Answer:
3.7 m
Explanation:
ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky
The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of
t = √ (2h/g) = √(2(15)/9.8) = 1.75 s
Without the tail wind, the ball travels horizontally
d = vt = 68(1.75) = 119 m
The tailwind exerts a constant acceleration on the ball of
a = F/m = 12/5.0 = 2.4 m/s²
The average horizontal velocity during the flight is
v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s
so the distance with tailwind is
d = v(avg)t = 70.1(1.75) = 122.675 m
The extra distance is 122.675 - 119 = 3.675 = 3.7 m
Answer:
A) 37 m
Explanation:
The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:
(1)
where
v = 0 m/s is the final velocity of the car
u = 24 m/s is the initial velocity
a is the acceleration
d is the length of the skid
We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

where
m = 1000 kg is the mass of the car
is the coefficient of friction
a is the deceleration of the car
g = 9.8 m/s^2 is the acceleration due to gravity
The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

And we can substitute it into eq.(1) to find d:

Impulse is the integral of a force, F.
Hope this helps.
(Please mark this brainliest, I would really appreciate it) Thanks!
A. a tsunami
(if the earthquake is hitting the ocean, the water will get effected)