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pickupchik [31]
3 years ago
6

A light wave encounters a partial physical barrier, such as a wall with a hole in it. What is MOST LIKELY to occur?

Physics
2 answers:
poizon [28]3 years ago
8 0
The light will be diffracted. 
diffraction - bending of a wave in a homogeneous medium around obstacles or through a narrow opening.

hope this helps :)


tatiyna3 years ago
5 0
Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).
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Sabiendo que el indice de refracción dela gua es 1,33 calcula el ángulo de refracción resultante para un rayo de luz que incide
Kaylis [27]

Answer:

35.16 degrees

Explanation:

Knowing that the index of refraction of the guide is 1.33, calculate the resulting angle of refraction for a ray of light that falls on a pool with an angle of incidence of 50º

Refractive index, n = 1.33

The angle of incidence, i = 50°

We need to find the angle of refraction. let it is r. It can be calculated using Snells law as follows:

n=\dfrac{\sin i}{\sin r}\\\\\sin r=\dfrac{\sin i}{n}\\\\r=\sin^{-1}(\dfrac{\sin i}{n})\\\\r=\sin^{-1}(\dfrac{\sin 50}{1.33})\\\\r=35.16^{\circ}

So, the angle of refraction is 35.16 degrees.

3 0
2 years ago
In 1865, jules verne proposed sending men to the moon by firing a space capsule from a 220-m-long cannon with final speed of 10.
Aneli [31]
I'm going to assume that the 10.97 km/s is the velocity at the end of the cannon.

vi = 0
d = 220m
vf = 10.97 km/s = 10970 m/s
a = ???

vf^2 = vi^2 + 2*a*d
10970^2 = 0 + 2*a*220
10970^2 / 440 = a You are right this is a pretty big number.
2.73 * 10^5 = a


5 0
3 years ago
Which direction will thermal energy flow if you pick up a snowball with your bare hand? Thermal energy will flow from the snowba
Lapatulllka [165]

Answer:

b. Thermal energy will flow from your hand to the snowball.

Explanation:

4 0
3 years ago
Read 2 more answers
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
2 years ago
A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87° from the direction of th
77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, v=8.2\times 10^3\ m/s

Angle between velocity and the magnetic field, \theta=87

Charge, q=5.7\ \mu C=5.7\times 10^{-6}\ C

Magnetic force, F = 0.002 N

The magnetic force is given by :

F=qvB\ sin\theta

B is the magnetic field  

B=\dfrac{F}{qv\ sin\theta}

B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0
3 years ago
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