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2 years ago

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An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole me

asured counterclockwise from the electric field direction for which the torque on the dipole is zero. B. Which part of orientation in part (a) is stable and which is unstable?

1 answer:

Tasya [4]2 years ago

3 0

Answer:

Explanation:

A) When a dipole is placed in an electric field , it experiences a torque equal to the following

torque = p x E = p E sinθ , where θ is angle between direction of p and E .

It will be zero if θ = 0

or if both p and E are oriented in the same direction.

It is the stable orientation of dipole.

If θ = 180° ,

Torque = 0

In this case both p and E are oriented in opposite direction .

It is the unstable orientation of the dipole because if we deflect the dipole by even small angle , it goes back to most stable orientation due to torque acting on it by electric field.

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g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the

Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

$R = \frac{u^{2}Sin2\theta }{g}$

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

$R = \frac{u^{2}Sin60 }{9.8}$

R = 0.088u^2

If the angle of projection is 45 degree.

$R = \frac{u^{2}Sin90 }{g}$

R = u^2 / g

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2 years ago

A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po

Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, $L_{a} = 1 m$

Frequency, $\vartheta = 1.5 MHz = 1.5\times 10^{6} Hz$

Power transmitted, $P_{t} = 4 W$

Now,

For a monopole antenna:

$\lambda_{a} = \frac{c}{\vartheta}$

where

$\lambda_{a}$ = wavelength transmitted by the antenna

c = speed of light in vacuum

$\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m$

Now,

Since, the value of $\lambda_{a}$ >> $L_{a}$ thus the monopole is a Hertian monopole.

The resistance is calculated as:

$R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}$

$R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega$

$P_{radiated} = P_{t}$

$P_{radiated} = \frac{R}{I^{2}}$

Now, the current I is given by:

$I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A$

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1 year ago

A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie

cluponka [151]

Answer:

a) v₃ = 19.54 km, b) 70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

cos 45 = x₁ / V₁

sin 45 = y₁ / V₁

x₁ = v₁ cos 45

y₁ = v₁ sin 45

x₁ = 26 cos 45

y₁ = 26 sin 45

x₁ = 18.38 km

y₁ = 18.38 km

Vector 2 moves 45 km north

y₂ = 45 km

Unknown 3 vector

x3 =?

y3 =?

Vector Resulting 70 km north of the starting point

R_y = 70 km

we make the sum on each axis

X axis

Rₓ = x₁ + x₃

x₃ = Rₓ -x₁

x₃ = 0 - 18.38

x₃ = -18.38 km

Y Axis

R_y = y₁ + y₂ + y₃

y₃ = R_y - y₁ -y₂

y₃ = 70 -18.38 - 45

y₃ = 6.62 km

the vector of the third leg of the journey is

v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

v₃ = √ (18.38² + 6.62²)

v₃ = 19.54 km

to find the angle let's use trigonometry

tan θ = y₃ / x₃

θ = tan⁻¹ (y₃ / x₃)

θ = tan⁻¹ (6.62 / (- 18.38))

θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

θ’= 180 -19.8

θ’= 160.19º

I mean the address is

θ’’ = 90-19.8

θ = 70.2º

70.2º north-west

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2 years ago

568 muons were counted by a detector on the top of Mount Washington in a one hour period of time. Assuming moving muons keep tim

AlladinOne [14]

The answer to this question is:

C-"That moving clocks run slower"

Your Welcome :)

6 0

2 years ago

Read 2 more answers

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

We need to calculate the velocity of the ball

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

$d =vt$

Where. d = distance

t = time

v = velocity

Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0

2 years ago