components of the speed of the coin is given as




now the time taken by the coin to reach the plate is given by



now in order to find the height



so it is placed at 1.52 m height
Answer:
<em>The distance the car traveled is 21.45 m</em>
Explanation:
<u>Motion With Constant Acceleration
</u>
It occurs when an object changes its velocity at the same rate thus the acceleration is constant.
The relation between the initial and final speeds is:
![v_f=v_o+at\qquad\qquad [1]](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat%5Cqquad%5Cqquad%20%5B1%5D)
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The distance traveled by the object is given by:
![\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3Dv_o.t%2B%5Cfrac%7Ba.t%5E2%7D%7B2%7D%5Cqquad%5Cqquad%20%5B2%5D)
Solving [1] for a:

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:


The distance is now calculated with [2]:

x = 21.45 m
The distance the car traveled is 21.45 m
Answer:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds
Answer:
As an object moves in a circle, it is constantly changing its direction. ... Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed.