If he leaves the ramp with a speed of 30.5 m/s and has a speed of 28.7 m/s at the top of his trajectory, determine his maximum h
1 answer:
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Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem
Now
The intensity at O when both speakers are on is given by
Here
- I is the intensity at O when both speakers are on which is given as 6
- I1 is the intensity of one speaker on which is 6
- δ is the Path difference which is given as
- λ is wavelength which is given as
Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.
where k=0,1,2
for minimum frequency , k=1
So the minimum frequency is 702.22 Hz
Answer:
a=0 v = v₀ + a t
a=0 line is horizontal
Explanation:
1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration
2, speed and relationship of a car is given by
v = v₀ + a t
where vo is the initial velocity, a is the acceleration and tel time
in this case I will calcograph velocity vs. time the constant acceleration is a straight line.
In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope
