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3 years ago

Two a.c V1=100sin(wt) and V2 = 150cos(wt) are fed into one circuit, determine the combined output of the two as a single a.c

Physics

1 answer:

Dominik [7]3 years ago

7 0

Answer:

Explanation:

V = 100sin(ωt) + 150cos(ωt)

let x = ωt

V = 100sin(x) + 150cos(x)

a maximum or minimum will occur when the derivative is zero

V' = 100cos(x) - 150sin(x)

0 = 100cos(x) - 150sin(x)

100cos(x) = 150sin(x)

100/150 = sin(x)/cos(x)

0.6667 = tan(x)

x = 0.588 rad

V = 100sin(0.588) + 150cos(0.588)

V = 180.27756

as the maximum will not occur until ωt = 0.588 radians, for a cosine function we subtract that amount as a phase angle φ

V = 180.3 cos(ωt - 0.588)

or as a sine function, the phase angle lags the cosine by a difference of π/2

V = 180.3sin(ωt - (0.588 - π/2)

V = 180.3sin(ωt + 0.983)

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4 years ago

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3 years ago

At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the she

insens350 [35]

Answer:

$1.34 * 10^{3}$ m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, $v_{x}$ = 990m/s

Surface charge density, σ = 2.34 x $10^{-9}$ C/ $m^{2}$

We need to calculate the speed at time, t = 7.0 * $10^{-8}$ s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed $v_{x}$ on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity $v_{y}$ in the y axis.

Hence, the resultant velocity of the proton is given by:

$v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v_{x}$ = 990m/s from the question. We need to find $v_{y}$ and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> $E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }$

E = 132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

$a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }$

a = 1.3 * $10^{10}$ m/ $s^{2}$

Therefore, at time, t = 7.0 * $10^{-8}$ , we can use one of the equations of linear motion to find the velocity in the y axis:

$a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at$

$v_{y}$ = 0 + (1.3 * $10^{10}$ * 7.0 * $10^{-8}$ )

$v_{y}$ = 910 m/s

∴ $v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v = \sqrt{990^{2} + 910^{2} }$

$v = \sqrt{1808200}$

v = 1344.69 m/s = $1.34 * 10^{3}$ m/s

8 0

4 years ago

By how much will the velocity of a vehicle change if it accelerates at 10 m/s/s for a period of 10 seconds

pishuonlain [190]

Answer:

Explanation:

we know that the time rate of change of velocity is called acceleration mathmetically;

a=Δv/Δt

Δv=a*Δt

a=10 m/s²

t=10 sec

Δv=10*10=100 m/s

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4 years ago

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Morgarella [4.7K]

Answer:

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One wave Crest can catch up to another wave Crest if the group of waves are from different places.

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