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aleksandr82 [10.1K]
2 years ago
14

Sabiendo que el indice de refracción dela gua es 1,33 calcula el ángulo de refracción resultante para un rayo de luz que incide

sobre una piscina con un ángulo de incidencia de 50º
Physics
1 answer:
Kaylis [27]2 years ago
3 0

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.

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2. A mass-spring system oscillates with a frequency of 20 Hz. What is the period?
Alex_Xolod [135]

Answer:

0.05s

Explanation:

f=20Hz

we know,

time period=t=1/f

t=1/20

=0.05s

5 0
3 years ago
A car travels 240km in 4h what’s the cars velocity
Tresset [83]

The car's speed is 240km/4hr= 60km/hr.

There's not enough information given in the question to determine its velocity.

6 0
3 years ago
Read 2 more answers
A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
3 years ago
20. Consider a model steel bridge that is 1/100 the exact scale of the real bridge that is to be built. a. If the model bridge w
Veseljchak [2.6K]
The model bridge captures all the structural attributes of the real bridge, at a reduced scale.

Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.

Because the model bridge weighs 50 N, the real bridge weighs
(50 N)*10^6 = 50 MN.

Part b.
The model bridge matches the structural characteristics of the actual bridge.
Therefore the real bridge will not sag either.
6 0
3 years ago
PLEASE HURRY 20 PTS
Crank

Answer:

A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

The electric force exerted on a charge by an electric field is given by:

where

F is the force

q is the charge

E is the electric field

We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.

5 0
3 years ago
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