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sergij07 [2.7K]
3 years ago
5

Define Refraction and give some knowlegde about it

Physics
1 answer:
shepuryov [24]3 years ago
7 0
Refraction is the change in direction of a wave, caused by the change in the wave's speed. Examples of waves include sound waves and light waves. Refraction is seen most often when a wave passes from one transparent medium to another transparent medium. Different types of medium include air and water. When a wave passes from one transparent medium to another transparent medium, the wave will change its speed and its direction. For example, when a light wave travels through air and then passes into water, the wave will slow and change direction.
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Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0
3 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
3 years ago
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are
adelina 88 [10]

Answer:

The final temperature of the two objects is the same.

Explanation:

The expression for the heat energy in terms of mass, specific heat and the change in the temperature is as follows:

Q=mc(T_{f} -T_{i})

Here, Q is the heat energy, m is the mass of the object, c is the specific heat and T_{f},T_{i} are the final temperature and initial temperature.

According to the given question, Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are added to each object.

Q=mc(T_{f}-T_{i})                                             ............(1)

Q=mc(T'_{f}-T_{i})                                            .............(2)

From (1) and (2),

mc(T_{f}-T_{i})=mc(T'_{f}-T_{i})

T_{f}-T_{i}=T'_{f}-T_{i}

T_{f}=T'_{f}

Therefore, the final temperature of the two objects is the same.

6 0
3 years ago
Which type of heat transfer takes place in a vacuum? (3 points)
tiny-mole [99]
Radiation is the only transfer in a vacuum
3 0
3 years ago
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