Answer:
Explanation:
The acceleration of gravity is 9.8m/s^2.
So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.
(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )
We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.
Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .
Answer - B. Betelguese.
I really hope this helps!!
Given a = 10 cm/s²
u = 0 cm/s
v = 50 cm/s
we know that
v²=u²+2aS
2500=2×10×S
2500÷20 = S
S= 125 cm
The ramp is 125 cm
Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
![\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]](https://tex.z-dn.net/?f=%5Cdelta%20p%20%3D%20m%20%5Csqrt%282g%20%2A%20y%2Fx%29%2A%20%5B%5Csqrt%20li%20%2B%20%5Csqrt%20lf%5D)
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
![1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]](https://tex.z-dn.net/?f=1.27%20%3D%200.4%5Csqrt%282%2A9.81%20%2A%2819%2F85%29%29%2A%20%5B%5Csqrt%201.9%20%2B%20%5Csqrt%20lf%5D)
final displacement lf = 0.39 m
Answer:
D
Explanation:
P=Work/Time
The rate at which work is done matches that.