For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as
F= 618.9 N
<h3>What is the centripetal
force?</h3>
Generally, the equation for the angular speed is mathematically given as
w = v/R
Therefore
w= 4.7/1.8
w= 2.611 rad/s
Where total momentum
Tm= 642.96 + 272.32
Tm= 915.28
and total inertia
Ti= 184 + 246.24
Ti= 430.24
In conclusion, centripetal force
F= mrw^2
F = m*R*w2^2
F = 76*1.8*2.127^2
F= 618.9 N
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a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?
Fibroblast, White Blood Cells, Plasma.
Answer:
Explanation:
V = J/C
V = 20/1
= 20 v
Option A is the correct answer
Answer:
1900 metres
Explanation:
Given that a train travels at a speed of 30 m/s. The train starts at an initial position of 1000 meters and travels for 30 seconds.
The parameters to be considered are:
Speed = 30 m/s
Time = 30 seconds
Speed = distance/time
Substitute the parameters into the formula
30 = distance / 30
Cross multiply
Distance = 30 × 30
Distance = 900 m
Since the train started from a position of 1000 m , the final position will be:
Final position = 1000 + 900
Final position = 1900 metres
