You said that she's losing 1.9 m/s of her speed every second.
So it'll take
(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)
to lose all of her initial speed, and stop.
<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
Answer:
- A vibrating object
- a medium to travel
HOPE IT HELPS :)
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Answer:
Explanation:
Gauge pressure at the bottom of the cylinder depends on the height of water in the cylinder
So here we can say that
now when liquid is filled to height "h" in base area "A" then gauge pressure of the liquid at the bottom is given as
now we put the whole liquid into another cylinder with twice radius of the first cylinder
So area becomes 4 times
now by volume conservation we can say that if area is increased by 4 times then height of liquid will decrease by 4 times
so we have
so gauge pressure is given as
For the scale od the paraboliczne curviczne Ur
