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4 years ago

14

Find the angle of refraction of a ray of light that enters a diamond (n=2.419) from air at an angle of 18.0° to the normal. ONLY

keep 1 decimal in your answer

1 answer:

Tpy6a [65]4 years ago

3 0

Answer:

n

Explanation:

hb

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It took 3 seconds for an object that was thrown up with velocity vo from

otez555 [7]

Answer:

ik u got it bc this was 2 weeks ago

Explanation:

but yes and yuea

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3 years ago

The density of aluminum is 2.70 g/ml. A piece of aluminum foil has a mass of 44 g. What is the volume of this piece of aluminum

anyanavicka [17]

Answer:

C) 16.3 ml

Explanation:

Density is equal to the ratio between the mass of an object and its volume:

$d=\frac{m}{V}$

where

m is the mass

V is the volume

In our problem, we know:

- density of aluminium: $d=2.70 g/mL$

- mass of the aluminium foil: $m=44 g$

So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

$V=\frac{m}{d}=\frac{44 g}{2.70 g/mL}=16.3 mL$

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4 years ago

A car was moving a 14 m/s. After 30 seconds, it’s speed increased to 20 m/s. What was its acceleration during this time?

Anuta_ua [19.1K]

Answer:

acceleration = 0.2 m/s/s

Explanation:

initial velocity u = 14

final velocity v = 20

time = 30

acceleration = ?

v = u + at

20 = 14 + 30a

30a = 6

a = 0.2 m/s/s

5 0

3 years ago

Read 2 more answers

Which plate is covering most of two continents ? what two continents?

ololo11 [35]

The plate that is covering most of the two continents is the Eurasian plate.
The continents are <span><span>
1. </span><span> Europe</span></span> <span><span>
2. </span><span> Asia</span></span>
The Eurasian plate is a tectonic plate that amasses the whole continent of what others call as “Eruasia”. It takes part of the oceanic crust that starts from the Mid-Atlantic Ridge and extends itself to northward of the Gakkel Ridge. The size of this plate is for about 67,800,000 km according to statistics and geography.

3 0

3 years ago

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An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in

asambeis [7]

Answer:

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Explanation:

$v= (19.0j+18.k)km/s$

$a=3.0x10^{13}m/s^2$

$\beta= 400x10{-6}T$

Electron information needed to solve the question:

$m_e=9.11x10^{-31}kg$

$q=-1.6x10{-19}C$

$F=F_E+F_B=q*(E+Vx \beta)$

$F=m*a$

$m*a=q*(E+Vx\beta)$

$E=\frac{m*a}{q}-(Vx\beta )$

$E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]$

$E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C$

$E=(-17.08 i +7.2 j -7.6 k )N/C$

6 0

3 years ago