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3 years ago

If the refractive index of benzere is 2.419, what is the speed of light in benzene?

Physics

1 answer:

Grace [21]3 years ago

4 0

Answer:

$v=1.24\times 10^8\ m/s$

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

$n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s$

So, the speed of light in bezene is $1.24\times 10^8\ m/s$ .

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An electric motor does 900j of work for 8 hours. Calculate the power used

nataly862011 [7]

Answer:

0.031 W

Explanation:

The power used is equal to the rate of work done:

$P=\frac{W}{t}$

where

P is the power

W is the work done

t is the time taken to do the work W

In this problem, we have:

W = 900 J is the work done by the motor

t = 8 h is the time taken

We have to convert the time into SI units; keeping in mind that

1 hour = 3600 s

We have

$t=8\cdot 3600 =28,800 s$

And therefore, the power used is

$W=\frac{900}{28800}=0.031 W$

6 0

3 years ago

The total yearly world consumption of energy is approximately 4.0 Ã— 1020 J. How much mass would have to be completely converted

BabaBlast [244]

Answer:

m = 4.4 × 10³ kg

Explanation:

Given that:

The total yearly energy is 4.0 × 10²⁰ J

The amount of mass that provides this energy can be determined by using the formula:

E = mc²

where;

c = speed of light in free space = (3 × 10⁸)

4.0 × 10²⁰ = m × (3 × 10⁸)²

$m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}$

m = 4.4 × 10³ kg

6 0

3 years ago

A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer to the nearest tenth.The

Galina-37 [17]

The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s

7 0

3 years ago

A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t

andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = $\frac{r}{2}$

f = $\frac{0.40}{2} = 0.20 m$

Now,

m = - $\frac{v}{u}$

- 2 = - $\frac{v}{u}$

$\frac{v}{u}$ = 2 (2)

Now, by lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

v = $\frac{uf}{u - f}$ (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = $\frac{uf}{u - f}$

2 = $\frac{f}{u - f}$

2(u - 0.20) = 0.20

u = 0.30 m

6 0

3 years ago

The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee

lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is $-27.3\hat{i}$

y component of vector is $43.6\hat{j}$

so position vector is

$r=-27.3\hat{i}+43.6\hat{j}$

Magnitude of vector is

$|r|=\sqrt{27.3^2+43.6^2}$

$|r|=\sqrt{2646.25}$

|r|=51.44 units

Direction

$tan\theta =\frac{43.6}{-27.3}=-1.597$

vector is in 2nd quadrant thus

$180-\theta =57.94$

$\theta =122.06^{\circ}$

4 0

3 years ago