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Sati [7]
2 years ago
11

A child jumps from a moving sled with a speed of 2.2 m/s and in the

Physics
1 answer:
butalik [34]2 years ago
4 0

The initial velocity of the sled will be 7.34 m/sec. V is the initial velocity of the sled.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁)  mass of child = 38 kg

(u₁) is the initial velocity child = 2.2 m/s

(m₂) is the mass of sled = 68 kg

(u₂) is the initial velocity of sled = ?

(v) is the velocity after collision = 5.5 m/s

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

\rm  m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\(38 \times  2.2) + (68 \times V) = 5.5 \times (38+68) \\\\ 83.6 + 68V = 5.5 \times 106 \\\\  83.6 + 68V =583 \\\\ \rm 68\ V= 583 -83.6 \\\\ \rm 68 \ V= 499.4\\\\ V= 7.34 \ m/sec

Hence,the initial velocity of the sled will be 7.34 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#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 velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electri
NeTakaya

Answer:

50k/h is the answer to iy

3 0
3 years ago
Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0
3 years ago
A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 n
RSB [31]

Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

Hope it helps

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2 years ago
acceleration is defined as the rate of change for which characteristic? a. displacement b. position c. velocity d. time
fredd [130]
The answer is velocity.
7 0
3 years ago
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