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just olya [345]
1 year ago
12

What is the net work doneon the object over the distance shown?

Physics
1 answer:
GuDViN [60]1 year ago
4 0

A)F_0d

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

so

Area:

\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}

so

\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}

therefore, the answer is

A)F_0d

I hope this helps you

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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
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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

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Explanation:

It is known that relation between pressure and density is as follows.

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     \rho = density

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            h = height

Putting the given values into the above formula as follows.

              P = \rho gh

                 = 1025 \times 9.8 \times 11000

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Now, relation between pressure and force is as follows.

                P = \frac{F}{A}

or,            F = PA

                F = 110495000 \times \pi \times (0.1)^{2}

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Thus, we can conclude that a force of 3.47 \times 10^{6} N can be  experienced at such depth.

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