Should i make my pet shark bark like a dog woof woof

Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 m

Rufina [12.5K]

<h2>Answer:</h2>

143μH

<h2>Explanation:</h2>

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

<em>From the question;</em>

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

<em>Substitute these values into equation (i) as follows;</em>

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

6 0

3 years ago

What is the SI Unit for amplitude?

seropon [69]

This distance is known as the amplitude of the wave, and is the characteristic height of the wave, above or below the equilibrium position. Normally the symbol A is used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).

4 0

4 years ago

A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?

jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina ( $Q$ ), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

$Q = \frac{\pi}{4}\cdot D^{2}\cdot v$ (1)

Donde:

$D$ - Diámetro de la manguera, medido en pies.

$v$ - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que $D = \frac{1}{12}\,ft$ y $v = 5\,\frac{ft}{s }$ , entonces el gasto de gasolina es:

$Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)$

$Q \approx 0.0273\,\frac{ft^{3}}{s}$

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0

3 years ago

A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.

ANTONII [103]

<u>Answer:</u> The number of photons are $3.7\times 10^8$

<u>Explanation:</u>

We are given:

Wavelength of microwave = $1.22\times 10^8nm=0.122m$ (Conversion factor: $1m=10^9nm$ )

To calculate the energy of one photon, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = 0.122 m

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

4 0

3 years ago

A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface

Blizzard [7]

Answer:

$\rho = 7.35\times \muC/m^3$

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

$E = \dfrac{Rl}{2\epsilon_0}$

ε₀ is the permitivity of free space

R is the radius of the rod

and also,

$\rho=\dfrac{2E\epsilon_0}{R}$

ρ is the volume charge density

$\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}$

$\rho = 7.35\times 10^{-6}\ C/m^3$

$\rho = 7.35\times \muC/m^3$

Hence, the volume charge density is equal to $\rho = 7.35\times \muC/m^3$

6 0

3 years ago

Should i make my pet shark bark like a dog woof woof​

Should i make my pet shark bark like a dog woof woof