Answer:
V = 2.87 m/s
Explanation:
The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.
Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:
Centripetal acceleration = V^2 / r
where r is the distance of water from the pivot or shoulder.
For our case, r will be 0.65 + 0.19 = 0.84 m
and solving the above equation we get:
9.81 = V^2 / 0.84
V^2 = 8.2404
V = 2.87 m/s
The train would need the greatest amount of force due to weight! If you think of it, a baseball won't need much force to stop it, but if you have a heavy train, it will need excessive force to stop the train. The answer would be #3
I hope this answer helps!
Sorry if it doesn't make sense, as I don't know that much about physics! I am just thinking of what makes sense.
Answer:
Explanation:
Given that
Initial velocity wo=0.210rev/s
Then, 1rev=2πrad
wo=0.21×2πrad/s
wo=0.42π rad/s
Given angular acceleration of 0.9rev/s²
α=0.9×2πrad/s²
α=1.8π rad/s²
Diameter of blade
d=0.75m,
Radius=diameter/2
r=0.75/2=0.375m
a. Angular velocity after t=0.194s
Using equation of angular motion
wf=wo+αt
wf=0.42π+ 1.8π×0.194
wf= 0.42π + 0.3492π
wf=1.319+1.097
wf= 2.42rad/s
If we want the answer in revolution
1rev=2πrad
wf= 2.42/2π rev/s
wf=0.385 rev/s
b. Revolution traveled in 0.194s
Using angular motion equation
θf - θi = wo•t + ½ αt²
θf - 0= 0.42π•0.194 + ½ × 1.8π•0.194²
θf = 0.256 + 0.106
θf = 0.362rad
Now, to revolution
1rev=2πrad
θf=0.362/2π=0.0577rev
Approximately θf= 0.058rev
c. Tangential speed? At time 0.194s
Vt=?
w=2.42rad/s at t=0.194s
Using circular motion formulae, relationship between linear velocity and angular velocity
V=wr
Vt=wr
Vt= 2.42×0.375
Vt=0.9075 m/s
Vt≈0.91m/s
d. Magnitude of resultant acceleration
Tangential Acceleration is given as
at=αr
at=1.8π× 0.375
at=2.12rad/s²
Now, radial acceleration is given as
ar=w²r
ar=2.42²×0.375
ar=2.196 m/s²
Then, the magnitude is
a=√ar²+at²
a=√2.196²+2.12²
a=√9.3171
a=3.052m/s²
a≈ 3.05m/s²
Answer:
<h2>
Epicenter</h2>
Explanation:
-The hypocenter is the point within the earth where an earthquake rupture starts. The <em>epicenter</em> is the point directly above it at the surface of the Earth. Also commonly termed the focus.
-The location below the earth's surface where the earthquake starts is called the hypocenter, and the location directly above it on the surface of the earth is called the <em>epicenter</em>. Sometimes an earthquake has foreshocks.
Pitch is related to frequency
