Answer:
The speed of the ball when it hits the ground is 102.1 m/s
Explanation:
Given;
initial velocity of ball, u = 25 m/s
distance traveled by the ball = height of the building = h = 500 m
when the ball hits the ground, the final velocity, v = ?
The final velocity of the ball is given by;
v² = u² + 2gh
where;
g is acceleration due to gravity = 9.8 m/s²
v² = (25)² + 2(9.8)(500)
v² = 10425
v = √10425
v = 102.1 m/s
Therefore, the speed of the ball when it hits the ground is 102.1 m/s
Answer:
the answer is A.) -1 * 10^3[N]
Explanation:
The solution consists of two steps, the first step is using the following kinematic equation:
![v=v_{i} +a*t\\where:\\v=final velocity [m/s]\\v_{i}=initial velocity [m/s]\\a=acceleration[m/^2]\\t=time[s]\\](https://tex.z-dn.net/?f=v%3Dv_%7Bi%7D%20%2Ba%2At%5C%5Cwhere%3A%5C%5Cv%3Dfinal%20velocity%20%5Bm%2Fs%5D%5C%5Cv_%7Bi%7D%3Dinitial%20velocity%20%5Bm%2Fs%5D%5C%5Ca%3Dacceleration%5Bm%2F%5E2%5D%5C%5Ct%3Dtime%5Bs%5D%5C%5C)
The initial velocity is 10 [m/s], and the final velocity is zero because the car stops in 0.5[s].
Replacing:
![0=10+a*(0.5)\\a=-20[m/s^2]](https://tex.z-dn.net/?f=0%3D10%2Ba%2A%280.5%29%5C%5Ca%3D-20%5Bm%2Fs%5E2%5D)
Now in the second part, we need to use the second law of Newton, this law relates the forces with the acceleration of a body.
In the moment when the car stops suddenly the driver will feel the force of the seatbelt acting in the opposite direction of the movement.
![F=m*a\\F=50[kg]*(-20[m/s^2])\\units\[kg]*[m/s^2]=[N]\\F=-1000[N] or -1*10^{3} [N]](https://tex.z-dn.net/?f=F%3Dm%2Aa%5C%5CF%3D50%5Bkg%5D%2A%28-20%5Bm%2Fs%5E2%5D%29%5C%5Cunits%5C%5Bkg%5D%2A%5Bm%2Fs%5E2%5D%3D%5BN%5D%5C%5CF%3D-1000%5BN%5D%20or%20-1%2A10%5E%7B3%7D%20%5BN%5D)
The minus sign means that the force is acting in the opposite direction of the movement.
3. 2 meters per second 4. i think object 7. i’ll try to figure it out 8. .77 9. 25km 10. 10m each second
Answer:
715 N
Explanation:
Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.
Let g = 9.8 m/s2
Gravity and equalized normal force is:
N = P = mg = 107*9.8 = 1048.6 N
Kinetic friction force and equalized tension force on the rope is

Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = 
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid