/s </span>
This is the rocket’s velocity at 10 seconds. The equation for its height at specific time is the integral velocity equation
<span>yf = yi + 0.5 * t^3, yi = 0 </span>
<span>yf = 0.5 * 10^3 = 500 meters </span>
<span>This is the rocket’s height at 10 seconds. </span>
<span>Part B </span>
<span>What is the speed of the rocket when it is 345 m above the surface of the earth? </span>
<span>Express your answer with the appropriate units. </span>
<span>Use the equation above to determine the time. </span>
<span>345 = 0.5 * t^3 </span>
<span>t^3 = 690 </span>
<span>t = 690^⅓ </span>
<span>This is approximately 8.837 seconds. Use the following equation to determine the velocity at this time. </span>
<span>v = 1.5 * t^2 = 1.5 * (690^⅓)^2 </span>
<span>This is approximately 117 m/s. </span>
<span>The graph of height versus time is the graph of a cubic function. The graph of velocity is a parabola. The graph of acceleration versus time is line. The slope of the line is the coefficient of t. This is a very different type of problem. For the acceleration to increase, the force must be increasing. To see what this feels like slowly push the accelerator pedal of a car to the floor. Just don’t do this so long that your car is speeding!!</span>
Answer:
a) Magnetic field strength, B = 2.397 * 10⁻⁷ T
b) Total energy density, U = 4.58 * 10⁻⁸ J/m³
c) Power flow per unit area, S = 13.71 W/m²
Explanation:
a) Electric field strength, E = 71.9 V/m
The relationship between the Electric field strength and the magnetic field strength in vacuum is:
E = Bc where c = 3.0 * 10⁸ m/s
71.9 = B * 3.0 * 10⁸
B = 71.9 / (3.0 * 10⁸)
B = 23.97 * 10⁻⁸
B = 2.397 * 10⁻⁷ T
b) Total Energy Density:

c)Power flow per unit area

Answer:
This is because the speed of a wave is determined by the medium through which it is passing. When light speeds up as it passes from one material to another, the angle of refraction is bigger than the angle of incidence. For example, this happens when light passes from water to air or from glass to water ❤
1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters