Answer:
Incomplete question
Check attachment for the given diagram
Explanation:
Given that,
Initial Velocity of drum
u=3m/s
Distance travelled before coming to rest is 6m
Since it comes to rest, then, the final velocity is 0m/s
v=3m/s
Using equation of motion to calculate the linear acceleration or tangential acceleration
v²=u²+2as
0²=3²+2×a×6
0=9+12a
12a=-9
Then, a=-9/12
a=-0.75m/s²
The negative sign shows that the cylinder is decelerating.
Then, a=0.75m/s²
So, using the relationship between linear acceleration and angular acceleration.
a=αr
Where
a is linear acceleration
α is angular acceleration
And r is radius
α=a/r
From the diagram r=250mm=0.25m
Then,
α=0.75/0.25
α =3rad/sec²
The angular acceleration is =3rad/s²
b. Time take to come to rest
Using equation of motion
v=u+at
0=3-0.75t
0.75t=3
Then, t=3/0.75
t=4 secs
The time take to come to rest is 4s
Because the air inside the tires is kept at high pressure.
In fact, the force applied by the tires upwards to counter-balance the weight of the car (pushing downwards) is

where p is the pressure of the air inside the tires and A is the area of contact between the tire and the car. Therefore, a higher pressure means a larger force F, and eventually if the pressure p is higher enough the force F will be large enough to counterbalance the weight of the car.
Answer:
Consider the diagram. We are effectively being asked to prove that $\alpha=i_1$, for any value of $i_1$. Now, from trigonometry,
Explanation:
We can use kinematics here if we assume a constant acceleration (not realistic, but they want a single value answer, so it's implied). We know final velocity, vf, is 1.0 m/s, and we cover a distance, d, of 0.47mm or 0.00047 m (1m = 1000mm for conversion). We also can assume that the flea's initial velocity, vi, is 0 at the beginning of its jump. Using the equation vf^2 = vi^2 + 2ad, we can solve for our acceleration, a. Like so: a = (vf^2 - vi^2)/2d = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, not bad for a flea!