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3 years ago

12

a 2 kg rabbit starts from rest and its moving at 6 m/s after 3 seconds . what force must the rabbit cause to makes this change i

n speed

2 answers:

cestrela7 [59]3 years ago

5 0

1. Acceleration= change of speed over the time = from 0 to 6m/s in 3sec= 6/3= 2m/s^2
2. Force=mass times acceleration = 2kg times 2m/s^2 = 4 N
Answer 4N

Alborosie3 years ago

3 0

Gravity pulling it down or the air cd (carbon dioxide)

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. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

Dmitriy789 [7]

Answer:

The magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

Explanation:

Given :

Magnitude of magnetic field $B = 0.078$ T

Radius of circle $r = 0.10$ m

Angle between field and surface normal $\theta =$ 25°

From the formula of flux,

$\phi = B.A$

$\phi = BA\cos \theta$

Where $\theta =$ angle between magnetic field line and surface normal, $A =$ area of circular surface.

$A = \pi r^{2}$

$A = 3.14 \times (0.10) ^{2}$

$A = 0.0314$ $m^{2}$

Magnetic flux is given by,

$\phi = 0.078 \times 0.0314 \times \cos 25$

$\phi = 2.22 \times 10^{-3}$ Wb

Therefore, the magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

6 0

4 years ago

A fixed coil of wire with 10 turns and an area of 0.055 m2 is placed in a perpendicular magnetic field. This field oscillates in

Blababa [14]

Answer:

Part a)

Average EMF for half cycle is

$E_{avg} = 2.64 V$

Part b)

For one complete cycle we will have

$E_{avg} = 0$

Part c)

Maximum induced EMF will be at

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = 0.05s and 0.1 s

Explanation:

As we know that magnetic field is oscillating in direction as well as magnitude

so induced EMF is given as

$E = NBA\omega sin(\omega t)$

Part a)

For average value of EMF from positive maximum to negative maximum which is equal to half cycle

so we have

$E_{avg} = NBA\omega \frac{2}{T}\int_0^{T/2} sin(\omega t) dt$

$E_{avg} = \frac{2NBA\omega}{\pi}$

$E_{avg} = \frac{2(10)(0.12)(0.055)(2\pi (10))}{\pi}$

$E_{avg} = 2.64 V$

Part b)

For one complete cycle we will have

$E_{avg} = NBA\omega \frac{1}{T}\int_0^T sin(\omega t) dt$

$E_{avg} = 0$

Part c)

Maximum induced EMF will be at

$t = \frac{T}{4} and \frac{3T}{4}$

here we know

$T = \frac{1}{f} = 0.1 s$

t = 0.025 s and 0.075 s

minimum induced EMF is at

$t = \frac{T}{2} and T$

so it is

t = 0.05s and 0.1 s

8 0

4 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

Bosons & Fermions A new type of quantum object has been discovered. When a large collection of these are cooled to almost ab

Lena [83]

Answer:

Only 2,3,4 are true

Explanation:

Bosons Particles are particles that condense to the same state. Bosons particle have integral spin like 0 , $\hbar$ , $2$\hbar$ , $3$\hbar$$ , etc. Bosons particles always have asymmetric wave function and there is exchange of particles.

1) It does not obey Fermi_ Dirac statistics

2) It obeys Bose-Einstein statistics

3) The object can have intrinsic spin $2$\hbar$

4) Yes the Bosons particle is always symmetric with exchange of particles

5) No Bosons particle are symmetric and not asymmetric

5 0

3 years ago

The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

$A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N$

the y comonent of the vecor is Ay = 49.92 N

3 0

3 years ago