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bonufazy [111]
3 years ago
7

The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun and Earth. Calculate the time taken

by Pluto to orbit the Sun in Earth days.
Physics
1 answer:
german3 years ago
7 0

Answer:

Explanation:

Given

Distance between Pluto and sun is 39.1 times more than the distance between earth and sun

According to Kepler's Law

T^2=kR^3

where k=constant

T=time period

R=Radius of orbit

Suppose R_1 is the radius of orbit of earth and sun

so Distance between Pluto and sun is R_2=39.1\cdot R_1

T_1 and T_2 is the time period corresponding to R_1 and R_2[/tex]

(T_1)^2=k(R_1)^3---1

(T_2)^2=k(R_2)^3---2

divide 1 and 2

(\frac{365}{T_2})^2=(\frac{R_1}{39.1})^3

T_2^2=365^2\times 39.1^3

T_2=89239.67\ Earth\ days                      

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The lowest possible temperature in outer space is 3.13 K. What is the rms speed of hydrogen molecules at this temperature? (The
Umnica [9.8K]

Answer:

v_{rms} =196.59 m/s

Given:

Temperature, T = 3.13 K

molar mass of molecular hydrogen, m = 2.02 g/mol = 2.02\times 10^{-3}kg/mol

Solution:

To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

v_{rms} = \sqrt{\frac{3TR}{m}}

where

R = rydberg's constant = 8.314 J/mol-K

Putting the values in the above formula:

v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}

v_{rms} =196.59 m/s

5 0
3 years ago
What is the speed of an electromagnetic wave with a frequency of 100 mhz as it travels through a vacuum?
SCORPION-xisa [38]

Answer:

The speed of the wave with a frequency 100 mhz will be  3\times 10^{8}m/sec

Explanation:

We have given that frequency of light is 100 mhz

We have to find the speed of light in vaccuum

We know that all electromagnetic waves travels in vaccum wth the same speed as the speed of light

And we know that speed of light is equal to 3\times 10^{8}m/sec

So the speed of the wave with a frequency 100 mhz will be  3\times 10^{8}m/sec

6 0
4 years ago
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
3 years ago
What is the speed of light​
Rainbow [258]
The speed of light is......299 792 458 m / s
3 0
3 years ago
Read 2 more answers
Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloo
bagirrra123 [75]
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span> =>v¹ = 2.25 m/s (East) </span>
5 0
3 years ago
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