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Stolb23 [73]
4 years ago
14

An electromagnetic wave in vacuum has an electric field amplitude of 470 V/m. Calculate the amplitude of the corresponding magne

tic field.
Physics
1 answer:
lapo4ka [179]4 years ago
6 0

Answer:

Magnetic field, B = B=1.56\times 10^{-6}\ T

Explanation:

It is given that,

The amplitude of an electromagnetic wave, E = 470 V/m

We need to find the amplitude of the corresponding magnetic field. The relation between electric and magnetic field is :

B=\dfrac{E}{c}

Where

c is the speed of light

B=\dfrac{470\ V/m}{3\times 10^8\ m/s}

B = 0.00000156

B=1.56\times 10^{-6}\ T

So, the amplitude of the corresponding magnetic field is 1.56\times 10^{-6}\ T. Hence, this is the required solution.

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Sedbober [7]

Answer:

2.5x10^-3 J

Explanation:

C=50mC

V=10V

U=1/2 C (V)^2

=1/2(50)(10)^2 x (10)^-6

=25(100)/1000000

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PLEASEEE HELP ME WITH THIS ALSO. I DONT WANT TO FAIL. You push a merry-go-round on which Kim and Katie are riding. Kim weighs 45
Serjik [45]

Answer:

The body weight

Explanation:

5 0
3 years ago
7) A crazy cat (yes, this is redundant) is running along the roof of a 60 m tall building. The cat is moving at a constant veloc
gregori [183]

Answer:

The distance from the base of the building to the landing site is 154 m.

The total flight time is 3.5 s.

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).

v = velocity vector of the cat at time t.

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

r1y = y0 + v0y · t + 1/2 · g · t²

-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

- 60 m = -4.9 m/s² · t²

-60 m / - 4.9 m/s² = t²

t = 3.5 s

The cat reaches the ground in 3.5 s

Now, we can calculate the horizontal component of r1:

r1x = x0 + v0 · t

r1x = 0 m + 44 m/s · 3.5 s

r1x = 154 m

The distance from the base of the building to the landing site is 154 m.

The total flight time was already calculated and is 3.5 s.

The velocity vector of the cat when it reaches the ground will be:

v = (v0x, v0y + g · t)

v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)

v = (44 m/s, -34.3 m/s)

The magintude of the vector "v" is calculated as follows:

|v| = \sqrt{(44 m/s)^{2}+(-34.3 m/s)^{2}} = 55.8 m/s

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

6 0
3 years ago
Pleaseeee help!!!! I will mark you as brainlinest for correct answer!!!!!!!
GrogVix [38]
I do believe that the answer is D.  

Hope this is right! Have a great day! :-)
7 0
3 years ago
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