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3 years ago

12

Weather balloons are filled with only a small amount of helium because the ____________________ of the balloon will increase as

the air pressure decreases at higher altitudes.

1 answer:

Shkiper50 [21]3 years ago

7 0

Weather balloons are filled with only a small amount of helium because the __Volume__. of the balloon will increase as the air pressure decreases at higher altitudes.

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What is the the steadiest firing position?

Paha777 [63]

The closer you are to the ground the more accurate you'll be. That's why most snipers are in the "prone" position.

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3 years ago

The gold has a density of 19300 kg/m3 calculate the mass of one gold bar 1= 2.54cm

icang [17]

Fair enough, but you'll have to tell us the volume of the bar first.

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3 years ago

Five groups of four vectors are shown below. All magnitudes of individual vectors are equal. Please rank the groups based on the

valkas [14]

With the addition of vectors we can find that the correct answer is:

C) Q> P > R = S > T

The addition of vectors must be done taking into account that they have modulus and direction. The analytical method is one of the easiest methods, the method to do it is:

Set a Cartesian coordinate system

Decompose vectors into their components in a Cartesian system

Perform the algebraic sums on each axis

Find the resultant vector using the Pythagoras' Theorem to find the modulus and trigonometry to find the direction.

In this exercise indicate that the modulus of all vectors is the same, suppose that the value of the modulus is A.

We fix a Cartesian coordinate system with the horizontal x axis and the vertical y axis, we can see that we do not need to perform any decomposition, so we perform the algebraic sums

Diagram P

x-axis

x = 2A

y-axis

y = 2A

The modulus of the resulting vector can be found with the Pythagorean Theorem

P = $\sqrt{x^2+y^2}$

P = $\sqrt{4A^2 +4A^2 }= \sqrt{8} \ A$

P = 2 √2 A

Diagram Q

x-axis

x = 3A

y-axis

y = A

Resulting

Q = $\sqrt{x^2+y^2}$

Q = $\sqrt{9A^2 + A^2 }$

Q = $\sqrt{10} \ A$

Diagram R

x- axis

x = 0

y-axis

y = 2 A

Resulting

R = $\sqrt{4A^2 + 0}$

R = $\sqrt{4} \ A$

Diagram S

x-axis

x = 2 A

y-axis

y = 0

Resulting

S = 2A

Diagram T

x- axis

x = 0

y-axis

y = 0

Resultant T = 0

We order the diagram from highest to lowest

Q> P> R = S> T

When reviewing the different answers, the correct one is:

C. Q> P> R = S> T

Learn more about adding vectors here:

brainly.com/question/14748235

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1 year ago

Radio waves transmitted through space at 3.00 ✕ 108 m/s by the Voyager spacecraft have a wavelength of 0.137 m. What is their fr

fredd [130]

Answer:

Frequency, $f=2.18\times 10^9\ Hz$

Explanation:

We have,

Speed of radio waves is $3\times 10^8\ m/s$

Wavelength of radio waves is $\lambda=0.137\ m$

It is required to find the frequency of the radio waves. The speed of a wave is given by :

$v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz$

So, the frequency of the radio wave is $2.18\times 10^9\ Hz$ .

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2 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:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

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

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