Speed = (distance covered) / (time to cover the distance) .
Distance covered = (mark at the end) minus (mark at the beginning)
= (64m - 4m) = 60 meters .
Speed = (60 meters) / (30 seconds) = 2 meters/second .
We don't care about the bridge or the other two elephants.
They don't have any effect on the question or the answer.
the most appropriate answer is A !! our mind automatically connects everything and thus make a story that we don't even familiar with in actual !!
Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here
Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)
Answer:
Resultant Force=33.8 lb
Angle=67.2°
Explanation:
Given data
Fa=22 lb
Fb=16 lb
Θ=55⁰
To find
(i) Resultant Force F
(ii)Angle α
Solution
First we need to represent the forces in vector form
![\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20F_%7B1%7D%3D22j%5C%5C%20F_%7B2%7D%3Du%2Bv%5C%5CF_%7B2%7D%3D16sin%2855%29i%2B16cos%2855%29j%5C%5CF_%7B2%7D%3D16%280.82%29i%2B16%280.5735%29j%5C%5CF_%7B2%7D%3D13.12i%2B9.176j)
Total Force
![F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j](https://tex.z-dn.net/?f=F%3DF_%7B1%7D%2BF_%7B2%7D%5C%5C%20F_%7B2%7D%3D22j%2B13.12i%2B9.176j%5C%5CF_%7B2%7D%3D13.12i%2B31.176j)
The Resultant Force is given as
![|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb](https://tex.z-dn.net/?f=%7CF%7C%3D%5Csqrt%7Bx%5E%7B2%7D%20%2By%5E%7B2%7D%20%7D%5C%5C%7CF%7C%3D%5Csqrt%7B%2813.12%29%5E%7B2%7D%20%2B%2831.176%29%5E%7B2%7D%20%7D%5C%5C%20%7CF%7C%3D33.8lb)
For(ii) angle
We can find the angle bu using tanα=y/x
So
![tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}](https://tex.z-dn.net/?f=tan%5Calpha%20%3D%5Cfrac%7B31.176%7D%7B13.12%7D%5C%5C%20%5Calpha%20%3Dtan%5E%7B-1%7D%20%28%5Cfrac%7B31.176%7D%7B13.12%7D%29%5C%5C%5Calpha%20%3D67.2%5E%7Bo%7D)
To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration
![\alpha = \frac{\omega_f - \omega_0}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%20-%20%5Comega_0%7D%7Bt%7D)
Here
= Angular acceleration
Initial and final angular velocity
There is not initial angular velocity,then
![\alpha = \frac{\omega_f}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%7D%7Bt%7D)
We know that the relation between the tangential velocity with the angular velocity is given by,
![v = r\omega](https://tex.z-dn.net/?f=v%20%3D%20r%5Comega)
Here,
r = Radius
= Angular velocity,
Rearranging to find the angular velocity
![\omega = \frac{v}{r}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7Br%7D%7D)
Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,
![\alpha = \frac{30}{0.20*6}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B30%7D%7B0.20%2A6%7D)
![\alpha = 25 rad /s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%2025%20rad%20%2Fs%5E2)
Therefore teh average angular acceleration of each wheel is ![25rad/s^2](https://tex.z-dn.net/?f=25rad%2Fs%5E2)
It will take "Exact 75 years", 'cause light takes 1 year for 1 light years.
Hope this helps!