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Katyanochek1 [597]
3 years ago
9

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

string. The load is not swinging, and the string is observed to make a constant angle of 45∘ with the horizontal. No other forces are acting on the load. Which of the following statements are correct? Check all that apply. 1 The train is an inertial frame of reference. 2 The train is not an inertial frame of reference. 3 The train may be instantaneously at rest. 4 The train may be moving at a constant speed in a straight line. 5 The train may be moving at a constant speed in a circle. 6 The train must be speeding up. 7 The train must be slowing down. 8 The train must be accelerating.
Physics
1 answer:
Triss [41]3 years ago
3 0

Here the weight is hung from the thread inside the ceiling of train

now the FBD of train is analysed

We can say

Tcos\theta = mg

Tsin\theta = ma

now by ratio of two equations we have

a = gtan\theta

this shows that train must be in accelerating condition due to which the thread is making an angle with the vertical

now we will have

2 The train is not an inertial frame of reference.

Since train is accelerating so it must be non inertial frame

5 The train may be moving at a constant speed in a circle.

Since its moving in circle so it is in accelerating condition

8 The train must be accelerating.

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The formula for the energy in a capacitor , u in terms of q and c is q²/2c

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The energy of a capacitor u = 1/2qv where

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Also, the capacitance of a capacitor c = q/v where

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So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

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= q²/2c

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96m

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A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .
Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}

Taking out constants from the integral:
B =\frac{\mu_0 i}{4\pi R^2}  \int ds

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

B =\frac{\mu_0 i}{4\pi R^2}  \int\limits^{2\pi R}_0 \, ds

Evaluate:
B =\frac{\mu_0 i}{4\pi R^2}  (2\pi R- 0) = \frac{\mu_0 i}{2R}

Plugging in our givens to solve for the magnetic field strength of one loop:

B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T

Multiply by the number of loops to find the total magnetic field:
B_T = N B = 0.00631 = \boxed{6.318 mT}

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.  

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}

Using the diagram, if 'z' is the point's height from the center:

r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}

Substituting this into our expression:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds

Evaluate:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Multiplying by the number of loops:
B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Plug in the given values:
B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ =  0.00006795 = \boxed{67.952 \mu T}

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