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3 years ago

Calculate the average value of an AC signal with a peak amplitude of 10V and a frequency of 10 kHz.

Physics

1 answer:

Solnce55 [7]3 years ago

3 0

Answer:

V(average)=6.37 V

Explanation:

Given Data

Peak Voltage=10V

Frequency=10 kHZ

To Find

Average Voltage

Solution

For this first we need to find Voltage peak to peak

Voltage (peak to peak)= 2× voltage peak

Voltage (peak to peak)= 2×10

Voltage (peak to peak)= 20 V

Now from Voltage (peak to peak) formula we can find the Average Voltage

Voltage (peak to peak)=π×V(average)

V(average)=Voltage (peak to peak)/π

V(average)=20/3.14

V(average)=6.37 V

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a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an

igor_vitrenko [27]

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is $m=230\ kg$

It reaches the steepest hill with speed of $u=6.2\ km/h\ or \ 1.72\ m/s$

Hill to bottom is 51 m long with inclination of $45^{\circ}$

Height of the hill is $h=51\sin 45^{\circ}=36.06\ m$

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

$\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ$

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2 years ago

True or false the hotter the star the higher is absolute magnitude?

liubo4ka [24]

Answer:

true

Explanation:

6 0

2 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

Nookie1986 [14]

The maximum amount of work performed is

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Explanation:

The efficiency of a real heat engine is given by the equation:

$\eta = 1-\frac{T_C}{T_H}$ (1)

where

$T_C$ is the temperature of the cold reservoir

$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

$\eta=\frac{W_{max}}{W_{max}+Q_C}$ (2)

By combining eq.(1) and (2) we get

$1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}$

And re-arranging the equation and solving for $W_{max}$ , we find

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

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3 years ago

7. When initially-unpolarized light passes through three polarizing filters, each oriented at 45 degree angles from the precedin

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Answer:

Explanation:

Let the intensity of unpolarised light be I₀ . After passing through the first polarising filter , the intensity is I₀ / 2 .

After second filter , the intensity will be I₀ / 2 x cos²45 = I₀ / 4

After third filter , the intensity will be I₀ / 4 x cos²45 = I₀ / 8 .

So,

1 / 8 the of initial light passes through the last filter .

7 0

3 years ago