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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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3 years ago

A ball is bouncing to a height that is 1/3 of the previous height after every bounce. What kind of sequence is this? How could y

tamaranim1 [39]

Answer and Explanation:

The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence

For example let the initial height is 243 fit

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3 years ago

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0

3 years ago

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