There are actually two different kinds of mirrors, and the answer is different
for each one.
-- Plain old everyday hand mirror, vanity mirror, bathroom mirror, makeup
mirror, etc.
Opaque, reflecting silver coating is on the back of the glass.
Light from your tongue or your teeth flows to the front surface of the glass,
through the glass, out of the back surface of the glass, bounces off of the silver
coating on the back, reverses its direction, enters the back surface of the glass,
comes back through the glass again, leaves the front of the glass, goes into your
eyes, and you can see your teeth or your tongue.
Both surfaces of the glass, as well as the glass in between the surfaces, are
transparent. The silver coating on the back is opaque. I know that, because
when I look at the back of a mirror, I can't see any light coming through it.
The coating on the back is also reflective ... a big part of the reason why
a mirror works.
-- Expensive mirrors used by astronomers and eye-doctors.
Known as "first surface" mirrors.
Opaque, reflecting silver coating is on the <em>front</em> of the glass.
Light
from your tongue or your teeth flows toward the front surface of the glass,
but never actually gets there. It bounces off of
the silver coating on the front of
the glass, reverses its direction, goes into your eyes, and you can see your teeth
or
your tongue.
The glass is transparent, but that doesn't matter, because the light never reaches
the glass. It only goes as far as the opaque silver coating on the front, and is
reflected from there.
Answer:
Explanation:
Tension provides centripetal force in the circular motion . In circular motion work done by force = torque x angle
torque is zero as , centripetal force passes through axis of rotation that is center.
So work done by centripetal force = 0
So work done by tension on M = 0
Answer:
ΔL = L0 C ΔT
We need to find C the constant of expansivity
C = ΔL / (L0 ΔT)
C = .96 / (15.04 * 65) = 9.82 * 10^-4 / deg C
Answer:
a)
a = 2 [m/s^2]
b)
a = 1.6 [m/s^2]
c)
xt = 2100 [m]
Explanation:
In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.
a)
When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:
where:
Vf = final velocity = 40 [m/s]
Vi = initial velocity = 0 (starting from rest)
a = acceleration [m/s^2]
t = time = 20 [s]
40 = 0 + (a*20)
a = 2 [m/s^2]
The distance can be calculates as follows:
where:
x1 = distance [m]
40^2 = 0 + (2*2*x1)
x1 = 400 [m]
Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.
v = x2/t2
where:
x2 = distance [m]
t2 = 30 [s]
x2 = 40*30
x2 = 1200 [m]
b)
Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.
Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.
0 = 40 - (a *25)
a = 40/25
a = 1.6 [m/s^2]
The distance can be calculates as follows:
0 = (40^2) - (2*1.6*x3)
x3 = 500 [m]
c)
Now we sum all the distances calculated:
xt = x1 + x2 + x3
xt = 400 + 1200 + 500
xt = 2100 [m]
