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3 years ago

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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Alexus [3.1K]

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

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) = 318.2 W/m^2

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 = 2.5 x 10^-4 J

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

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3 years ago

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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

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3 years ago

Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high

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2 years ago

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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 mass of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the lab cart

$v_{2}_{i}$ : is the initial velocity of the object = 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the lab cart

$v_{2}_{f}$ : is the final velocity of the object

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 will be the same, 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

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2 years ago

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Answer:

False

Explanation:

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

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