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fomenos
4 years ago
13

If the oil thickness is 200 nm and the index is 1.5, what (vacuum) wavelength (in nm) does a diver see looking at the same posit

ion as the scientist, but from under water?
Physics
1 answer:
gavmur [86]4 years ago
6 0

Answer: 133.33nm

Explanation: refractive index, n = real depth/apparent depth

Oil thickness= real depth = 200nm

Wavelength from under the water = apparent depth = a

n = real depth/a

1.5 = 200/ a

a = 200/ 1.5

a = 133.33nm

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What effect does the distance between Earth and the Sun have on Earth’s seasons?
antiseptic1488 [7]

Answer:

So actually, it is not so much the distance at a particular time that affects the seasons. It is a common misconception. It is in-fact the earths tilt. When we rotate around the sun, due to our tilt, the North and South of earth have different seasons. This is because when light is shining on to earth, because of Earths tilt, on one half the light shines more directly and on the other it scatters thus the different temperatures, causing season. Hope this helps.

4 0
3 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
Read 2 more answers
We observe the wheels of buses and trucks are heavier than the wheels of cars and scooters why.​
scoray [572]

Answer:

they carry a heavier load

Explanation:

4 0
2 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
4 years ago
Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter
kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

T_e=2\pi \sqrt{\frac{L}{g_e}}

where

g_e = 9.8 m/s^2 is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

T_m=2\pi \sqrt{\frac{L}{g_m}}

where

g_m = 1.6 m/s^2 is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0
3 years ago
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