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3 years ago

A ray in benzene has a critical angle of 41.8 deg when trying to enter air. What is the index of refraction for benzene?

Physics

1 answer:

ziro4ka [17]3 years ago

3 0

Answer:

1.5

Explanation:

Data obtained from the question include:

C (critical angle) = 41.8°

n (refractive index od benzene) =?

Using the formula n = 1/sineC, the refractive index of benzene can be achieved as shown below:

n = 1/sineC

n = 1/sine 41.8°

n = 1.5

Therefore, the refractive index of benzene is 1.5

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A student has connected two generator/motors together. The student turns the crank 20 times and noticed the other crank only tur

Rufina [12.5K]

Answer:

eu não consigo entender essa situação o Fim dessa história só pode ter confusão dos iguais com cabeças diferentes a invés de amigas são concorrentes

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3 years ago

A 5.0 kg wooden block Is 100 meters from the ground. What is the gravitational potential energy for the block?

iren [92.7K]

Answer:

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 5 × 9.8 × 100

We have the final answer as

Hope this helps you

7 0

3 years ago

Two cylinders with the same mass density rC = 713 kg / m3 are floating in a container of water (with mass density rW = 1025 kg /

cluponka [151]

Answer:

$\dfrac{h_2}{h_1}=\dfrac{1}{2}$

Explanation:

Lets take h is height of cylinder immersed in the water

We know that for floating body

$h=\dfrac{\rho_cL}{\rho_l}$

Where $\rho_c$ density of cylinder

$\rho_l$ density of water

For both cylinder fluid is same also density of cylinders are also same

$\dfrac{h_1}{L_1}=\dfrac{h_2}{L_2}$

$\dfrac{h_1}{h_2}=\dfrac{L_1}{L_2}$

$\dfrac{h_1}{h_2}=\dfrac{20}{10}$

$\dfrac{h_2}{h_1}=\dfrac{1}{2}$

3 0

3 years ago

The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string

Katyanochek1 [597]

Answer:

$\dfrac{\mu_A}{\mu_G}=0.197$

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

$v_A = \sqrt{\dfrac{F_A}{\mu_A}}$ .......(1)

speed for string G

$v_G = \sqrt{\dfrac{F_G}{\mu_G}}$ ........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}$

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{\mu_A}{\mu_G}=0.197$

5 0

3 years ago

1. A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ba

MariettaO [177]

Momentum = (mass) x (velocity)

Original momentum before the hit =

   (0.16 kg) x (38 m/s) this way <==

   =    6.08 kg-m/s this way <==

Momentum after the hit =

   (0.16) x (44 m/s) that way ==>

   = 7.04 kg-m/s that way ==>

Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .
-----------------------------------------------

Change in momentum = impulse.

   Impulse = (force) x (time the force lasted)

13.12 kg-m/s = (force) x (0.002 sec)

(13.12 kg-m/s) / (0.002 sec) = Force

   6,560 kg-m/s² = 6,560 Newtons = Force

    ( about 1,475 pounds ! ! ! )

8 0

3 years ago