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ollegr [7]
3 years ago
6

What is the orbital period in years of a planet with a semi major axis of 35 au

Physics
1 answer:
mezya [45]3 years ago
6 0

Answer:

Orbital period of the planet will be 207.06 year                      

Explanation:

We have given the planet have the semi major axis as 35 au

We have to find the orbital period of the planet

From Keplar's third  law there is relation between the orbital period and semi major axis which is t T^2=R^3

So T^2=35^3

T^2=42875

T=207.06year

So orbital period of the planet will be 207.06 year

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
3 years ago
Based on the data, which prediction should he expect to occur? A2 repels B1. C2 attracts B2. B1 repels C1. A1 attracts C2.
VladimirAG [237]
I believe the answer is a1
6 0
3 years ago
Who was the carthaginian general who used elephants to cross the alps in the second punic war
goldfiish [28.3K]
<span>the answer is Hannibal</span>
3 0
3 years ago
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a car accelerates from rest at 3.6m/s^2. how much time does it need to attain a speed of 7 m/s? awnser in units of s.
Ivahew [28]
Initial velocity: 0
final velocity: 7 m/s
a = 3.6
t = ?
x = ?

(7-0)/3.6 = t
t = 1.94 s
6 0
3 years ago
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A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

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3 years ago
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