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BaLLatris [955]
2 years ago
7

A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their v

elocities remain constant during the chase?
Physics
1 answer:
dedylja [7]2 years ago
7 0

The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

<h3>What time will the dog catch the rabbit?</h3>

The time that the dog will catch up with the rabbit is given as follows:

Let the distance covered by the rabbit be x.

Distance covered by dog = x + 30

  • Time taken = distance/speed

The time taken will be the same T

  1. Time taken by dog, T = (x + 30)/10
  2. Time taken by rabbit, T = x/5

Equating both times.

(x + 30)/10 = x/5

x = 30 m

Solving for T in equation (ii);

T = 30/5 = 6 minutes

In conclusion, time is obtained as a ratio of distance and speed.


Learn more about time and speed at: brainly.com/question/26046491

#SPJ1

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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
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
The SI unit for energy is Newton true or false
ipn [44]

Explanation:

false....it is joules

..........

6 0
3 years ago
What are some ways houses along the coastlines can protect themselves from storm surges?
BARSIC [14]
Build walls around the coast
8 0
3 years ago
A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre
AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz

3 0
3 years ago
A small 25 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe’s kinetic energy?
Helen [10]
Kinetic energy = 1/2mv^2

=1/2(25)(1^2)
= 12.5J
4 0
3 years ago
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