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3 years ago

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Physical

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zysi [14]3 years ago

4 0

Answer: 2.5

Explanation:

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Marvin uses a long copper wire with resistivity 1.68 x 10^-8 Ω⋅m and diameter 1.00 x 10^−3 m to create a solenoid that has a 3

Mazyrski [523]

Answer with Explanation:

We are given that

Resistivity of copper wire= $\rh0=1.68\times 10^{-8}\Omega m$

Diameter=d= $1.00\times 10^{-3} m$

Radius of copper wire= $r=\frac{d}{2}=\frac{1}{2}\times 10^{-3} m$

Radius of solenoid=r' $=3 cm=3\times 10^{-2} m$

1 m=100 cm

a.Length of wire=l=11.3 m

Area of wire=A= $\pi r^2$

Where $\pi=3.14$

A= $3.14\times (\frac{1}{2}\times 10^{-3})^2$

Resistance, R= $\rho \frac{l}{A}$

Using the formula

$R=1.68\times 10^{-8}\times\frac{11.3}{3.14\times (\frac{1}{2}\times 10^{-3})^2}$

$R=0.24\Omega$

B.Length of solenoid= $2\pi r'=2\times 3.14\times 3\times 10^{-2}=0.188$ m

Number of turns= $n_0=\frac{l}{2\pi r'}=\frac{11.3}{0.188}$

$n_0$ =60

C.Potential difference,V=3 V

Current,I= $\frac{V}{R}$

I= $\frac{3}{0.24}=12.5 A$

D.Total length =0.1 m

Number of turns per unit length,n= $\frac{60}{0.1}=600$

Magnetic field along central axis inside of the solenoid,B= $\mu_0 nI$

$B=4\pi\times 10^{-7}\times 12.5\times 600=9.42\times 10^{-3} T$

4 0

3 years ago

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¿Qué diferencia una magnitud fundamental de una derivada?

artcher [175]

Answer:

Explanation:

Una magnitud fundamental es aquella que se define por si misma y es independiente de las demás (masa, tiempo, longitud, etc.). magnitud derivada. Una magnitud derivada es aquella que se obtiene mediante expresiones matemáticas a partir de las magnitudes fundamentales (densidad, superficie, velocidad).

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3 years ago

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What is the speed of sound dependent on?

mojhsa [17]

Depends on the elasticity and density of the medium through what it is traveling <span> </span>

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3 years ago

A 20.0-µF capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280

natali 33 [55]

Answer

given,

Capacitance of capacitor = 20.0-µF

Voltage = 150.0-V

inductance = 0.280 m H

a) the oscillation frequency of circuit

$f = \dfrac{1}{2\pi}\ sqrt{\dfrac{1}{LC}}$

$f = \dfrac{1}{2\pi}\ sqrt{\dfrac{1}{0.280 \times 10^{-3}\times 20 \times 10^{-6}}}$

f = 2126.9 Hz

b) $U = \dfrac{1}{2}CV^2$

$U = \dfrac{1}{2}(20\times 10^{-6})(150)^2$

U = 0.225 J

c)Current in the inductor

$V = L \dfrac{dI}{dt}$

$150 = 0.28 \times 10^{-3} \dfrac{dI}{dt}$

$\dfrac{dI}{dt} = 535714$

instantaneous rate of change of current is equal to 535714 A/s

7 0

3 years ago

Helpppppppp please !!!!

agasfer [191]

Answer:

second one is correct that is right

7 0

4 years ago