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aleksandr82 [10.1K]
3 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]3 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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A certain pendulum with a 1.00 kg bob has a period of 3.50 s. What will happen to the period of the pendulum if the 1.00 kg bob
Serggg [28]

Answer:

The period will be doublef because increasing the mass increases the period linearly.

Explanation:

A stiffer spring with a constant mass decreases the period of oscillation. Increasing the mass increases the period of oscillation

8 0
3 years ago
I have an object that I want to know when it was made. I calculate that the object contains 0.12g of Iodine-131 but started with
Sav [38]

Answer: 71.72 days

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

A=A_{o}.2^{\frac{-t}{h}} (1)  

Where:  

A=0.12 g is the final amount of Iodine-131

A_{o}=60 g is the initial amount of Iodine-131

t is the time elapsed  

h=8 days is the half life of Iodine-131

Knowing this, let's substitute the values and find t from (1):

0.12 g=(60 g)2^{\frac{-t}{8 days}} (2)  

\frac{0.12 g}{60g}=2^{\frac{-t}{8 days}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.12 g}{60g})=ln(2^{\frac{-t}{8 days}}) (4)  

-6.21=-\frac{t}{8 days}ln(2) (5)

Finding t:

t=71.72 days

7 0
4 years ago
Can someone please answer this pleaseee!!!!!!!!
Andru [333]

Answer:

With a slower speed-perhaps 5 cm/s answer is c

5 0
3 years ago
A 16 kg mass suspended from a light spring is replaced by a 4 kg mass. What factor changes the frequency of the oscillation? (a)
AnnZ [28]

Answer:

Frequency change by a factor of 2.

(b) is correct option.

Explanation:

Given that,

Mass = 16 kg

Replaced mass = 4 kg

We need to calculate the frequency

Using formula of frequency  

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Put the value into the formula

f_{1}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{16}}

f_{1}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{4}}....(I)

f_{2}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{4}}

f_{2}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{2}}...(II)

f_{2}=2f_{1}

Hence, Frequency change by a factor of 2.

5 0
3 years ago
A cow is given some corn as part of its diet. The corn mostly contains complex carbohydrates like starch. What will happen to th
belka [17]

Answer:

The corn is mixed with saliva in the mouth, travels through the oesophagus to the rumen where it is fermented by enzymes from microorganisms in the rumen and they are converted to acids which are absorbed from the wall of the rumen into the liver where they are used as a source of energy for the cow.

Explanation:

The digestion in cows differs from the digestion in humans.

Cows are know as RUMINANTS.

The digestive system in cows are divided into 5 parts.

a) The mouth

b) The oesophagus

c) The stomach, which is divided into 4 compartments:

i. Rumen

ii Recticulum

iii Omasum

iv Abomasum (“true stomach”)

d. The small intestine

e. The large intestine

For ruminants like cows, they don't chew the food in their mouth so much, just a little and then they swallow.

The half chewed corn mixes with saliva in the mouth and it is swallowed. From there it travels through the oesophagus and enters the first compartment of the stomach which is the Rumen.

The digestion of carbohydrates such as corn in ruminant animals occurs in the rumen through a process known as microbial fermentation.

Microorganisms that can be found in the rumen of a cow are: bacteria, protozoa and or fungi. These microorganisms break down the carbohydrates. The starch in which is the major molecule in the corn is broken down into fatty acids which are called volatile fatty acids. Examples include: Acetic acid, Butyric acid.

These acids are transferred to the liver of the cow through a portal vein in the wall of the rumen using the process of absorption where they serve as a source of energy for the cow.

5 0
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