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belka [17]
3 years ago
5

The law of conservation of mass states that in a chemical reaction matter is not created or destroyed true or false

Physics
1 answer:
lisov135 [29]3 years ago
4 0
True it only changes form. Like when water evaporates goes from liquid to a gas and is not destroyed.
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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
2 years ago
Read 2 more answers
A mass of 5kg accelerates at 3m/s/s, how much force was put on it?
goldfiish [28.3K]

Answer:

15N

Explanation:

According to Newton's Second Law of Motion

F = m*a

mass = m = 5Kg

acceleration = a = 3m/s^2

=> F = 5kg * 3m/s^2

=> F = 15 N

5 0
3 years ago
Read 2 more answers
If a 3.5 gram ping pong ball were traveling to the right horizontally at 12 m/s, and a larger 12 g super ball were thrown direct
algol [13]

Answer:

v = 14.32 m/s

Explanation:

According to the principle of conservation of linear momentum, both the momentum and kinetic energy of the system are conserved. Since the two balls are in the same direction of motion before collision, then;

m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v

0.035 × 12 + 0.120 × 15 = (0.035 + 0.120) v

0.420 + 1.800 = (0.155) v

2.22 = 0.155 v

⇒ v = \frac{2.22}{0.155}

      = 14.323

The velocity of the balls after collision is 14.32 m/s.

3 0
3 years ago
Which body could NOT be called a gas giant?
Dafna11 [192]
Sorry I don’t know but good luck and hope you find it
4 0
3 years ago
Read 2 more answers
!Necesito ayuda:Un movil parte del reposo y en 5seg adquiere una rapidez de 12m/seg.Calcular la distancia recorrida.
KatRina [158]

Answer:

Distancia = 60 metros

Explanation:

Dados los siguientes datos;

Velocidad inicial = 0 m/s (ya que comienza desde el reposo)

Velocidad final = 12 m/s

Por lo tanto, velocidad total = velocidad inicial + velocidad final

Velocidad total = 0 + 12 = 12 m/s

Tiempo = 5 segundos

Para encontrar la distancia recorrida;

Distancia = velocidad total * tiempo

Sustituyendo en la fórmula, tenemos;

Distancia = 12 * 5

Distancia = 60 metros

7 0
3 years ago
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