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anastassius [24]
2 years ago
12

How has completing the advance cross- training program affective you in a positive way?

Physics
1 answer:
kondaur [170]2 years ago
8 0

Hakeem fees ya utensilios just hat

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What does a trench form?What happens here?
Vinvika [58]
A hole, and people die if they fall in there.
8 0
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}

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
I need help, ASAP, I know what thermal energy is, but how does a skater use it when on the track?
Flauer [41]

Answer: Looked it up but

Explanation:

When the skater lands on the track, the vertical component of his kinetic energy is converted to thermal energy. You can do experiments where there is no loss to thermal energy (only PE and KE conversions) by turning friction off and by making sure the skater doesn't leave the track.

4 0
3 years ago
You place a solid cylinder of mass M on a ramp that is inclined at an angle β to the horizontal. The coefficient of static frict
zlopas [31]

Answer:

Explanation:

Let the frictional force required be f .

frictional force is responsible in creating rotational motion in the cylinder.

torque created by frictional force = f R

if angular acceleration be α

α = f R / I , I is moment of inertia of cylinder .

α = a / R , a is linear acceleration.

f R / I = a / R

a = f R² / I

linear acceleration a of cylinder down the slope

ma = mgsinθ - f  , ( f force is acting upwards and mgsinθ is acting downwards )

mf R² / I = mgsinθ -f

f ( m R² / I + 1) = mgsinθ

f = mgsinθ / ( m R² / I + 1)

= mgsinθ / ( m R² / mk² + 1) , k is radius of gyration of cylinder.

= mgsinθ / (  R² / k² + 1)

Putting the given values

f = Mgsinβ /(  R² / k² + 1)

for cylinder ,  R² / k² = 2

f =  Mgsinβ /3

6 0
3 years ago
Doppler effect is caused by sound or light wave being pushed together and spread apart due to motion. True or False.
pychu [463]

Answer: false.

Explanation:

Doppler effect is caused to the relative motion of the source of light/sound with respect to the observer.

If the source is moving towards you, you will perceive the frequency to be larger (also called a blue shift), while if the wave is moving away, you will perceive the frequency to be smaller (also called a red shift).

Then the statement:

"Doppler effect is caused by sound or light wave being pushed together and spread apart due to motion."

Is false

3 0
2 years ago
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