Answer:
The input force (effort) is the amount of effort used to push down on a rod, or pull on a rope in order to move the weight. In this example, the force the little guy is using to pull the elephant is the input force.
Explanation:
The line that should be secured first in pushing the boat
away from the dock in preparation to dock is the bow line. When the bow line is
secured, it is best to reverse it and turn to the dock, this will engage the
line to tighten in a way that will help it swing back in the dock.
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.
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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CQ
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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?
Answer:
19.2 m/s
Explanation:
The train is moving at 18 m/s and you are walking in the same direction (east) so the speeds are added
18 + 1.2 = 19.2
If you were walking backwards (west) your velocity with respect to the ground would be
18 - 1.2 = 16.8
