How do you know which magnitude is higher or how do you compare them?

A uniform metal rod, with a mass of 2.6 kg and a length of 1.5 m, is attached to a wall by a hinge at its base. A horizontal wir

gregori [183]

Answer:

the horizontal force acting on rod is - 25.909 N

veritical force is 25.48 N

Explanation:

Given data:

mass of rod is m 2.6 kg

length of rod is 1.5 m

Hinge distance from the wire is 0.70 m

Apply torque law about hinge

$T = (\frac{L}{2} cos 28^o) mg$

Tension in the wire is

$T = \frac{[\frac{1.5}{2} cos 28 ]2.6 \times 9.8}{0.70}$

= -25.909 N

the horizontal force acting on rod is

Fx = T

FX = - 25.909 N

vertical force is

Fy = mg

$= 2.6 \times 9.8 = 25.48 N$

3 0

3 years ago

A 100-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

olchik [2.2K]

Answer:

The force required is equal to $30\pi N$ .

Explanation:

We know that

Torque = force × perpendicular distance

= F × R = I× $\alpha$

I = M× $\frac{R^{2} }{2}$

From above equation $\alpha = \frac{FR}{I}$ .............. 1

We know that

$w=w_{o} +\alpha t$ ...........................2

Given that t= 2 sec

$w_{o} = 0$

$w=0.4 rev/s = 0.8\pi rad/s$

From equation 1 and 2 , we get

F = $\frac{wI}{RT}$ = $\frac{w\times m \times r}{2t}$

Upon substituting the above values we will be getting

F = $30\pi N$

3 0

3 years ago

which of the following is accurate in describing the placement and classification of iodine? a. iodine is located in period 5, g

Elden [556K]

A. iodine is located in period 5, group 17 and is classified as a nonmetal

Group is the column, period is the raw.

8 0

3 years ago

The plate area is doubled, and the plate separation is reduced to half its initial separation. What is the new charge on the neg

Norma-Jean [14]

Answer:

Q = 4 Q₀

Explanation:

This is an exercise on capacitors, where the capacitance is

C = $\epsilon_{o} \ \frac{A}{d}$

if we apply the given conditions

C = \epsilon_{o} \ \frac{2A}{0.5d}

C = 4 \epsilon_{o} \ \frac{A}{d}

let's call the capacitance Co with the initial values

C₀ = \epsilon_{o} \ \frac{A}{d}

C = 4 C₀

The charge on each plate of a capacitor is

Q = C ΔV

If the potential difference is maintained, the new charge is

Q = 4 C₀ ΔV

let's call

Q₀ = C₀ ΔV

we substitute

Q = 4 Q₀

5 0

2 years ago

4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o

goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through a 10-to-1 step-down transformer. It uses a silicon diode that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

(i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0

3 years ago