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Rudiy27
3 years ago
10

Which wave has a higher frequency than microwaves but lower frequency than UV waves?

Physics
1 answer:
Effectus [21]3 years ago
3 0

Answer:

Infrared

Visible light

Explanation:

Electromagnetic waves are waves consisting of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion of the wave.

Electromagnetic waves travel in a vacuum at the speed of light, which is a constant value:

c=3.00\cdot 10^8 m/s

Also, electromagnetic waves are classified into 7 different types, depending on their wavelength and frequency. From highest to lowest frequency, we have:

gamma-rays 10^{20}-10^{24} Hz

x-rays 10^{17}-10^{20}Hz

ultraviolet 10^{15}-10^{17} Hz

visible [4-7.5]\cdot 10^{14} Hz

infrared      10^{14} Hz

microwaves 10^{11}-10^{13} Hz

radio waves

Therefore, from the table we see that both infrared and visible light have higher frequency than microwaves, but lower frequency than UV (ultraviolet).

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Read 2 more answers
A water skier lets go of the tow rope upon leaving the end ofa
ANTONII [103]

Answer:

1.35m

Explanation:

At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.

The vertical speed at the jumping point can be computed:

v_v^2 + v_h^2 = v^2

v_v^2 + 13^2 = 14^2

v_v^2 = 196 - 169 = 27

v_v = \sqrt{27} = 5.2 m/s

When the skier jumps to the its potential energy is converted to kinetic energy:

E_p = E_k

mgh = mv_v^2/2

where m is the skier mass and h is the vertical distance traveled, v_v is the vertical velocity at jumping point, and h is the highest point.

Let g = 10m/s2

We can divide both sides of the equation by m:

gh = v_v^2/2

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3 0
3 years ago
Determine the minimum angle at which a roadbedshould be banked
poizon [28]

To solve this problem, apply the concepts related to the relationship given between the centripetal Force and the Weight.

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T cos\theta = mg \rightarrow T = \frac{mg}{cos\theta}

Tsin\theta = \frac{mv^2}{r} \rightarrow T = \frac{mv^2/r}{sin\theta}

Equating both equation we have that,

\frac{mv^2}{r} = mgtan\theta

tan\theta = \frac{v^2}{rg}

Rearranging to find the angle we have that,

\theta = tan^{-1} (\frac{v^2}{rg})

Our values are given as,

r = 2.00*10^2m

v = 20m/s

\theta = tan^{-1}(\frac{20^2}{(2*10^{2})(9.8)})

\theta = 11.53 \°

Therefore the minimum angle will be 11.53°

5 0
3 years ago
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