It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
Frequency, f = 3.73Hz
Explanation:
The frequency of a simple harmonic 6is given by:
f = w/2pi
But w= Sqrt( k/m)
Where k is the spring constant
And m is the mass
Given:
Mass=0.20kg
Spring constant, k=130N/m
w= Sqrt(130/0.20)
w= Sqrt(650)
w= 25.50 m
Frequency, f = w/2pi
f = 25.50/(2×3.142)
f = 3.73Hz
Answer:
Video
Explanation:
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