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Dominik [7]
2 years ago
8

In an electromagnetic wave in free space, the electric and magnetic fields are.

Physics
1 answer:
Vlad [161]2 years ago
6 0

Answer:

PERPENDICULAR

Explanation:

<h3>HOPE IT HELPS </h3>
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A brown bear runs at a speed of 9.0\,\dfrac{\text m}{\text s}9.0 s m ​ 9, point, 0, start fraction, start text, m, end text, div
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Explanation:

Below is an attachment containing the solution.

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3 years ago
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In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met
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Answer:

The kinetic energy of the ejected electrons increases.

Explanation:

As we know that electrons are only ejected from a metal surface if the frequency of the incident light increases the work function of the metal. If the frequency of the incident light is less than the work function of the metal no matter how intense the beam the electrons will not be ejected from the surface.

Using conservation of energy principle we have

E_{incident}=h\nu +\frac{1}{2}mv^{2}

If we increase the intensity  of incident light the term on the LHS of the above equation increases this increase appears in the kinetic energy term in RHS of the equation since h\times \nu remains constant.

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4 years ago
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Which statement best describes the skier?
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its c!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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3 years ago
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A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=
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The frequency of the radio station is
f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength \lambda is
\lambda= \frac{c}{f}
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m
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4 years ago
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A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

3 0
3 years ago
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