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1 year ago

In a vacuum all electromagnetic waves have the same?

1 answer:

7 0

Electromagnetic waves do not need a material medium for its movement. They all travel at the same speed in a vacuum

What is Electromagnetic waves ?

Electromagnetic waves are those that do not require a material medium for their propagation. They are transverse waves which made up of electric and magnetic waves

In a vacuum all electromagnetic waves have the same speed. That is, all types of electromagnetic waves travel at the same speed in a vacuum which is equal to 299792458 m/s

Examples are;

radio wave

infrared

ultraviolet

x - rays

Gamma ray

visible light

microwave

Therefore, in a vacuum, all electromagnetic waves have the same speed.

Learn more about Waves here: brainly.com/question/25699025

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when a tuning fork vibrates over an open pipe and the air in the pipe starts to vibrate, the vibrations in the tube are caused b

malfutka [58]

Sound waves....................................

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4 years ago

Newton's _____law states that when one object exerts a force on a second object, the second object always exerts a force back on

Nina [5.8K]

<h2>Answer:Third</h2>

Explanation:

Newton's third law states that every action force has a equal and opposite reaction force.

Example 1:

Consider yourself hitting a wall with your hand,you are exerting a force on the wall but still you feel hurt because the wall exerts same force on your hand in opposite reaction.

Example 2:

Consider yourself standing on ground.You are exerting force on ground.But how are you able to stand on ground even if your mass is forcing you to go down?The answer is normal reaction force.The ground exerts a force on your feet which makes you to stand still.

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3 years ago

If a lever lifts a load four times the effort applied and effort distance is 5 times the load distance, calculate its efficiency

noname [10]

Answer:

If effort distance was 4 times, efficiency would be 100%.

Since it takes 5 times for effort distance, efficiency drops to output/input

output is 1*F

input is (1/4*F)*5

so: F/1/5*F/4 = 4F/5F = .8 or 80%

8 0

3 years ago

How fast must a 0.142-kg baseball move in order to have the same momentum as the bus?

Otrada [13]

Momentum is the product of mass and velocity of a body. In this case, the bus has a mass of 18200 kg and moving with a velocity of 10 m/s.
Therefore, the momentum of the bus will be, 18200×10 = 182,000 kgm/s.
Thus a ball of mass 0.142 kg will require to travel with a speed of;
= 182000/0.142
= 1.282 ×10∧6 m/s to have the same momentum with the bus.

6 0

4 years ago

A ladder whose length L is 12 m and mass m is 45 kg rests against a wall. Its upper end is a distance h of 9.3 m above the groun

amm1812

Answer:

FN₁ = 1146.6N : Force exerted on the ladder by the floor , vertical and upward

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

Explanation:

The equilibrium equation are:

∑Fx=0

∑Fy=0

∑M = 0

M = F*d

Where:

∑M : Algebraic sum of moments

M : moment ( N*m)

F : Force ( N)

d :Perpendicular distance of the force to the point ( meters )

Data

m =45 kg : mass of the ladder

M =72 kg : mass of the fire fighter

g = 9.8 m/s²: acceleration due to gravity

L = 12 m : ladder length

h = 9.3 m: ladder height

L/3 = 12/3 = 4m Location of the center of mass of the ladder of the way up

L/2 = 12/2 = 6m Location of the center of mass of the fire fighter

µ = 0 : coefficient of friction between the ladder and the wall

θ : angle that makes the ladder with the floor

sinθ = h/L = 9.3 m/12 m

θ =sin⁻¹( 9.3 / 12)

θ = 50.8°

Forces acting on the ladder

W₁ =m*g = 45 kg* 9.8 m/s² = 441 N: Weight of the ladder (vertical downward)

W₂ =M*g = 72 kg * 9.8 m/s² = 705.6 N : Weight of the fire fighter(vertical downward)

FN₁ :Normal force that the floor exerts on the ladder (vertical upward) (point A)

fs : friction force that the floor exerts on the ladder (horizontal and opposite the movement )(point A)

FN₂ : Normal Force that the wall exerts on the ladder ( horizontal and opposite to friction force between the floor and the ladder)

∑Fy=0

FN₁ -W₁ -W₂= 0

FN₁ = W₁ + W₂

FN₁ = 441N+ 705.6N

FN₁ = 1146.6N : Force exerted on the ladder by the wall (vertical and upward)

Calculation of the distances of the forces at the point A (contact point of the ladder on the floor)

d₁ = 4*cos 50.8° (m) = 2.53 m: Distance from W₁ to the point A

d₂ =6*cos 50.8° (m)= 3.79 m : Distance from W₂ to the point A

d₃ = 9.3 m : Distance from FN₂ to the point A

The equilibrium equation of the moments at the point A (contact point of the ladder with the floor)

∑MA = 0

FN₂(d₃) - W₁( d₁) - W₂(d₂) = 0

FN₂(d₃) = W₁(d₁) + W₂(d₂)

FN₂(9.3) = (441 )(2.53) + (705.6)( 3.79 )

FN₂(9.3) = 1115.73 + 2674.2

FN₂ = (3790) / (9.3)

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

8 0

3 years ago