Let $R_{1}=1.0$ ohms, $R_{2}=3.0$ ohms and $R_{3}=6.0\:$ ohms. Since $R_{2}$ and $R_{3}$ are in parallel, their combined resistance $R_{23}$ is given by

$\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

$R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms$

The total current flowing through the circuit <em>I</em> is given

$I=\dfrac{V_{s}}{R_{Total}}$

where

$R_{Total}=R_{1}+R_{23}= 3.0\:ohms$

Therefore, the total current through the circuit is

$I=\dfrac{30\:V}{3.0\:ohms}=10\:A$

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor $V_{1}$ :

$V_{1}=(10\:A)(1.0\:ohms)=10\:V$

This means that voltage drop across the 6-ohm resistor $V_{3}$ is 20 V. The power dissipated <em>P</em> by the 6-ohm resitor is given by

$P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W$

4 0

3 years ago

A flat uniform circular disk (radius = 2.00 m, mass= 100

ExtremeBDS [4]

Answer:

0.5 rad / s

Explanation:

Moment of inertia of the disk I₁ = 1/2 MR²

M is mass of the disc and R is radius

Putting the values in the formula

Moment of inertia of the disc I₁ = 1/2 x 100 x 2 x 2

= 200 kgm²

Moment of inertia of man about the axis of rotation of disc

mass x( distance from axis )²

I₂ = 40 x 1.25²

= 62.5 kgm²

Let ω₁ and ω₂ be the angular speed of disc and man about the axis

ω₂ = tangential speed / radius of circular path

= 2 /1.25 rad / s

= 1.6 rad /s

ω₁ = ?

Applying conservation of angular moment ( no external torque is acting on the disc )

I₁ω₁ = I₂ω₂

200 X ω₁ = 62.5 X 1.6

ω₁ = 0.5 rad / s

7 0

3 years ago

Name 5 machine principles

lidiya [134]

The Wheel & Axle. The wheel has always been considered a major invention in the history of mankind

The Inclined Plane

The Wedge

The Pulley

The Screw

3 0

4 years ago

You illuminate a slit with a width of 70.3 μm with a light of wavelength 719 nm and observe the resulting diffraction pattern on

Zielflug [23.3K]

Answer:

4.3 cm

Explanation:

We are given that

Width,d=70.3 $\mu m=70.3\times10^{-6} m$

$1\mu m=10^{-6} m$

Wavelength, $\lambda=719 nm=719\times 10^{-9} m$

$1nm=10^{-9} m$

$r=2.11 m$

We have to find the width in cm of the pattern.

The angle for the first minimum m=1

$sin\theta_{min}\approx \theta=\frac{\lambda}{d}=\frac{719\times 10^{-9}}{70.3\times 10^{-6}}=0.0102 rad$

$y=r\theta=2.11\times 0.0102=0.0215 m$

The width of the pattern= $2y=2\times 0.0215=0.043 m=0.043\times 100=4.3 cm$

8 0

3 years ago

In physics, explain electricity.​

In physics, explain electricity.