Answer:
a) The magnitude of the speed of the ball at 2s is 19.62 m/s.
b) The ball has traveled 19.62 m after 2 seconds
c) The ball reaches the ground in 4.24 s.
Explanation:
a) The speed of the ball at 2 s after it was dropped is:
![v_{f} = v_{0} - gt](https://tex.z-dn.net/?f=%20v_%7Bf%7D%20%3D%20v_%7B0%7D%20-%20gt%20)
Where:
: is the final speed =?
: is the initial speed = 0 (it is dropped)
g: is the gravity = 9.81 m/s²
t: is the time = 2 s
![v_{f} = -9.81 m/s^{2}*2 s = -19.62 m/s](https://tex.z-dn.net/?f=%20v_%7Bf%7D%20%3D%20-9.81%20m%2Fs%5E%7B2%7D%2A2%20s%20%3D%20-19.62%20m%2Fs%20)
Then, the speed of the ball at 2s is -19.62 m/s. The minus sign is because the speed is in the negative direction (down).
b) The height at which is the ball after 2 seconds is:
![y_{f} = y_{0} + v_{0}t - \frac{1}{2}gt^{2}](https://tex.z-dn.net/?f=%20y_%7Bf%7D%20%3D%20y_%7B0%7D%20%2B%20v_%7B0%7Dt%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D%20)
Taking y₀ = 0 we have:
The ball has traveled 19.62 m after 2 seconds. The minus sign is because the height is in the negative direction.
c) The time at which the ball reaches the ground is:
Therefore, the ball reaches the ground in 4.24 s.
I hope it helps you!
Kinetic energy<span> increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the </span>stopping distance<span> should increase by a factor of four, assuming that the driver is </span>can<span> apply the brakes with sufficient precision to almost lock the brakes.</span>
17 Sin (45) in the vertical and 17 Cos
(45)
Im sorry i dont know but j hope you have a great day and you find your answers
Answer:
3. Wg is positive and WT is negative.
Explanation:
The work done by a force is given by:
![W=Fdcos \theta](https://tex.z-dn.net/?f=W%3DFdcos%20%5Ctheta)
where
F is the force
d is the displacement of the object
is the angle between the directions of F and d
This means that:
- When force and displacement are parallel,
, so the work done is positive
- When force and displacement are anti-parallel,
, so the work done is negative
In this case, the crane is moving downward. The force of gravity is also downward, while the tension in the cable is upward. so we have:
- Wg is positive, because gravity is parallel to the displacement
- Wt is negative, because the tension is opposite to the displacement