<span>When the question says the ball lands a distance of 235 meters from the release point, we can assume this means the horizontal distance is 235 meters.
Let's calculate the time for the ball to fall 235 meters to the ground.
y = (1/2)gt^2
t^2 = 2y / g
t = sqrt{ 2y / g }
t = sqrt{ (2) (235 m) / (9.81 m/s^2) }
t = 6.9217 s
We can use the time t to find the horizontal speed.
v = d / t
v = 235 m / 6.9217 s
v = 33.95 m/s
Since the horizontal speed is the speed of the plane, the speed of the plane is 33.95 m/s</span>
Answer:
The correct choice is A. Yes, the more exercise the better.
I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :


W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.