Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.
What gas? Gas is an invisible form so I don't see how it would have a "defined structure".
Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325
as the volume remains constant therefore
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass 
Final mass 
Therefore 
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
Answer:
1224km/hr
Explanation:
To convert from m/s to km/hr
1000m = 1km
Divide both sides by 1000
1m = 1/1000 km................. (1)
60×60 seconds = 1 hr
3600s = 1hr
Divide both sides by 3600
1s = 1/3600 .............(2)
Divide (2) by (1)
1m/s = 1/1000 ÷ 1/3600 km/hr
1m/s = 1/1000 × 3600/1 km/hr
1m/s = 3600/1000 km/hr
1m/s = 3.6 km/hr .............(3)
To convert 340m/s to km/hr
Multiply (3) by 340
1× 340m/s = 3.6 × 340 km/hr
340m/s = 1224km/hr
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Answer:
340.67 kgm²/s
Explanation:
R = Radius of merry-go-round = 1.9 m
I = Moment of inertia = 209 kgm²
= Initial angular velocity = 1.63 rad/s
m = Mass of person = 73 kg
v = Velocity = 4.8 m/s
Initial angular momentum is given by
The initial angular momentum of the merry-go-round is 340.67 kgm²/s
