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Harrizon [31]
3 years ago
10

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of

2.70 x 10^-3 V/m and is parallel to the x axis. What are the frequency and the magnetic-field amplitude of this wave? In calculating the amplitude of the magnetic field, you divide the electric-field amplitude by the speed of light, which requires you to convert some units. Which of the following is equivalent to 1 T?
a. 1 (N/A.m)
b. 1 (N/c.m)
c. 1 (N.s/C.m)
d. 1 (V.s/m^2)
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

a) 6.9*10^14 Hz

b) 9*10^-12 T

Explanation:

Given that

The wavelength of the wave, λ = 435 nm

Amplitude of the electric field, E(max) = 2.7*10^-3 V/m

a)

The frequency of the wave can be found by using the formula

c = fλ, where c = speed of light

f = c/λ

f = 3*10^8 / 435*10^-9

f = 6.90*10^14 Hz

b)

E(max) = B(max) * c, magnetic field amplitude, B(max) =

B(max) = E(max)/c

B(max) = 2.7*10^-3 / 3*10^8

B(max) = 9*10^-12 T

c)

1T = 1 (V.s/m^2)

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As we know, Effort×effort distance = load × load distance

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So when the spanner’s handle is long

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Which is not an issue with waste from nuclear energy?
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3 years ago
The nucleus of most atoms is composed of which of the following sub-atomic particles?
Blababa [14]

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protons (+ charge) & neutrons (neutral charge)

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3 years ago
A rock is lifted by a machine to a height of 10m. If it has a mass of 22 kilograms
zmey [24]

Answer:

2156J

Explanation:

Given parameters:

Height of lift  = 10m

Mass  = 22kg

Unknown:

Work done by the machine  = ?

Solution:

Work done is the force applied to move a body through a certain distance.

So;

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Here;

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5 0
3 years ago
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

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2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
2 years ago
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