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netineya [11]
3 years ago
15

The drawings show two examples in which a ray of light is refracted at the interface between two liquids. In each example the in

cident ray is in liquid
A and strikes the interface at the same angle of incidence. In one case the ray is refracted into liquid
B, and in the other it is refracted into liquid
C. The dashed lines denote the normals to the interfaces.
D.Rank the indices of refraction of the three liquids in descending order (largest first).
Physics
1 answer:
Nataly [62]3 years ago
6 0

Answer:

Possible options:

A. nA, nC, nB

B. nA, nB, nC

C. nC, nB, nA

D. nC, nA, nB

E. nB, nA, nC

Answer is D

Explanation:

When the light is refracted into liquid B it is bent away from the normal, so that nA > nB. When the light is refracted into liquid C it is bent toward the normal, so that nC > nA. Therefore, we conclude that nC > nA > nB

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4 years ago
a racing car undergoes a uniform acceleration of 4.00m/s2. if the net force causing the acceleration is 3.00 times 10^3 N, what
yKpoI14uk [10]

Answer: 750Kg

Explanation:

Recall that force is the product of the mass M, of an object moving at a uniform acceleration.

i.e Force = Mass x Acceleration

In this case, Mass = ?

Force = 3.00 x 10^3 N = (3.00 x 1000N)

= 3000N

Uniform acceleration = 4.00m/s^2

Force = Mass x Acceleration

3000N = Mass x 4.00m/s^2

Mass = (3000N/4.00m/s^2)

Mass = 750Kg (The SI unit of mass is kilograms)

Thus, the mass of the car is 750Kg

4 0
3 years ago
A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?
MakcuM [25]

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2.  This means that you have to multiply 5 by 5.  5 x 5= 25.  

Edit: Also, because the surface is frictionless, it will make the object go faster too.  Nothing can really slow it down unless something blocks it.

8 0
3 years ago
Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th
lapo4ka [179]

Answer:

T_f=5.0116^{\circ}C

Explanation:

Given:

  • mass of water, m_w=0.25\ kg
  • initial temperature of water, T_i_w=20^{\circ}C
  • initial temperature of pan, T_i_p=173^{\circ}C
  • mass of pan, m_p=0.6\ kg
  • mass of water evapourated, m_v=0.03\ kg
  • specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}
  • specific heat of aluminium pan, c_a=900\ J.kg^{-1}.K^{-1}
  • latent heat of vapourization, L=2256000\ J.kg^{-1}

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})

0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)

T_f=5.0116^{\circ}C

5 0
3 years ago
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