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atroni [7]
3 years ago
8

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

or at the point P, a distance d above the midpoint of the wire. The field E-vector points along one of the primary axes, yWhat is the magnitude E of the electric field at point P? Throughout this part, express your answers in terms of the constant k, defined by k=1/(4πε)

Physics
1 answer:
Stels [109]3 years ago
4 0

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

So

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now by integrating above equation

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

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2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos
Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh

there, no vertical movement

So, on neglect of height term

P_{1}=P+\dfrac{1}{2}\rho v_{1}^2

Where, P= atmospheric pressure

\rho = air density

v = speed

Put the value in the equation

P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)

P_{1}=2126.0\ lb/ft^2

P_{1}=\dfrac{2126.0}{144}

P_{1}= 14.76\ psi

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{2}=P+\dfrac{1}{2}\rho v_{2}^2

Put the value in the equation

P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)

P_{2}=2246.17\ lb/ft^2

P_{2}=\dfrac{2246.17}{144}

P_{2}= 15.59\ psi

(c).  Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{3}=P+\dfrac{1}{2}\rho v_{3}^2

Put the value in the equation

P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)

P_{3}=2257.93\ lb/ft^2

P_{3}=\dfrac{2257.93}{144}

P_{3}= 15.68\ psi

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

5 0
3 years ago
A ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second2. What is the final velocity of the
leva [86]
Used an app called Mephyso to do the calculation.

8 0
3 years ago
Read 2 more answers
A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.
Alex Ar [27]

Good.  You can do some very interesting experiments with that equipment.

3 0
3 years ago
if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be
Kisachek [45]

Answer:

I_{rms}=2.12\ A

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

I=3\sin\theta\ A

We need to find the rms value of the current.

The general equation of current is given by :

I=I_o\sin\theta

It means, I_o=3\ A

We know that,

I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A

So, the rms value of current is 2.12 A.

4 0
3 years ago
If tides occur twice a day in most places. How much time is there between a morning high tide and the next high tide?
algol13

About 12 hours is the time between a morning high tide and the next high tide

Explanation:

The Earth’s rotation happens between two tidal bulges  

The “periodic rise and fall” of the surface water levels of the ocean is called tides. The gravitational action and interaction on the earth by the sun and the moon causes these tides. Different regions of the World experiences different patterns of tides like the diurnal, semi-diurnal etc.

When there is one high and one low tide occurring on a lunar day, then it is diurnal pattern. Semi-diurnal pattern occurs when there are two equal high and low tides on a single lunar day.

Since the Earth’s rotation happens between two tidal “bulges” on each lunar day, the coastal areas can experience two high and two low tides in every 24 hours plus 50 minutes.

Accordingly the time between two high tides would be 12 hours plus 25 minutes. Similarly, the time gap between a high to low tide would be 6 hours plus 12.5 minutes.

8 0
3 years ago
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