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Ad libitum [116K]
3 years ago
15

A block lies on a horizontal frictionless surface and

Physics
1 answer:
zhenek [66]3 years ago
6 0

Answer:

0.1 m

Explanation:

F = Force exerted on spring = 3 N

k = Spring constant = 60 N/m

x = Displacement of the block

As the energy of the system is conserved we have

Fx=\dfrac{1}{2}kx^2

\\\Rightarrow x=\dfrac{2F}{k}

\\\Rightarrow x=\dfrac{2\times 3}{60}

\\\Rightarrow x=0.1\ m

The position of the block is 0.1 from the initial position.

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The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current
irina1246 [14]

Answer:

a  average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by  P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

6 0
3 years ago
Read 2 more answers
Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190
julia-pushkina [17]

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C

8 0
3 years ago
Phys-1A Horizontal Practice 2
krek1111 [17]
So basically you, then, finally, you
7 0
2 years ago
speed of sound is 343 Ms at 20 degrees Celsius. The frequency heard from the sound is 256 Hz. what is the sounds wavelength?
Lina20 [59]
S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

 = 1.339844m     
7 0
3 years ago
What happens to the particles of the liquid inside a thermometer when the thermometer is heated?
aev [14]
The particles of the liquid inside a thermometer speed up and spread apart when the thermometer is heated. In short, the particles expand from one another when they're heated, and become condensed and compact when chilled.

5 0
4 years ago
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