Answer:
Scenario 2
Explanation:
The momentum of an object is a vector quantity given by:
where
m is the mass of the object
v is the velocity of the object
From the equation, we see that the momentum is directly proportional to the velocity of the object.
In this problem, the velocities of the student in the 3 situations are:
v1 = 3 m/s
v2 = 0.5 m/s
v3 = 7 m/s
And since momentum is directly proportional to the momentum, the student has least momentum when he has least velocity: therefore, in situation 2.
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
A mixture IS NOT a chemical combining of substances. If substances
combine chemically, then they form a compound, not a mixture.