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3 years ago

In which era did the universe’s clouds start to condense and the universe became transparent for the first time?

Physics

1 answer:

alukav5142 [94]3 years ago

8 0

Answer:

The Universe became transparent to the light left over from the Big Bang when it was roughly 380,000 years old

Explanation:

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Ocean breezes are due to which method of heat transfer?

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Ocean breezes are due to the convection method of heat transfer

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2 years ago

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What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance

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Answer:

Explanation:

it just a

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3 years ago

If a marble is released from a height of 10 meters how long would it take to hit the ground?

zlopas [31]

S = ut + 0.5at^2

10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} 

t^2 ~ 2.04 

t ~ 1.43 seconds

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3 years ago

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

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2 years ago

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A rock thrown with speed 8.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m bef

frez [133]

Draw a diagram to illustrate the problem as shown below.

The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s

Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s

The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m

Answer: The height is 21.7 m (nearest tenth)

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3 years ago