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UNO [17]
3 years ago
6

HELP ASAP!!

Physics
2 answers:
Arada [10]3 years ago
7 0
Percentage error:

1.55 – 1.53 ÷ 1.53
0.02 ÷ 1.53
.013 x 100
1.3 % error

I hope this is right.
Elis [28]3 years ago
3 0

Let's deal with the question one piece at a time:

-- "What value did you calculate for the index of refraction of the glass block in Part 2 ?" . . . . . We were not present for the calculation in Part-2, and have no idea what was going on in that part.

-- "How does your value compare to the accepted value of 1.53 ?" . . . . .  Since we were not present for Part-2 and didn't do any calculation, we have nothing to compare with the accepted value of 1.53 .

-- "Calculate the percentage error." . . . . . Since we have no record of a calculation pursuant to Part 2, and are unable to compare it to any other number, we have no way to describe any relationship between them.

-- "Different materials have distinct indexes of refraction. Explain how you might identify a material based on experiments like this one." . . . . . We have no description of the protocol or results of the experiment, so we have no way to explain anything based on it.

-- "Mention at least one of the difficulties in identifying materials based on their indexes of refraction." . . . . . One such drawback is the difficulty of obtaining a sample of the material that is sufficiently, sufficiently homogeneous, and of a useful size and shape, so that a beam of light can be directed through the sample, and accurate representative measurements made of the sample's influence on the light.

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
3 years ago
I need a correct answer plzzzzzzzzzzzzzzzzzzzz
olya-2409 [2.1K]

Answer:

option 1

Explanation:

i just used the SOH CAH TOA, and since the given is tan=opposite/adjacent, that should be the answer

4 0
3 years ago
Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high
konstantin123 [22]
  • L=8m
  • B=4m
  • H=300cm=3m

\\ \bull\tt\longmapsto Volume=LBH

\\ \bull\tt\longmapsto Volume=8(4)(3)

\\ \bull\tt\longmapsto Volume=96m^3

6 0
2 years ago
Read 2 more answers
A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m
lbvjy [14]

The equation 15v_{i} + 2*0 = (15 + 2)v_{f} (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.  

The horizontal momentum is given by:

p_{i} = p_{f}

m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}

Where:

  • m₁: is the mass of the lab cart = 15 kg
  • m₂: is the <em>mass </em>of the object dropped = 2 kg
  • v_{1}_{i}: is the initial velocity of the<em> lab cart </em>
  • v_{2}_{i}: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
  • v_{1}_{f}: is the final velocity of the<em> lab cart </em>
  • v_{2}_{f}: is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

15v_{1}_{i} + 2*0 = v_{f}(15 + 2)

Therefore, the equation 15v_{i} + 2*0 = (15 + 2)v_{f} represents the horizontal momentum (option 3).

Learn more about linear momentum here:

  • brainly.com/question/2141713?referrer=searchResults
  • brainly.com/question/2400186?referrer=searchResults

I hope it helps you!            

4 0
2 years ago
A current cannot produce a magnetic field. *<br><br> True or false
GalinKa [24]

Answer:

False

Explanation:

False, as a magnetic field is generated whenever current travels through a conductor.

An electromagnet consists of a coil of wire wrapped around a bar of iron. The coil and iron bar get magnetized when electric current flows through the wire. An electromagnet also has north and south magnetic poles. The magnetic field is strongest at either pole of the magnet.

8 0
3 years ago
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