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Len [333]
3 years ago
15

A light ray in medium 1 striking a boundary at an angle of theta Subscript 1 Baseline. In medium 2 is a second light ray that is

at a larger angle from the normal with angle theta Subscript 2 Baseline. Medium 1 has the index of refraction of n Subscript 1 Baseline and in medium 2 has the index of refraction of n Subscript 2 Baseline.
Which medium (1 or 2) is more dense?
Medium

Physics
1 answer:
notsponge [240]3 years ago
6 0

Answer:

The correct answer is 1

Explanation:

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A cube of gold weights1.89N when suspended in air from a spring scale. When suspended in molasses, it appears to weight 1.76 N.
Eddi Din [679]

smaller depending on friction and aiming of the gold cube

3 0
3 years ago
A cylindrical container closed of both end has a radius of 7cm and height of 6cm A.)find the total surface area of the container
Natalka [10]

Given:

A cylindrical container closed of both end has a radius of 7cm and height of 6cm.

Explanation:

A.) Find the total surface area of the container.

  • A = 2πrh + 2πr²
  • A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
  • A = 263.76 + 307.72
  • A = 571.48

B.) Find the volume of the container.

  • V = πr²h
  • V = (3.14)(7×7)(6)
  • V = 923.16

Not sure huhuness.

#CarryOnLearning

8 0
3 years ago
Read 2 more answers
If planet A is three times as far from planet C, then the period of its orbit will be __ times as long
liubo4ka [24]
I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.

If that's the real question, then the answer is that the period of Orbit-A
is about<em>  5.2</em>  times as long as the period of Orbit-C .

Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.

This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.


6 0
3 years ago
What is 16.558 m/s rounded to three significant figures?
In-s [12.5K]

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

3 0
4 years ago
A highway patrolman traveling at the speed limit is passed by a car going 20 mph faster than the speed limit. After one minute,
Oksi-84 [34.3K]

Answer:

It will take 40 seconds to catch the speeding car

Explanation:

Initial speed of the patrolman = 55 mph

after one minute speed of patrol man = 115 mph

now acceleration of patrolman is given by

a = \frac{v_f - v_i}{t}

a = \frac{115 - 55}{1/60} = 3600 m/h^2

now at this acceleration the distance covered by patrolman in "t" time is given as

d = v_i t + \frac{1}{2}at^2

d = 55t + 1800t^2

now we know the speed of the speeding car is given as

v' = (55+20) mph

now in the same time distance covered by it

d = 75 t

now since the distance covered is same

75 t = 55t + 1800 t^2

t = \frac{20}{1800} = \frac{1}{90} h

t = 40 seconds

7 0
4 years ago
Read 2 more answers
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