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.
True. Being an electrical engineer, you learn very quickly that current will take All paths of resistance. But, the higher the resistance the lower the voltage. So, if a high resistance is shorted and the current flows through the short, there will be some small voltage across it, so some small amount of current will still flow through the high resistance.
Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve 
is 
.
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
Where
, 
- Initial and final position, respectively, measured in meters.
- Force as a function of position, measured in newtons.
Given that 
and the fact that 
when 
, the spring constant (
), measured in newtons per meter, is:
Now, the work function is obtained:
![W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%28250%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%5Cright%29%5Ccdot%20%5B%280.05%5C%2Cm%29%5E%7B2%7D-%280.00%5C%2Cm%29%5E%7B2%7D%5D)
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be . The area of the region enclosed by one loop of the curve is given by the following integral:
![A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta](https://tex.z-dn.net/?f=A%20%3D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_0%20%7B%5Br%28%5Ctheta%29%5D%5E%7B2%7D%7D%20%5C%2C%20d%5Ctheta)
By using trigonometrical identities, the integral is further simplified:
The area of the region enclosed by one loop of the curve is .
The correct answer would be the last one
Answer:
3.7 m/s
Explanation:
M = 444 kg
U = 5 m/s
m = 344 kg
u = - 5 m/s
Let the velocity of train is V and the car s v after the collision.
As the collision is elastic
By use of conservation of momentum
MU + mu = MV + mv
444 x 5 - 344 x 5 = 444 V + 344 v
500 = 444 V + 344 v
125 = 111 V + 86 v .... (1)
By using the formula of coefficient of restitution ( e = 1 for elastic collision)
-5 - 5 = V - v
V - v = - 10
v = V + 10
Substitute the value of v in equation (1)
125 = 111 V + 86 (V + 10)
125 = 197 V + 860
197 V = - 735
V = - 3.7 m/s
Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.
