Answer:
<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>
Explanation:
This is definitely the correct answer because kinetic friction acts when an object is in motion and it allows the object to move without slipping, etc
<em>ALSO</em><em>,</em><em> </em><em>PLEASE DO</em><em> </em><em>MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>
<em>Bonne</em><em> </em><em>journée</em><em> </em><em>;</em><em>)</em><em> </em>
Answer:
230
Explanation:
= Rotational speed = 3600 rad/s
I = Moment of inertia = 6 kgm²
m = Mass of flywheel = 1500 kg
v = Velocity = 15 m/s
The kinetic energy of flywheel is given by

Energy used in one acceleration

Number of accelerations would be given by

So the number of complete accelerations is 230
P = mv
p = 3.5 × 5
p = 17.5 kg .m/s
Hope this helps!
Answer: 1 m/s
Explanation:
We have an object whose position
is given by a vector, where the components X and Y are identified by the unit vectors
and
(where each unit vector is defined to have a magnitude of exactly one):
![r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j](https://tex.z-dn.net/?f=r%3D%5B2%20m%20%2B%20%282%20m%2Fs%29%20t%5D%20i%20%2B%20%5B3%20m%20-%20%281%20m%2Fs%5E%7B2%7D%29t%5E%7B2%7D%5D%20j)
On the other hand, velocity is defined as the variation of the position in time:

This means we have to derive
:
![\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B2%20m%20%2B%20%282%20m%2Fs%29%20t%5D%20i%20%2B%20%5Cfrac%7Bd%7D%7Bdt%7D%5B3%20m%20-%20%281%20m%2Fs%5E%7B2%7D%29t%5E%7B2%7D%5D%20j)
This is the velocity vector
And when
the velocity vector is:

This is the velocity vector at 2 seconds
However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed
:


Finally:
This is the speed of the object at 2 seconds
Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m