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Ksenya-84 [330]

2 years ago

6

A truck was carrying a substance in a tank. The molecules of that substance were moving away from each other. The truck parked o

vernight in a place where energy transferred out of the substance. In the morning, the substance was a gas. How were the molecules moving in the morning? Explain why the molecules were moving that way after energy was transferred out of them.
(Make it ask long as you need)
Or (short)

1 answer:

DochEvi [55]2 years ago

5 0

Answer:

In the morning the molecules were moving away from each other with a smaller speed than when the truck was carrying the substance.

Explanation:

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

5 0

1 year ago

Read 2 more answers

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b

nekit [7.7K]

<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=\frac{1}{T}\intop_{0}^{T}p(t)dt$

But:

$v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)$

So:

$P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}$

$P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}$

In terms of RMS values:

$V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI$

7 0

3 years ago

A 72.0 kg stuntman jumps from a moving car to a 2.50 kg skateboard at rest. If the velocity of the car is 15.0 m/s to the east w

yawa3891 [41]

Answer:

B 14.5 m/s to the east

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, if the system is isolated, the total momentum of the system must be conserved.

Here the total momentum before the stuntman reaches the skateboard is:

$p_i = Mv$

where

M = 72.0 kg is the mass of the stuntman

v = 15.0 m/s is his initial velocity (to the east)

The total momentum after the stuntmen reaches the skateboard is:

$p_f = (M+m)v'$

where

m = 2.50 kg is the mass of the skateboard

v' is the final velocity of the stuntman and the skateboard

Since momentum must be conserved, we have

$p_i = p_f\\Mv=(M+m)v'$

And solvign for v',

$v'=\frac{Mv}{M+m}=\frac{(72.0)(15.0)}{(72.0+2.50)}=14.5 m/s$

And since the sign is the same as v, the direction is the same (to the east).

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3 years ago

Which increases the work output of a machine?

Dafna11 [192]

A reducing friction

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What kind of pollution causes nitrogen and phosphorous to enter bodies of water?

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Nutrient pollution. I just took the quiz and got it right

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