Answer:
First tell what are the directions and what we've to do?? add subtract or multiply??
The frequency of the activity <span />
1) exhalation
2)neutralization reaction of antacids
3)washing clothes with detergents release a small amount of heat
4) degradation of biowaste
5) adding water to calcium oxide
Hope this answer helps you...
Answer:
197 hits/s
Explanation:
If the record is turning at the rate of 33 rev/min that means it's turning
![\omega = 33rev/min*2\pi rad/rev*\frac{1}{60}min/sec = 3.456 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%2033rev%2Fmin%2A2%5Cpi%20rad%2Frev%2A%5Cfrac%7B1%7D%7B60%7Dmin%2Fsec%20%3D%203.456%20rad%2Fs)
If the groove is being played at radius of 10 cm (or 0.1m), that means the record is rotating with the speed of:
![\omega*r = 3.456*0.1 = 0.346 m/s](https://tex.z-dn.net/?f=%5Comega%2Ar%20%3D%203.456%2A0.1%20%3D%200.346%20m%2Fs)
The goove bumps is uniformly separated at 1.75 mm(or 0.00175m/bump) bumps. So the number of hits per second it should be
![\frac{0.346}{0.00175} = 197.47 \approx 197 hits/s](https://tex.z-dn.net/?f=%5Cfrac%7B0.346%7D%7B0.00175%7D%20%3D%20197.47%20%5Capprox%20197%20hits%2Fs)
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)