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emmasim [6.3K]
4 years ago
7

Does visible light make up a relatively large part or a relatively small part of the electromagnetic spectrum?

Physics
1 answer:
Yakvenalex [24]4 years ago
8 0

Answer:

Relatively small part of the electromagnetic spectrum

Explanation:

Let us take a look at the electromagnetic spectrum

Radio waves

Wavelength = 10^3\ m

Microwaves

Wavelength = 10^{-2}\ m

Infrared waves

Wavelength = 10^{-5}\ m

Visible waves

Wavelength = 0.5\times 10^{-6}\ m

Radio waves

Wavelength = 10^3\ m

Ultraviolet waves

Wavelength = 10^{-8}\ m

X ray waves

Wavelength = 10^{-10}\ m

Gamma ray waves

Wavelength = 10^{-12}\ m

So, it can be seen that the visible light makes up a relatively small part of the electromagnetic spectrum.

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A 100 kg person rides in an elevator moving at a constant speed of
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Explanation:

All this information only applies to the person. There is an extra tension force if we are talking about the elevator, but we are not. Dont forget to apply the units

Acceleration means change in speed or velocity. The elevator is moving at a constant speed of 3 meters. You wont even know you are moving because there is no change in acceleration. It equals 0

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1) Do liquids and gases also exert pressure?​
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A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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