Answer:
By a factor of 1/4.
Explanation:
The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,
in which ,
represent the change in momentum and the time taken for that change.
If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by , the following manipulation confirms the answer to this question.
Here is the force that was applied to the object previously.
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Answer:
w=3.05 rad/s or 29.88rpm
Explanation:
k = coefficient of friction = 0.3900
R = radius of the cylinder = 2.7m
V = linear speed of rotation of the cylinder
w = angular speed = V/R or to rewrite V = w*R
N = normal force to cylinder
N=
These must be balanced (the net force on the people will be 0) so set them equal to each other.
There are 2*pi radians in 1 revolution so:
So you need about 30 RPM to keep people from falling out the bottom
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Reference the diagram below for clarification.
1.
We must begin by knowing the following rules for resistors in series and parallel.
In series:
In parallel:
We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.
Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
2.
We can use Ohm's law to solve for the current in the circuit.
3.
For resistors in series, both resistors receive the SAME current.
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.
4.
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.
Answer:
part (a)
Part (b)
Explanation:
Given,
Let I be the moment of inertia of the both disk after the welding,
part (a)
A block of mass m is hanging on the smaller disk,
From the f.b.d. of the block,
Let 'a' be the acceleration of the block and 'T' be the tension in the string.
Net torque on the smaller disk,
From eqn (1) and (2), we get,
part (b)
In this case the mass is rapped on the larger disk,
From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,
Let '' be the acceleration of the block in the second case,
From the above expression,
