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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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How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a pu
notka56 [123]

Answer:

Newton's first law

Explanation:

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. Therefore, when the ice is smooth, friction gets lesser, and the force acted on that Puck will be decreased.

6 0
3 years ago
Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?​
Sliva [168]

Answer:

\huge\star{\underline{\mathtt{\blue{Answer}}}}\huge\star...

<h2>PE=<em>work done</em></h2><h2><em>m</em><em>gh</em><em>=</em><em>2</em><em>0</em><em>×</em><em>1</em><em>0</em><em>×</em><em>2</em><em>0</em><em>.</em><em>.</em></h2>

\huge\boxed{\fcolorbox{white}{blue}{mgh=4000}}

.

<em>I </em><em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

7 0
3 years ago
A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free
Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α   .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R      from equation (1)

T = I x a / R²      .... (2)

mg - T = ma    .... (3)

Substitute the value of T from equation (2) in equation (3)

mg - \frac{Ia}{R^{2}}=ma

a=\frac{mg}{m+\frac{I}{R^{2}}}

T is the acceleration in the system

Substitute the value of a in equation (2)

T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}

T=\frac{I\times mg}{I+mR^{2}}

This is the tension in the string.

4 0
3 years ago
What kind of research in teaching of physical science can be done???​
11Alexandr11 [23.1K]

Answer:

Physical sciences are those academic disciplines that aim to uncover the underlying laws of nature - often written in the language of mathematics. It is a collective term for areas of study including astronomy, chemistry, materials science and physics.

Explanation:

3 0
3 years ago
A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
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