The approximate orbital period of this star is 13 years.
<h3>What is Kepler's third law?</h3>
The square of a planet's period of revolution around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, states Kepler's law of periods.
T² ∝ a³
The time it takes for one rotation to complete depends on how closely the planet orbits the sun. With the use of the equations for Newton's theories of motion and gravitation, Kepler's third law assumes a more comprehensive shape:
P² = 4π² /[G(M₁+ M₂)] × a³
where M₁ and M₂ are the two circling objects' respective masses in solar masses.
Learn more about Kepler's third law here:
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Answer:
The burden distance is 7 ft
Solution:
As per the question:
Specific gravity of package emulsion, 
Specific gravity of diabase rock, 
Diameter of the packaged sticks, d = 3 in
Now,
To calculate the first trail shot burden distance, B:
![B = [\frac{2SG_{E}}{SG_{R}} + 1.5]\times d](https://tex.z-dn.net/?f=B%20%3D%20%5B%5Cfrac%7B2SG_%7BE%7D%7D%7BSG_%7BR%7D%7D%20%2B%201.5%5D%5Ctimes%20d)
![B = [\frac{2\times 1.25}{2.76} + 1.5]\times 3 = 7.22](https://tex.z-dn.net/?f=B%20%3D%20%5B%5Cfrac%7B2%5Ctimes%201.25%7D%7B2.76%7D%20%2B%201.5%5D%5Ctimes%203%20%3D%207.22)
B = 7 ft
Answer: 6.12 kg
Explanation:
Since Mass of ball = ? (let the unknown value be Z)
Acceleration due to gravity, g= 9.8m/s^2
Height, h = 1.5 metres
Gravitational potential energy GPE = 90J
Gravitational potential energy depends on the weight of the ball, the action of gravity and height.
Thus, GPE = Mass m x Acceleration due to gravity g x Height h
90J = Z x 9.8m/s^2 x 1.5m
90 = Z x 14.7
Z = 90/14.7
Z = 6.12 kg
Thus, the bowling ball weigh 6.12 kilograms
Answer:
by moving your hands up and down your we're creating transverse wave which travel in a right angle shape towards your friend
Answer:
Explanation:
The distance travelled in the free fall is H - h
Since the apple originally started from rest we can use v^2 = u^2 + 2 x g x s where v is the final velocity, g the accln due to gravity and s the distance travelled and u is the initial velocity = 0
So the velocity just before it enters the grass is sq rt [2 x g x (H - h)]
Once in the grass, it slows down at a constant rate which means that the acceleration (a) during this period is constant.
So once again using the same formula we have v = O and u = sq rt[2 x g x (H-h)]
so since v^2 = u^2 + 2 x a x s then
O^2 = 2 x g x (H-h) + 2 x a x h
{O^2 - 2 x g x (H - h)}/(2 x h) = a
