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Blababa [14]
3 years ago
8

Light traveling through air at 3.00 · 10^8 m/s reaches an unknown medium and slows down to 2.00 · 10^8 m/s. What is the index of

refraction of that medium? n =
Physics
2 answers:
nataly862011 [7]3 years ago
5 0
V 1= 3.00 · 10^8 m/s
v 2 = 2.00 · 10^8 m/s
The index of refraction:
n = v 1 / v 2 = 3.00 · 10^8 m/s : 2.00 · 10^8 m/s = 1.5
Answer:
The index of refraction of that medium is 1.5
lesya692 [45]3 years ago
4 0

Answer

The index of refraction of  medium is 1.5 .

Explanation:

Equation for  index of refraction is given by .

n = \frac{c}{v}

Where n is index of refraction in medium , v is speed of light in the medium and c is the speed of light in air .

As given

Light traveling through air at 3.00\times 10^8 m/s reaches an unknown medium and slows down to 3.00\times 10^8 m/s.

c = 3.00\times 10^{8}

v = 2.00\times 10^{8}

Putting the values in the formula

n = \frac{3.00\times 10^{8}}{2.00\times 10^{8}}

Solving the above

n = 1.5

Therefore the index of refraction of  medium is 1.5 .


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The weight should be shared between the two string equally. Therefore, tension in each string, T is;

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A 60kg bicyclist (including the bicycle) is pedaling to the
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a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

3 0
3 years ago
Which of your groups should you NOT change anything for? (in other
ella [17]

Answer:

B

Explanation:

The control is something that is meant to not be changed, the control is a comparison of the experimental.

7 0
3 years ago
A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st
wel

Answer:

the tension T2 when the rock is completely immersed is T2 =  29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where  ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V=  T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced  

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V  =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

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3 0
3 years ago
A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

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Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

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           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
3 years ago
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