A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from
the block? Speed is less than when in glass. Speed is less than before it entered glass. Speed is same as that in glass. Speed is same as that before it entered glass.
1 answer:
Answer:
Speed is same as that before it entered glass.
Explanation:
Given:
A light ray enters and passes through the glass as shown in the diagram.
We have to analyze its speed.
Speed of light in air is and speed of light in glass is
Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.
So speed of light will decrease in glass medium and again it passes to the air.
So
Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.
Finally
Speed is same as that before it entered glass as it in the same medium (air).
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Answer:
246.28 K
Explanation:
The total energy of one mole of gas molecules can be calculated by the formula given below
E =
Where R is gas constant and T is absolute temperature.
Put the value of R as 8.314 and temperature as 245 , we get
E =
= 3055.4 J
Add 16 j to it
Total energy of gas molecules = 3055.4 + 16 = 3071.4 J.
If T be the temperature after addition of energy then
= 3071.4
T =
T = 246.28 K
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: