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Tju [1.3M]
4 years ago
5

As light shines from air to another medium, i = 30.0o, the light bends toward the normal and refracts at 22.0o. What is the inde

x of refraction? n = _____
Physics
2 answers:
Art [367]4 years ago
8 0
The answer is 1.33 ...i think

iVinArrow [24]4 years ago
7 0
The answer is n = 1.3347
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3 years ago
An object is realsed From rest . how far does it fall during the second second of fall?​
Savatey [412]

Answer:

14.715 m

Explanation:

Assume that the acceleration due to gravity is 9.81 m/s^2 downwards, take downwards as positive

First second:

v = u + at

v = 9.81 m/s

Second second:

s = ut + (1/2)at^2

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3 years ago
PLEASE I NEED HELP IS DUE IN A FEW MINUTES
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3 years ago
An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether
Anna [14]

Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

I = \frac{P}{A}

where :

\frac{I}{c}= \frac{F}{A}

Replacing that into the above previous equation; we have:

\frac{P}{Ac}=\frac{F}{A}

F = \frac{P}{c}

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the  astronaut and a = its acceleration

we as well say;

\frac{P}{c} = ma

a = \frac{P}{mc}

Replacing P with 1000 W ; m with 80 kg and 3*10^{8} \  m/s for  c

Then; a = \frac{1000 \ W}{(80)(3.0*10^8)}

a = 4.2*10^{-8} \ m/s

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

v_1 = at_1

where ;

v_1 = final \ velocity

replacing t_1 = 1.0 \ h and a =  4.2*10^{-8} \ m/s; Then:

v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})

v_1 =  1.51 *10^{-4} \ m/s

Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:

To find the final distance traveled by the astronaut ;we have:

\Delta x_2 = d - \Delta x_1

where;

\Delta x_2 = the final distance

d = total distance

So;

\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m

The time taken to reach the final distance can be calculated as:

t_2 = \frac{\Delta x_2 }{v_1}

where;

t_2 = is the  time to  reach the final distance

Replacing 4.73 for {\Delta x_2 } and  1.51*10^{-4} m/s for v_1

t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}

t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )

t_2 = 8.8 \ h

We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to  reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.

Hence ; the time taken by the astronaut to reach safety = 9.8 hr

8 0
3 years ago
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