I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.
If that's the real question, then the answer is that the period of Orbit-A
is about<em> 5.2</em> times as long as the period of Orbit-C .
Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.
This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.
Answer:
F = 183.153 N
Explanation:
given,
mass of the toothpick = 0.12 g = 0.00012 kg
initial velocity = 227 m/s
final velocity = 0 m/s
penetration depth = 16 mm = 0.016 m
using the equation of motion
v² - u² = 2 a s
0 - u² = 2 a s
- 221² = 2 × a × 0.016
a = 1526281.25 m/s²
Force is equal to
F = m a
= 0.00012 × 1526281.25
F = 183.153 N
(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
