Q3. (a) 0m/s, as they are asking for initial velocit.
(b)(i) The paper has a large surface area or weighs less than the coin,thus,falls smoothly.
(ii)The coin has more mass than the paper.
(c) They fall at the same acceleration and hit the bottom at the same time.
Their mass doesn't matter in the vacuum.
Answer:
Explanation:
<h3>Given Data:</h3>
Mass = m = 68 kg
Velocity = v = 30 m/s
Time = 2 hours = 2 × 60 × 60 = 7200 s
<h3>Required:</h3>
Force = F = ?
<h3>Formula to be used:</h3>
<h3>Solution:</h3>
![\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%20%3D%20%5Cfrac%7B%2868%29%2830%29%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B2040%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%200.28%20N%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Answer:90N
Explanation:
Mass=30kg
Centripetal acceleration=3m/s^2
centripetal force=mass x centripetal acceleration
Centripetal force=30 x 3
Centripetal force =90
Centripetal force =90N
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.
D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .
-- They don't change by the same factor, because 1/g is inside
the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
