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Alexxx [7]
3 years ago
9

An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 13 V

. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning
Physics
1 answer:
MrMuchimi3 years ago
4 0

Answer:

9.2 V

Explanation:

The RMS value of an AC is the effective value of a varying voltage or current in DC, that is the equivalent value of the AC which produces the same effect as an DC. For example if a motor is supplied by a 9V RMS voltage, it will rotate as if the voltage applied was 9V DC.

The RMS value is given by:

RMS voltage = Peak voltage * 1/√2

Given that the maximum voltage should not exceed 13 V, this means that the peak voltage is 13 V. The maximum RMS voltage is:

RMS voltage = Peak voltage * 1/√2 = 13 * 1/√2 = 9.2 V

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The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s
alex41 [277]
Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
8 0
3 years ago
a person is sitting on the last bench can see clearly see things written on book but cannot see them distinctly on board. what t
marysya [2.9K]

Answer:

I think concave

Explanation:

7 0
2 years ago
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the
Zepler [3.9K]

Answer:

51.94°

Explanation:

I_0 = Unpolarized light

I_2 = Light after passing though second filter = 0.19I_0

Polarized light passing through first filter

I_1=\frac{I_0}{2}

Polarized light passing through second filter

I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}

The angle between the two filters is 51.94°

5 0
2 years ago
Someone please help !!
rodikova [14]
What is the Investigation about!
5 0
3 years ago
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons
yarga [219]

Answer:

q\approx 6.6\cdot 10^{13}~electrons

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

\displaystyle F=k\frac{q^2}{d^2}

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

\displaystyle q=\sqrt{\frac{F}{k}}\cdot d

Substituting values:

\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1

q = 1.05\cdot 10^{-5}~c

This charge corresponds to a number of electrons given by the elementary charge of the electron:

q_e=1.6 \cdot 10^{-19}~c

Thus, the charge of any of the spheres is:

\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}

\mathbf{q\approx 6.6\cdot 10^{13}~electrons}

5 0
2 years ago
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