What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when release
1 answer:
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Answer:
a)
b)
c)
d) or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:
Part b
For this case first we can convert the spring constant to N/m like this:
And the work donde by the spring on this case is given by:
Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:
And if we solve for the initial velocity we got:
Part d
Let d1 represent the new maximum distance, in order to find it we know that :
And replacing we got:
And we can put the terms like this:
If we multiply all the equation by 2 we got:
Now we can replace the values and we got:
And solving the quadratic equation we got that the solution for or 18.3 cm because the negative solution not make sense.
Answer:
Option D: 21.8 degrees
Explanation:
In a parallel RL circuit, the current in the resistor R and that in the inductor L are separated among themselves 90 degrees as illustrated in the attached image. In the image the current in the resistor is represented in orange, that of the inductor in blue, and the total current (vector addition of the previous two) is represented in red, forming a certain angle (theta) with respect to the current in the resistor. The output voltage is the same as the input voltage as measured over the resistor R.
Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = , therefore the angle is the arctangent of 4/10:
degrees.
