A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The
1 answer:
Answer:
The value to be reported is 5.48V
Explanation:
The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.
The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.
It is expressed as =
where Vm is the maximum or peak value of the voltage
In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)
=
=
= 5.48 V
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Answer:
a) the three longest wavelengths = 4.8m, 2.4m, 1.6m
b) what is the frequency of the third-longest wavelength = 75Hz
Explanation:
The steps and appropriate formula and substitution is as shown in the attached file.
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: