consider an electric dipole lying along x-axis with mid-point O as the origin of coordinate system.find the electric potential V at point P due to dipole.in addition to this,find electric potentials for special cases. when P lies on the axial line and when P lies on the equatorial line.

Answer:

Nicolaus Copernicus

Explanation:

Nevertheless, Copernicus began to work on astronomy on his own. Sometime between 1510 and 1514 he wrote an essay that has come to be known as the Commentariolus (MW 75–126) that introduced his new cosmological idea, the heliocentric universe, and he sent copies to various astronomers

Answer:

Explanation:

a)

A = 3m , B = 4m .

(A-B)² = A² + B² - 2ABcosθ where θ is angle between A and B.

4 = 9 + 16 - 2. 3.4 cosθ

cosθ = .875

θ = 29° .

b ) Unit vector in the direction of A X B will be k vector because A and B are in X-Y plane and A X B lies perpendicular to both A and B .

Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F =  average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

Answer:

reflected angle - secod mirror = 60°

Explanation:

I attached an image with the solution to this problem below.

In the solution the reflection law, incident angle = reflected angle, is used. Furthermore some trigonometric relation is used.

You can notice in the image that the angle of reflection in the second mirror is 60°

Answer:

a) [tex]t \approx 9.879\,s[/tex], b) [tex]v = 20.518\,\frac{m}{s}[/tex], c) [tex]t = 16.368\,s[/tex], d) [tex]v = 19.545\,\frac{m}{s}[/tex]

Explanation:

a) Since train is only translating in a straight line and experimenting a constant deceleration throughout the station, whose length is 210 meters. The time required for the nose of the train to reach the end of the station can be found with the help of the following motion formula:

[tex]210\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

The following second-order polynomial needs to be solved:

[tex]-0.075\cdot t^{2} + 22\cdot t - 210 = 0[/tex]

Whose roots are presented herein:

[tex]t_{1}\approx 283.455\,s[/tex] and [tex]t_{2} \approx 9.879\,s[/tex]

Both solutions are physically reasonable, although second roots describes better the braking process of the train.

b) The speed of the nose leaving the station is given by this expression:

[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (9.879\,s)[/tex]

[tex]v = 20.518\,\frac{m}{s}[/tex]

c) First, it is required to calculate the time when nose of the train reaches a distance of 130 meters.

[tex]130\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]-0.075\cdot t^{2} + 22\cdot t - 130 = 0[/tex]

Roots of the second-order polynomial are:

[tex]t_{1} \approx 287.300\,s[/tex] and [tex]t_{2} \approx 6.033\,s[/tex]

Both solutions are physically reasonable, although second roots describes better the braking process of the train. Now, the speed experimented by the train at this instant is:

[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (6.033\,s)[/tex]

[tex]v = 21.095\,\frac{m}{s}[/tex]

The distance traveled by the end of the train throughout station is modelled after the following equation:

[tex]210\,m = \left(21.095\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]-0.075\cdot t^{2} + 21.095\cdot t - 210 = 0[/tex]

Roots of the second-order polynomial are:

[tex]t_{1} \approx 270.932\,s[/tex] and [tex]t_{2} \approx 10.335\,s[/tex]

Both solutions are physically reasonable, although second roots describes better the braking process of the train. The instant when the end of the train leaves the station is:

[tex]t = 6.033\,s + 10.335\,s[/tex]

[tex]t = 16.368\,s[/tex]

d) The velocity experimented by the end of the train is:

[tex]v = 21.095\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}} \right)\cdot (10.335\,s)[/tex]

[tex]v = 19.545\,\frac{m}{s}[/tex]

The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all. The negative sign indicates that the train is moving in the opposite direction of its initial velocity. The end of the train leaves the station after approximately 14.98 seconds. The velocity of the end of the train as it leaves the station is approximately 19.753 m/s.

(a) Using the equation :

v² = u² + 2as

0 = (22.0)² + 2 × (-0.150 ) × s

s = -(22.0 )^2 / (2 × -0.150)

s = 325.3 m

The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all.

(b) The velocity of the train when the nose leaves the station is given by:

v = u + at

v = 22.0 + (-0.150 ) × 210

v ≈ 22.0 - 31.5

v ≈ -9.5 m/s

The negative sign indicates that the train is moving in the opposite direction of its initial velocity.

(c)

s = ut + (1/2)at²

210= 22.0 × t + (1/2) × (-0.150) × t²

0.075 t² - 22.0 t + 210 = 0

we find two solutions for t: t ≈ 14.98 s and t ≈ 3.01 s.

The end of the train leaves the station after approximately 14.98 seconds.

(d) The velocity of the end of the train as it leaves the station is given by:

v = u + at

v = 22.0 + (-0.150) × 14.98

v = 22.0 - 2.247

v = 19.753 m/s

So, the velocity of the end of the train as it leaves the station is approximately 19.753 m/s.

To know more about the velocity:

https://brainly.com/question/17127206

#SPJ6

Answer:

a)  T = 1342.6 cos θ, b)  v = 13.93 √(sin θ tan θ)

Explanation:

We can solve this problem using Newton's second law

Let's fix a reference system with a horizontal axis and the other vertical, therefore the only force to decompose is the tension, in these problems the most common is to measure the angle with respect to the vertical. Let's use trigonometry to find the components of the dispute

      cos θ = [tex]T_{y}[/tex] / T

      T_{y} = T cos tea

     sin θ = Tₓ / T

     Tₓ = T sin θ

let's write Newton's second law

axis and vertical

      T cos θ - W = 0

       T = mg / cos θ

let's calculate

      T = 137  9.8 cos θ

       T = 1342.6 cos θ

unfortunately there is no drawing or indication of the angle

Axis x Horizontal

       T sin θ = m a

acceleration is centripetal

        a = v² / R

        T sin θ = m v² / R

        v² = (g / cos θ) R sin θ

        v = √ (gR tan θ)

let's use trigonometry to find radius of gyration

          sin θ = R / L

          R = L sin θ

         v = √ (g L  sin θ tan θ)

let's calculate

         v = √(9.8 19.8 sin θ tant θ)

         v = 13.93 √(sin θ tan θ)

they do not give the angle for which the calculation cannot be finished

Answer:

6.61 m/s²

Explanation:

Given:

Δx = 0.40 km = 400 m

v₀ = 0 m/s

t = 11.0 s

Find: a

Δx = v₀ t + ½ at²

400 m = (0 m/s) (11.0 s) + ½ a (11.0 s)²

a = 6.61 m/s²

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

[tex]T \approx e^{-2CL}[/tex]

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

[tex]0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)[/tex]

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

[tex]ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)[/tex]

Next, you divide the equation (3) into (4), and finally, you solve for L':

[tex]\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L[/tex]

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

Answer:

Current, I = 0.073 A

Explanation:

It is given that,

Number of turns in a long solenoid is 1080

Length of the solenoid is 0.420 m

It produces a magnetic field of [tex]10^{-4}\ T[/tex] at its center.

We need to find the current is required in the winding for that to occur. The magnetic field at the center of the solenoid is given by :

[tex]B=\mu_0 NI[/tex]

I is current

[tex]I=\dfrac{B}{\mu_o N}\\\\I=\dfrac{10^{-4}}{4\pi \times 10^{-7}\times 1080}\\\\I=0.073\ A[/tex]

Answer:

v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]

Explanation:

Use Bernoulli's equation:

P₁ + ½ ρ v₁² + ρgz₁ = P₂ + ½ ρ v₂² + ρgz₂

Since there's no elevation change between points 1 and 2, z₁ = z₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Assuming incompressible fluid, the volumetric flow rate is the same at points 1 and 2.

Q₁ = Q₂

v₁ A₁ = v₂ A₂

v₂ = v₁ A₁ / A₂

Substituting:

P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ A₁ / A₂)²

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (A₁ / A₂)²

P₁ − P₂ = ½ ρ v₁² (A₁ / A₂)² − ½ ρ v₁²

P₁ − P₂ = ½ ρ v₁² ((A₁ / A₂)² − 1)

v₁² = 2 (P₁ − P₂) / (ρ ((A₁ / A₂)² − 1))

v₁² = 2 (ρgh) / (ρ ((A₁ / A₂)² − 1))

v₁² = 2gh / ((A₁ / A₂)² − 1)

v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

[tex]B=\frac{\mu_0I}{2\pi d}[/tex]-----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

[tex]\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A[/tex]

Therefore, the magnitude of current I is 200A

Answer:

c. 25,000 frames/s

Explanation:

For computing the minimum frame rate for high speed first we have to determine the time by applying the following equation

[tex]t = \frac{d}{s}[/tex]

[tex]= \frac{0.2\ mm}{4.6\ m/s }[/tex]

[tex]= \frac{0.2 \times 10 ^{-3}}{4.6\ m/s }[/tex]

[tex]= 4.347 \times 10^{-5} sec[/tex]

Now the frame rate is

[tex]Frame\ rate = \frac{1}{t}[/tex]

[tex]= \frac{1}{4.347 \times 10^{-5} sec}[/tex]

= 23,000 frame per sec

≈ 25,000 frame per sec

First we have find the time then after finding out the time we calculate the frame time by applying the above formula so that the minimum frame rate could come

Answer:

Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt

Explanation:

From Maxwell's fourth equation

Curl B = μ₀J + μ₀ε₀dE/dt (1) where the second term is the displacement current.

If the oscillation conduction current in the conductor is much larger than the displacement current then, the displacement current goes to zero. So we have

Curl B = μ₀J  (2)(since μ₀ε₀dE/dt = 0)

From maxwell's third equation

Curl E = -dB/dt  (3)  

taking curl of the above from the left

Curl(Curl E) = Curl(-dB/dt)

Curl(Curl E) = (-d(CurlB)/dt)  (4)

Substituting for Curl B into (4), we have

Curl(Curl E) = -dμ₀J/dt

Del(DivE) - (Del)²E = -dμ₀J/dt    (5)

From Maxwell's first equation,

DivE = ρ/ε₀

Substituting this into (5), we have

Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt

Answer:

1.97 m/s.

Explanation:

From the question,

Using the law of conservation of energy,

The energy stored in the charged plate = Kinetic energy of the mass

1/2(qV) = 1/2mv².......................... Equation 1

Where q = charge, V = voltage, m = mass, v = velocity.

make v the subject of the equation

v = √(qV/m)......................... Equation 2

Given: q = 6.5×10⁻⁶ C, V = 12000 Volts, m = 0.02 kg

Substitute these values into equation 2

v = √(6.5×10⁻⁶×12000 /0.02)

v = √3.9

v = 1.97 m/s.

Answer:

option A.

Explanation:

Since the two lumps collide together, it is an inelastic collision.  An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction. Part of the kinetic energy is changed to some other form of energy in the collision

Momentum is conserved in inelastic collisions. This means the correct option is option A.

Answer: It was Democritus, in fact, who first used the word atomos to describe the smallest possible particles of matter.

Explanation: hope this helped

Answer:

By exact formula

5076.59N/C

And by approximation formula

5218.93N/C

Explanation:

We are given that

Length of rod,L=2.6 m

Charge,q=98nC=[tex]98\times 10^{-9} C[/tex]

[tex]1nC=10^{-9} C[/tex]

a=13 cm=0.13 m

1 m=100 cm

By exact formula

The magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{a^2+\frac{L^2}{4}}}[/tex]

Where k=[tex]9\times 10^9[/tex]

Using the formula

The magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{(0.13)^2+\frac{(2.6)^2}{4}}}=5076.59N/C[/tex]

In approximation formula

a<<L

[tex]a^2+(\frac{L}{2})^2=\frac{L^2}{4}[/tex]

Therefore,the magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{\frac{L^2}{4}}}[/tex]

The magnitude of  the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{\frac{(2.6)^2}{4}}}=5218.93N/C[/tex]

Answer: 0.3 m/sec

Explanation:

Vrel = (Vs-Vm) = (0.8-0.5) = 0.3 m/sec

Answer:

0.3 m/s east

Explanation:

Answer:

a. 451.72 mi/ga

b. 1078.33 mi/ga

Explanation:

The computation is shown below:

a. The fuel economy for a person walking is

Given that

Walking at 3.20 mi/h requires about 220 kcal/h so it is equal to

[tex]= 220 \times 4186[/tex]

= 920920 j/hr

Now

[tex]= \frac{mi}{j}[/tex]

[tex]= \frac{3.2}{920920}[/tex]

So,

[tex]= \frac{mi}{ga}[/tex]

[tex]= \frac{3.2}{920920}\times 1.3 \times 100000000[/tex]

= 451.72 mi/ga

b. Now

Bicycling 12.5 mi/h requires about 360 kcal/h  energy so it is equal to

[tex]= 360 \times 4186[/tex]

= 1506960 j/hr

So,

[tex]= \frac{mi}{j}[/tex]

[tex]= \frac{12.5}{1506960}[/tex]

Now

[tex]= \frac{mi}{ga}[/tex]

[tex]= \frac{12.5}{1506960} \times 1.3 \times 100000000[/tex]

= 1078.33 mi/ga

We simply applied the above formula

Answer:

Explanation:

The maximum efficient power plant will be the plant based on carnot cycle whose efficiency is given by the following formula

Efficiency = (T₁ - T₂)  / T₁

T₁ is temperature of hot reservoir and T₂ is temperature of cold reservoir.

Putting the given values

efficiency of power plant = (35 - 5) / (273 + 35 )

= 30 / 308

= .097

= 9.7 %

Answer:

microwave

Explanation:

Answer:

a) v = 3.71m/s

b) U = 616.71 J

Explanation:

a) To find the speed of the skier you take into account that, the work done by the friction surface on the skier is equal to the change in the kinetic energy:

[tex]-W_f=\Delta K=\frac{1}{2}m(v^2-v_o^2)\\\\-F_fd=\frac{1}{2}m(v^2-v_o^2)[/tex]  

(the minus sign is due to the work is against the motion of the skier)

m: mass of the skier = 58.0 kg

v: final speed = ?

vo: initial speed = 6.00 m/s

d: distance traveled by the skier in the rough patch = 3.65 m

Ff: friction force = Mgμ

g: gravitational acceleration = 9.8 m/s^2

μ: friction coefficient = 0.310

You solve the equation (1) for v:

[tex]v=\sqrt{\frac{2F_fd}{m}+v_o^2}=\sqrt{\frac{2mg\mu d}{m}+v_o^2}\\\\v=\sqrt{-2g\mu d+v_o^2}[/tex]

Next, you replace the values of all parameters:

[tex]v=\sqrt{-2(9.8m/s^2)(0.310)(3.65m)+(6.00m/s)^2}=3.71\frac{m}{s}[/tex]

The speed after the skier has crossed the roug path is 3.71m/s

b) The work done by the rough patch is the internal energy generated:

[tex]U=W_fd=F_fd=mg\mu d\\\\U=(58.0kg)(9.8m/s^2)(0.310)(3.50m)=616.71\ J[/tex]

The internal energy generated is 616.71J

Answer:

Q = - 4312 W = - 4.312 KW

Explanation:

The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:

Q = - kA dt/dx

Integrating from one side of the slab to other along the thickness dimension, we get:

Q = - kA(T₂ - T₁)/L

Q = kA(T₁ - T₂)/t

where,

Q = Rate of Heat Loss = ?

k = thermal conductivity = 1.4 W/m.k

A = Surface Area = (11 m)(8 m) = 88 m²

T₁ = Temperature of Bottom Surface = 10°C

T₂ = Temperature of Top Surface = 17° C

t = Thickness of Slab = 0.2 m

Therefore,

Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m

Q = - 4312 W = - 4.312 KW

Here, negative sign shows the loss of heat.

Answer:

a) 0.5985 V

b) 0.05985 V

c) 0.00684 V

Explanation:

Given that

Number of turn in the coil, N = 60 turns

Magnetic field of the coil, B = 0.6 T

Diameter of the coil, d = 0.13 m

If area is given as, πd²/4, then

A = π * 0.13² * 1/4

A = 0.0133 m²

The induced emf, ε = -N(dΦ*m) /dt

Note, Φm = BA

Substituting for Φ, we have

ε = -NBA/t.

Now, we substitute for numbers in the equation

ε = -(60 * 0.6 * 0.0133)/0.8

ε = 0.4788/0.8

ε = 0.5985 V

at 8s,

ε = -(60 * 0.6 * 0.0132)/8

ε = 0.4788/8

ε = 0.05985 V

at 70s

ε = -(60 * 0.6 * 0.0132)/70

ε = 0.4788/70

ε = 0.00684 V

Answer:

Index of Refraction = 1.52

Explanation:

The index of refraction or the refractive index is given as:

Index of Refraction = Sin i/Sin r

where,

i = angle of incidence

r = angle of refraction

In this case of karo syrup, we have the following data:

i = angle of incidence  = 30°

r = angle of refraction = 19.24°

Therefore, substituting these values in the equation, we get:

Index of Refraction = Sin 30°/Sin 19.24°

Index of Refraction = 0.5/0.3295

Index of Refraction = 1.52

Hence, the refractive index or the index of refraction of the Karo Syrup is found to be 1.52.

Answer:

The speed will be "1.06 m/s".

Explanation:

The given values are:

Momentum,

m1 = 244 g

m2 = 45.2 g

On applying momentum conservation ,

Let v2 become the final golf's speed.  

From Momentum Conservation

⇒  [tex]Total \ initial \ momentum = Total \ final \ momentum[/tex]

⇒  [tex]m1\times u1 + m2\times u2 = m1\times v1 + m2\times v2[/tex]

On putting the estimated values, we get

⇒  [tex]0.244\times 57.6+0=0.244\times 39.9+45.2\times v2[/tex]

⇒  [tex]57.844+0=9.7356+45.2\times v2[/tex]

⇒  [tex]48.1084=45.2\times v2[/tex]

⇒  [tex]v2=\frac{48.1084}{45.2}[/tex]

⇒  [tex]v2=1.06 \ m/s[/tex]

Answer:

[tex]F=5.16\times 10^{15}\ N[/tex]

Explanation:

We have,

Mass of Mars is, [tex]m_M=6.42\times 10^{23}\ kg[/tex]

Mass of its moon Phobos, [tex]m_P=1.06\times 10^{16}\ kg[/tex]

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

[tex]F=G\dfrac{m_Mm_P}{d^2}[/tex]

Plugging all values, we get :

[tex]F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N[/tex]

So, the gravitational force is [tex]5.16\times 10^{15}\ N[/tex].