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Andrej [43]
3 years ago
12

Yellow light (600 nm) passes through a

Physics
1 answer:
I am Lyosha [343]3 years ago
6 0

Answer:

19.284

Explanation:

The answer is right on Accellus.

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Similarities and differences between solar power and coal.
ira [324]
Solar power is renewable, which means it can be used again. However, coal is non-renewable, meaning that once it is used, it can't be used again.
3 0
3 years ago
Now, let's look at how the distance from the charge affects the magnitude of the electric field. Select Values on the menu, and
marin [14]

Electric field strength decreases as the distance from the source increases.

<h3>How the distance from the charge affects the magnitude of the electric field?</h3>

The strength of an electric field is inversely related to square of the distance from the source. This means that the electric field strength decreases when the distance from the source increases.

So we can conclude that Electric field strength decreases as the distance from the source increases.

Learn more about source here: brainly.com/question/25578076

#SPJ9

6 0
2 years ago
Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please) :D
kap26 [50]
12.could be D
13.A
14.D
15.C
16.A
3 0
3 years ago
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
ValentinkaMS [17]

Answer:

60 Ω

Explanation:

R(com) = 15 Ω

1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

1/15 = 1/20 + 1/R2

1/R2 = 1/15 - 1/20

1/R2 = (4 - 3) / 60

1/R2 = 1/60

R2 = 60 Ω

así, la combinada de resistencia necesaria es 60 Ω

5 0
3 years ago
A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on
muminat

Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

x=ucos\alpha t-(1/2)gt^{2} but the acceleration in the horizontal path is zero, hence we have

x=ucos\alpha t

Since the horizontal distance covered is 155m at 7.6secs, we have ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1

Also from the vertical path, the distance covered is expressed as

y=usin\alpha t-(1/2)gt^{2}

since the horizontal distance covered in 7.6secs is 195m, then we have

y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2

Hence if we divide both equation 1 and 2 we arrive at

\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\

Hence if we substitute the angle into the equation 1 we have

ucos72.02=20.38\\u=66.02m/s

Hence the initial velocity is 66.02m/s

3 0
4 years ago
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