What is the velocity of all electromagnetic waves in a vacuum?
1 answer:
Answer:
Explanation:
All electromagnetic waves travel in a vacuum at the same speed, regardless of their frequency. The magnitude of their velocity is
This value is one of the universal constant and it is called speed of light.
According to Einstein's theory of special relativity, the value of c is the same for all inertial frames (it means that we measure always the same value of c in a vacuum, even if we are moving with respect to the light).
However, the speed of the electromagnetic waves decreases as they move through a medium. In particular, their speed decreases according to the equation:
where n is called index of refraction of the medium.
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The final speed of the block after the collision with the obstacle is .
Further Explanation:
Given:
The mass of the block is .
The initial momentum of the block is .
The impulse imparted by the obstacle is .
Concept:
The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.
According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.
The expression for the impulse momentum theorem is.
…… (1)
Substitute for
and
for
in equation (1).
The final momentum of the block can be expressed as:
…… (2)
Substitute for
and
for
in equation (2).
Thus, the final speed of the block after the collision with the obstacle is .
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Answer Details:
Grade: High School
Chapter: Impulse-momentum theorem
Subject: Physics
Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
