First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed.
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use:
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
Answer:
ugmd = 1/2 kx²
d = (1/2 kx²) / (ugm)
= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)
= 7.4 m
ugmd = 1/2 mv²
v = √2ugd
= √(2(0.23)(9.81 m/s²)(7.4 m)
= 5.8 m/s
Explanation:
Answer:
i) 0.9504
ii) 0.0452
Explanation:
Given data: reliability of hydraulic brakes= 0.96
reliability of mechanical brakes = 0.99
So the probability of stopping the truck = 0.96×0.99= 0.9504
At low speed
case: A works and B does not
= 0.96×(1-0.99) = 0.0096
case2 : B works and A does not
= 0.99×(1-0.96) = 0.0396
Therefore, probality of stopping = 0.0096+0.0396 = 0.0492
Answer;
It's about acceleration, right?
Answer:
The centripetal acceleration of the car is 
.
Explanation:
Let the mass of the car, 
Diameter of the circular path, d = 100 m
Speed of car, v = 20 m/s
Radius, r = 50 m
When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :
So, the centripetal acceleration of the car is . Hence, this is the required solution.
