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lbvjy [14]
3 years ago
8

Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through

?
A. index of refraction
B.index of reflection
C.angle of reflection
D.angle of incidence
Physics
2 answers:
Eddi Din [679]3 years ago
7 0
<span>The ratio of the speed of light in a vacuum to the
speed of light in the material it is passing through
is the index of refraction of that material. (A)</span>
melisa1 [442]3 years ago
6 0

Answer:

index of refraction

Explanation:

A number that represents how refractive a medium; a ratio of the speed of light in a vacuum to the spee of light in a medium (n=c/v). The larger the index of refraction, the slower light travels.

-the larger the index of refraction, the more refractive it is

-materials with a larger refractive index will bend light more

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A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
uranmaximum [27]

Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

7 0
3 years ago
Relative density has no unit .why ?​
Klio2033 [76]

Answer:

Relative density is a ratio of the density of a certain material to the density of a reference material. Relative density has no unit, because the units of the material in question cancel out with the units of the reference material. For example, say a cooking oil’s density is 0.9 g/cm^3.

Explanation:

7 0
3 years ago
n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

3 0
3 years ago
Read 2 more answers
When would the carrying capacity of an area be most likely to change? Choose the correct answer.
pickupchik [31]

Answer:

when resources remain the same

Explanation:

The capacity of an area will change if the available resources remain the same, especially if the population is changing.

8 0
3 years ago
Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 20° above the horizontal. Ball 2 has twi
Nutka1998 [239]

Answer:

E) d_{2} = 4d_{_{1}}

Explanation:

We know that the range of a projectile is given by:

R=\frac{u^2.sin\ \2\theta}{g}

where:

u = initial velocity of the projectile

\theta = angle of projection of projectile

<u>Now, range for ball 1 :</u>

d_1=\frac{u^2.sin\ \2\theta}{g}

<u>Now, range for ball 2 with double velocity and all the other parameters being the same:</u>

d_2=\frac{(2u)^2.sin\ \2\theta}{g}

d_2=4\times \frac{u^2.sin\ \2\theta}{g}

d_2=4\times d_1

5 0
3 years ago
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