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mr_godi [17]
3 years ago
8

The estimated distance of the earth from the sun is 149000000km. The speed of the light is 300000km/s. What is the time taken fo

r the light to travel from the sun to the earth
Physics
1 answer:
tester [92]3 years ago
6 0

Answer:

The time taken for the light to travel from the earth to the sun = 496.67 seconds

Explanation:

Estimated distance of the earth to the sun = 149000000km

Speed of light =  300000km/s

Time taken for the light to travel from the sun to the earth = ?

Speed = Distance/time

time = Distance/speed

time = 149000000/300000

time = 496.67 s

The time taken for the light to travel from the earth to the sun = 496.67 seconds

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Saturn’s rings are composed of __________.
sineoko [7]

Answer:

C

Explanation:

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3 years ago
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A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee
Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2}  = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\

<u>v = 7.69 x 10³ m/s</u>

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v =  \frac{2\pi r}{T}\\\\

where,

T = Time Period of Satellite = ?

Therefore,

T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\

<u>T = 5500 s = 91.67 min = 1.53 h</u>

7 0
3 years ago
You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
3 years ago
Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o
Ierofanga [76]
<span>C. Sample C would be best, because the percentage of the energy
in an incident wave that remains in a reflected wave from this material
is the smallest.

The coefficient of absorption is the percentage of incident sound
that's absorbed.  So the highest coefficient of absorption results in
the smallest </span><span>percentage of the energy in an incident wave that remains.
That's what you want. </span>
6 0
3 years ago
Kon'nichiwa~<br>please help me with this question!!​
andrezito [222]

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

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Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

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