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2 years ago

Analyze the relationship between total internal reflection and the critical angle.

Physics

1 answer:

padilas [110]2 years ago

8 0

Answer:

5.15 when the angle of incidence is equal to critical angle the angle of reflection is equal to 90 if the angle of incidence is bigger than the critical anger the reflected ray will not emerge from the medium but will be reflected back into the medium this is called total internet reflection

Explanation:

hope it's helpful

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Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit

MrMuchimi

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, $g=-9.81 m/s^2$ .

So, the final velocity is

$v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4$

$\Rightarrow v=-15.24$ m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

8 0

3 years ago

Which one of newton's laws does a doll riding a dog represent

strojnjashka [21]

I believe it would best represent Newton’s first law; an object tends to stay at rest and an object tends to stay in motion unless acted upon by an unbalanced force. When the dog stops walking, the doll will continue to go forward because there is no unbalanced force acting in it.

4 0

3 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s

velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2 = 2r

Centripetal force is given by

$F= \frac{mv^{2}}{r}$

Where v be the orbital velocity, which is given by

$v=\sqrt{gr}$

So, the centripetal force is given by

$F= \frac{mgr}}{r}}=mg$

where, g bet the acceleration due to gravity

$g=\frac{GM}{r^{2}}$

So, the centripetal force

$F= \frac{GMm}}{r^{2}}}$

Gravitational force on the satellite having larger orbit

$F= \frac{GMm}{4r^{2}}$ .... (1)

Gravitational force on the satellite having smaller orbit

$F'= \frac{GMm}{r^{2}}$ .... (2)

Comparing (1) and (2),

F' = 4 F

So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0

4 years ago

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

$\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}$

Put the value into the formula

$525=\sqrt{\dfrac{c+v}{c-v}}\times950$

$\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2$

$\dfrac{c+v}{c-v}=0.305$

$c+v=0.305\times(c-v)$

$v(1+0.305)=c(0.305-1)$

$v=\dfrac{0.305-1}{1+0.305}c$

$v=−0.532c$

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0

3 years ago