Answer:
Keeping the speed fixed and decreasing the radius by a factor of 4
Explanation:
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :
We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"
It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,
R' = R/4
New centripetal acceleration will be,
So, the centripetal acceleration of the ball can be increased by a factor of 4.
The mass is still 10 kg. But instead of weighing 98N as it does on Earth, it weighs 245N on Jupiter.
(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
<span>If you have only two
data from two recording stations then you will be having a hard time finding
the correct location of the epicenter. This is because triangulation method requires
3 recording station. If you have 2 recording station, the 2 circles will
intersect at 2 points giving you 2 locations that could possibly be the
epicenter.</span>
