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lidiya [134]
3 years ago
5

The sun is 1.5 × 108 km from Earth. The index of refraction for water is 1.349. How much longer would it take light from the sun

to reach Earth if the space between them were filled with water rather than a vacuum?
Physics
1 answer:
tankabanditka [31]3 years ago
4 0

Answer:

175s

Explanation:

time it takes sunlight to reach the earth in  vacuum

C=light speed=299792458m/s

X=1.5x10^8km=1.5x10^11m

c=X/t

T1=X/c

T1=1.5X10^11/299792458=500.34s

time it takes sunlight to reach the earth in  water:

First we calculate the speed of light in water taking into account the refractive index

Cw=299792458m/s/1.349=222233104.5m/s

T2=1.5x10^11/222233104.5m/s=675s

additional time it would take for the light to reach the earth

ΔT=T2-T1=675-500=175s

You might be interested in
One speaker generates sound waves with amplitude A.
raketka [301]

Answer:

iv) It is 9x bigger than before

Explanation:

As the amplitudes of the new speakers add directly with the original one, taking into account the phase that they have, the composed amplitude of the sound wave is as follows:

At = A + 4A -2A = 3 A

The intensity of the wave, assuming it propagates evenly in all directions, is constant at a given distance from the source, and can be expressed as follows:

I = P/A

where P= Power of the wave source, A= Area (for a point source, is equal to the surface area of a sphere of radius r, where is r is the distance to the source along a straight line)

For a sinusoidal wave, the power is proportional to the square of the amplitude, so the intensity is proportional to the square of the amplitude also.

If the amplitude changes increasing three times, the change in intensity will be proportional to the square of the change in amplitude, i.e., it will be 9 times bigger.

So, the statement iv) is the right one.

7 0
3 years ago
What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?
Assoli18 [71]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

4 0
3 years ago
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
3 years ago
During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61
bonufazy [111]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

7 0
3 years ago
In which situations is elastic potential energy present? Check all that apply. A bow is drawn back from its equilibrium position
Simora [160]
1,4,6
a bow is drawn back
a gun is loaded w/ a dart
a bungee cord is stretched
6 0
3 years ago
Read 2 more answers
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