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2 years ago

The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?

1 answer:

Klio2033 [76]2 years ago

7 0

The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:
$n= \frac{c}{v}$
where the speed of light in vacuum is $c=3 \cdot 10^8 m/s$ . The speed of light in benzene is $v=2.00 \cdot 10^8 m/s$ , so we can use the previous relationship to find the refractive index of benzene:
$n= \frac{3 \cdot 10^8 m/s}{2.00 \cdot 10^8 m/s}=1.5$

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mash [69]

Answer:

$\Delta x = v_0 t + \frac{1}{2}at^2$

Explanation:

To solve this problem, we can use the following suvat equation:

$\Delta x = v_0 t + \frac{1}{2}at^2$

where

$\Delta x$ is the vertical displacement of the frog

$v_0$ is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from $v_0$ , all the other quantities are known. In fact:

$\Delta x =0.1 m$ is the vertical displacement

t = 2 s is the total time of flight

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

Therefore, solving for $v_0$ , we find the initial velocity of the frog:

$v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s$

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2 years ago

Nc-1 has the same dimension as

lisabon 2012 [21]

Answer:

Explanation:

The answer is electric field intensity. Electric field intensity is the force per unit positive charge which the charge exerts at any point.

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2 years ago

you are flying a kite that has two strings attached to either side of it you are holding those strings in your left and right ha

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The kite would move to the right as well

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Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant sp

Savatey [412]

km x h = km/h

First trial: 6 x 1 = 6km/h

Second trial: 9 x 2 = 18km/h

6 + 18 = <u>24km/h</u> (Total)

6 + 9 = 15 km

2 + 1 = 3h

15 + 3 = 18

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2 years ago

An object attached to an ideal massless spring is pulled across a frictionless surface. If the spring constant is 45 N/m and the

Nataly [62]

Answer:

The mass of the object is 49.5kg which is approximately 50kg

Explanation:

Given that

Spring constant (k)=45N/m

The extension (e)=0.88m

Also given that the acceleration is 0.8m/s²

Force by the spring is given as

Using hooke's law

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

F=ke

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a is the acceleration = 0.8m/s²

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k is the spring constant = 45N/m

F=45×0.88

F=39.6N

Now this force will set the object in motion, now using newton second law of motion

F=ma

Then, m=F/a

m=39.6/0.8

m=49.5kg

The mass of the object is 49.5kg which is approximately 50kg

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