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Ugo [173]
3 years ago
12

An electromagnetic wave is transporting energy in the negative y direction. At one point and one instant the magnetic field is i

n the positive x direction. The electric field at that point and instant is:
Why is (D) the answer? I thought it's (C)

a. positive y direction
b. negative y direction
c. positive z direction
d. negative z direction
Physics
1 answer:
natali 33 [55]3 years ago
4 0

Answer:. Option c

Explanation: the speed of an electromagnetic wave is simply the vector product of the magnetic field and the electric field.

The direction of the velocity is the direction of the electromagnetic wave.

The wave is already moving towards the negative y axis (-j) and the magnetic field is already pointing towards the positive x axis (i)

From cross product of unit vectors

i × j = k

i × k = - j

With the second identity, we can see that the electric field will be pointing towards the positive of the x axis (k).

Option c is validated

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Please help me
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Answer:

0.3956

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0.43*0.92 = 0.3956.

0.4 if you round

5 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f
Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

s=ut+ \frac{1}{2} at^2

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0
3 years ago
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