A horizontal spring with spring constant 200N/m is compressed by 15cm and used to launch a 2kg box across a frictionless horizon
1 answer:
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Answer :
(a) The acceleration of the car is,
(b) The distance covered by the car is, 150 m
Explanation :
By the 1st equation of motion,
...........(1)
where,
v = final velocity = 5 m/s
u = initial velocity = 25 m/s
t = time = 10 s
a = acceleration of the car = ?
Now put all the given values in the above equation 1, we get:
The acceleration of the car is,
By the 2nd equation of motion,
...........(2)
where,
s = distance covered by the car = ?
u = initial velocity = 25 m/s
t = time = 10 s
a = acceleration of the car =
Now put all the given values in the above equation 2, we get:
By solving the term, we get:
The distance covered by the car is, 150 m
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: