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3 years ago

A person suffering from anaemia gets tired after a short walk

2 answers:

6 0

Yes, again, it is a blood disorder, making oxygen travel to the brain difficult, especially when exasperated such in physical activities.

FrozenT [24]3 years ago

5 0

Answer:

yes

Explanation:

you will feel weary after shorter times

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According to Newton's law of universal gravitation, which statement is true?

valkas [14]

Answer:

Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

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3 years ago

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Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia

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Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

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3 years ago

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:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

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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1 year ago

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Which planets are visible without a telescope

Annette [7]

The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.

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3 years ago

A gas is compressed from an initial volume of 5.75 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the

nataly862011 [7]

Answer:

<em>The internal energy change is 330.01 J</em>

Explanation:

Given

$v_{1}$ the initial volume = 5.75 L

$v_{2}$ the final volume = 1.23 L

$P_{e}$ is the external pressure = 1.00 atm

q the heat energy removed = -128 J (since is removed from the system)

expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;

W = - $P_{e}$ ΔV

W = -1.00 x(1.23 - 5.75)

W = -1.00 x -4.52

W = 4.52 L atm

converting to joules we have

W = 4.52 L atm x 101.33 J/ L atm = 458.01 J

The internal energy change during compression can be calculated thus;

ΔU = q + W

ΔU = -128 J + 458.01 J

ΔU = 330.01 J

Therefore the internal energy change is 330.01 J

8 0

3 years ago

A person suffering from anaemia gets tired after a short walk​

A person suffering from anaemia gets tired after a short walk