Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
2 answers:
Answer:
2.83
Explanation:
Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:
(1)
with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):
(2)
Because in the right side of the equation (2) we have only constant quantities, that implies the ratio is constant for all the planets orbiting the same sun, so we can said that:
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Answer:
The displacement is
The distance is
Explanation:
From the question we are told that
The height from which the ball is dropped is
The height attained at the first bounce is
The height attained at the second bounce is
The height attained at the third bounce is
Note : When calculating displacement we consider the direction of motion
Generally given that upward is positive the total displacement of the ball is mathematically represented as
Here the 0 show that there was no bounce back to the point where Billy released the ball
=>
Generally the distance covered by the ball is mathematically represented as
The 2 shows that the ball traveled the height two times
=>
Answer:
The distance between the tip of her nose from the center of the crystal ball is 14.3 cm.
Explanation:
Given that,
Charge,
The end of her nose experiences an electric field strengths of 2200 N/C
It is required to find the distance between the tip of her nose from the center of the crystal ball. The formula of the electric field is given by :
r is distance
So, the distance between the tip of her nose from the center of the crystal ball is 14.3 cm.
Answer:
electricfield at (0,0,0) is Et = 2 k q / a²
Explanation:
For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity
For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation
E = k q / r²
Point (0,0,0)
We calculate for the charge -q which is at a distance R = a
E1 = k (-q) / a²
E1 = - kq / a²
As the test charge is positive in the field it goes to the left, attractive force
We calculate for the charge that is also at R = a
E2 = k q / a²
This field goes to the left, repulsive force
We find the total electric field
Et = E1 + E2
Et = kq / a² + kq / a²
Et = 2 k q / a²
Point (0,0, R)
We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a
E1 = k q / (R + a)²
E2 = kq / (R-a)²
Et = kq [1 / (R + a)² + 1 / (R-a)²]
Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}
Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}
If we use the condition that R> a we can despise in the patents "a"
(R² + a²) = R² (1+ a² / R²) ≈ R²
(R + a)² = R² (1 + a / R)² ≈ R²
(R- a)² = R² (1-a / R)² ≈ R²
Substituting in the total electric field
Et = kq {2 R²) / [R²R²]}
Et =kq 2 / R²
