Answer:
d = 69 .57 meter
Explanation:
First case
Speed of car ( v ) = 20.5 mi/h = 9.164 M/S
distance ( d ) = 11.6 meter ( m = mass of the car )
Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J
Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N
Second case
v = 50.2 mi/h = 22.44135 m/s
d = ?
Work done = 0.5 * 22.44² * m J = 251.7768 * m J
Since the braking force remains the same .
3.619 m = ( 251.7768 m / d )
d = 69 .57 meter
Plant trees around the perimeter of his fields
The answer is 2400 centimeters
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
![\Phi = BA Cos \theta](https://tex.z-dn.net/?f=%5CPhi%20%3D%20BA%20Cos%20%5Ctheta)
Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
![\epsilon= -N \frac{d\Phi }{dt}](https://tex.z-dn.net/?f=%5Cepsilon%3D%20-N%20%5Cfrac%7Bd%5CPhi%20%7D%7Bdt%7D)
![\epsilon = -N \frac{(BAcos\theta)}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-N%20%5Cfrac%7B%28BAcos%5Ctheta%29%7D%7Bdt%7D)
![\epsilon = -NAcos\theta \frac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-NAcos%5Ctheta%20%5Cfrac%7BdB%7D%7Bdt%7D)
![\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-N%5Cpi%20r%5E2%20cos%5Ctheta%20%5Cfrac%7Bd%7D%7Bdt%7D%20%28%20%28%203.75%20T%20%29%20%2B%20%28%203.05T%2Fs%20%29%20t%20%2B%20%28%20-6.95%20T%2Fs%5E2%20%29%20t%5E2%29)
![\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-N%5Cpi%20r%5E2%20cos%5Ctheta%28%20%283.05T%2Fs%29-%2813.9T%2Fs%29t%20%29)
At time
, Induced emf is,
![\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20-%281%29%20%5Cpi%20%280.220m%29%5E2%20cos%2819.5%5C%C2%B0%29%28%20%20%283.05T%2Fs%29-%2813.9T%2Fs%29%285.71s%29%29)
![\epsilon = 10.9V](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%2010.9V)
Therefore the magnitude of the induced emf is 10.9V
Answer:
a) S = 2.35 10³ J/m²2
,
b)and the tape recorder must be in the positive Z-axis direction.
the answer is 5
c) the direction of the positive x axis
Explanation:
a) The Poynting vector or intensity of an electromagnetic wave is
S = 1 /μ₀ E x B
if we use that the fields are in phase
B = E / c
we substitute
S = E² /μ₀ c
let's calculate
s = 941 2 / (4π 10⁻⁷ 3 10⁸)
S = 2.35 10³ J/m²2
b) the two fields are perpendicular to each other and in the direction of propagation of the radiation
In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.
the answer is 5
C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis