Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m
The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²
The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s
The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s
Answer: 12.5 m/s
The answer
2y + 14 = 17
The 17 is to the right of the = sign
It is also the answer
<span>x=((12.3/100)m)cos[(1.26s^−1)t]
v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
v=<span>
<span>-0.13261622 m/s
</span></span>the object moving at 0.13 m/s <span>at time t=0.815 s</span>
Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by
where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by
for n=0
for n=2
Thus there is one additional resonance corresponding to n=1 , between 
and 
