Answer:
d) is the same as when it started from rest
Explanation:
using equation of motion
v = u + at
second law of momentum defines
F = ma
a = F /m
the equation becomes
v = u + (F/m)t
from hear
since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.
Can have any number of exceptions as long as we know about them. Hope this helps!!! Brainleist Please!!
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
We will use this equation:
s = 1/2*a*t^2 + v0*t + s0
where:
s = space traveled
a = acceleration
t = time
v0 = initial speed
s0 = initial space
In this case::
v0 = 0
s0 = 0
So our equation will look like that now:
s = 1/2 * a * t^2
let's calculate the acceleration first of all:
a = (vf - vi) / t
where vf is the final speed and vi is the initial speed. t is the time.
a = (25m/s) / 10s = 2.5 m/s^2
Now we can calculate the space:
s = 1/2 * (2.5 m/s^2) * (10s)^2 = 125m
---
Hope it was helpful! Have a great day.
