What is your operational definition for a fast reaction time?
1 answer:
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Answer:
9 cm
-36 cm
Explanation:
u = Object distance
v = Image distance
f = Focal length = 12
m = Magnification = 4
Lens equation
Object distance is 9 cm
Image distance is -36 cm (other side of object)
, where
the mass of the planet, and
the universal gravitation constant.
Explanation:
Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.
<em>Equation 1</em> (see below) relates net force the object experiences, to its orbit velocity
and its mass
required for it to stay in orbit :
<em>(equation 1)</em>
The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force acting on the soccer ball shall equal to its weight,
where
the gravitational acceleration constant. Thus
<em>(equation 2)</em>
Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for ; note how the mass of the soccer ball,
, cancels out:
<em>(equation 3)</em>
<em>Equation 4 </em> gives the value of gravitational acceleration, , a point of negligible mass experiences at a distance
from a planet of mass
(assuming no other stellar object were present)
<em>(equation 4)</em>
where the universal gravitation <em>constant</em>
Thus