Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Answer:
t = 2.2 s
Explanation:
Given that,
Height of the roof, h = 24.15 m
The initial velocity of the pumpkin, u = 0
We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :
Here, u = 0 and a = g
So, it will take 2.22 s for the pumpkin to hit the ground.
Answer:
<u>Conventions used in SI to indicate units are as follows:</u>
- Only singular form of units are used. for example: use kg and not kgs.
- Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
- Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
- Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
- While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
- Units named after a scientist should be written in small letters. for example: newton, henry.
- Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k
Answer:
619.8 N
Explanation:
The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:
where
T is the tension
m is the mass of the rock
v is the speed
r is the radius of the circular path
At the beginning,
T = 50.4 N
v = 21.1 m/s
r = 2.51 m
So we can use the equation to find the mass of the rock:
Later, the radius of the string is decreased to
r' = 1.22 m
While the speed is increased to
v' = 51.6 m/s
Substituting these new data into the equation, we find the tension at which the string breaks:
