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2 years ago

What is the amplitude of an AC voltage waveform, in units of Volts, if the RMS value is 369 V?

Physics

1 answer:

Blizzard [7]2 years ago

4 0

Answer:

521.8 V

Explanation:

The RMS value of the voltage of an AC signal is given by

$V_{rms}=\frac{V_0}{\sqrt{2}}$

where

$V_0$ is the peak voltage, which corresponds to the amplitude of the AC waveform

In this problem, we know the RMS voltage

$V_{rms}=369 V$

Therefore, we can re-arrange the previous equation to find the peak voltage (the amplitude of the waveform):

$V_0 = \sqrt{2}V_{rms}=\sqrt{2}(369 V)=521.8 V$

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The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper

-BARSIC- [3]

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.

Let X be the output voltage of power supply

X∼N $(5,0.02^2)$

The lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively

$P(4.95$

Hence probability that a power supply selected at random will conform to the specifications on voltage is 0.9876

8 0

3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 2 is tripled, then the new

Leya [2.2K]

Answer:

The force will be 54.0 units

Explanation:

The magnitude of the electrostatic force between two charged objects is given by Coulomb's Law:

$F=\frac{kq_1 q_2}{r^2}$

where

k is Coulomb's constant

q1, q2 are the magnitude of the two charges

r is the separation between the two charges

From the equation, we see that the magnitude of the force is directly proportional to the charge of object 2:

$F\propto q_2$

In this problem, the initial force between the two objects is

F = 18.0 N

And so, when the charge on object 2 is tripled,

$q_2'=3q_2$

The new electrostatic force will be

$F'\propto q_2' = (3q_2) = 3F$

So, the force will also triple: since the original force was 18.0 units, the new force will be

$F'=3F=3(18.0)=54.0$

5 0

2 years ago

Please help me so I don't fail.

NikAS [45]

Answer:

Explanation:

4 0

2 years ago

Using the graph, describe what is happening between 4 and 6 seconds. The object is moving away from the origin at a constant vel

Hunter-Best [27]

The position in the graph between 4 - 6 seconds is the region of constant velocity.

<h3>What is constant velocity?</h3>

The term constant velocity refers to the period in the graph when the velocity is not changing with time. As such, the graph is shown to be a flat portion at such a point.

Hence, it follows that the position in the graph between 4 - 6 seconds is the region of constant velocity.

Learn more about constant velocity:brainly.com/question/17014780

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8 0

2 years ago