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Alex
3 years ago
14

If the index of refraction of a lens is 1.5, how fast does light travel in the lens?

Physics
2 answers:
lina2011 [118]3 years ago
7 0

Answer:

C.

Explanation:

Len [333]3 years ago
5 0
The speed of light in a material is given by:
v= \frac{c}{n}
where
c=3 \cdot 10^8 m/s is the speed of light in vacuum
n is the refractive index of the material

The lens in this problem has a refractive index of n=1.50, therefore the speed of light in the lens is
v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.50}=2 \cdot 10^8 m/s

And the correct answer is C).
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A 1 kg particle moves upward from the origin to (23) m. Wit is work done by the force of gravity which is in - y direction s B.
sveta [45]

Answer:

Explanation:

mass, m = 1 kg

Position (2, 3 ) m

height, h = 2 m

acceleration due to gravity, g = 9.8 m/s^2

Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.

Force  mass x acceleration due to gravity

F = 1 x 9.8 = 9.8 N

Work = force x displacement x CosФ

Where, Ф be the angle between force vector and the displacement vector.

Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward

So, W = 9.8 x 2 x Cos 180°

W = - 19.6 J

Thus, option (A) is correct.

4 0
3 years ago
Sally goes on a hike. The distance she walks compared to the amount of time she walks is shown on the graph. Which statement is
Neko [114]
According to the task there should be the graph that supports Sally's hike, but after looking on the options it seems that Sally doesn't walks at a constant rate and there is the negative option that coincides with my thoughts. So, I bet the false statement is the third option represented in the scale above.
5 0
3 years ago
An astronaut drops a rock from the top of a crater on the moon. When the rock is halfway down to the bottom of the crater, its s
Alexxx [7]

Answer: vf1/vf2= 1/ sqrt(2)

Explanation :on the moon no drag force so we have only the  force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics

if the rock travels H to the bottom we can calculate velocity:

vo=0m/s (drops the rock)  , yo=0

vf*vf= vo*vo+2g(y-yo)

when the rock is halfway  y = H/2 so:

vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)

when the rock reach the bottom y=H so:

vf2*vf2=2*g*H so vf2 = sqrt(2gH)

so vf1/vf2= 1/ sqrt(2)

good luck from colombia

8 0
3 years ago
Read 2 more answers
You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building.
katen-ka-za [31]

Answer:

22.505 seconds

Explanation:

V =19.8m/s

V = a*to

t1 = 19.8/3.3

= 6seconds

Distance travelled during acceleration

= 1/2 x 3.3 x 6²

= 59.4m

X_total = x1 + x2

X2 = 373-59.4

X2 = 313.6m

t2 = x2/v

= 313.6/19.8

= 16.505

Total = 16.505 + 6

= 22.505 seconds

the minimum time in which an elevator can travel the 373 m from the ground floor is 22.505 seconds.

7 0
3 years ago
An electric motor rotating a workshop grinding wheel at 1.06 102 rev/min is switched off. Assume the wheel has a constant negati
kvasek [131]

Answer:

t = 106π / 30*2.1

Explanation:

w_{i} = 1.06*10^{2}    => 106

    => 106 x 2π/60

    => 106/30π

∝ = -2.1 rad/sec²

w_{f} => 0

w_{f} = w_{i} + ∝t

∴ (w_{f} - w_{i}) / ∝ = t

t = 106π / 30*2.1

6 0
3 years ago
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