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3 years ago

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We have a long wire with a circular cross section and radius a = 2.40 cm. The current density in this wire is uniform, with a to

tal current of 3.00 A. Find the magnitude of the magnetic field at a distance of 0.72 cm from the center axis. Treat the wire as a cylinder.

1 answer:

Alika [10]3 years ago

7 0

Answer:

The magnitude of the magnetic field is 7.49x10⁻⁶T

Explanation:

The magnetic field is:

$B=(\frac{\mu i}{2\pi R^{2} } )r$

Where

i = current = 3 A

R = radius = 2.4 cm = 0.024 m

r = distance = 0.72 cm = 7.2x10⁻³m

Replacing:

$B=(\frac{4\pi x10^{-7} *3}{2\pi *0.024^{2} } )*7.2x10^{-3}= 7.49x10^{-6} T$

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the force between a planet and a spacecraft is 1 million newtons. whst will the force be iv the spacecraft moves to halfg its or

dybincka [34]

The forces of gravity between two objects are inversely proportional to
the square of the distance between them. So reducing the distance
by 1/2 means increasing the gravitational force by 2² = 4 times.
The 1 million newtons becomes 4 million newtons.

Note that this does NOT mean the satellite's altitude above the surface.
When you're calculating gravitational forces, it's the distance between
the centers of the objects. So the question is a meaningful exercise
only if we use the distance between the satellite and the planet's center.

6 0

3 years ago

Determine la resistencia equivalente de la "escalera" de

defon

Las respuestas a cada inciso son:

a) La resistencia equivalente de la "escalera" de resistores iguales de 125 Ω que se muestra en la figura adjunta es:

$R_{t} = 341.7 \: \Omega$

b) La corriente a través de cada uno de los tres resistores de la izquierda si se conecta una batería de 50.0 V entre los puntos A y B es:

Correspondiente a R8 y R9 es 0.23 A
Correspondiente a R7 es 0.17 A.

a) En la imagen adjuntada correspondiente a la Figura 26-40, podemos observar que las resistencias 1, 2 y 3 están en serie, por lo tanto la ressitencia equivalente entre estas 3 es:

$R' = R_{1} + R_{2} + R_{3}$

De aquí en adelante tendremos presente que las todas las resistencias son iguales entre sí y por ende igual a 125 Ω. Las notaciones del 1 al 9 son para poder mostrar la resolución del problema.

Entonces:

$R' = 3R$

Ahora, esta resistencia está en paralelo con la resistencia R₄, por lo tanto la resistencia equivalente entre estas dos es:

$\frac{1}{R''} = \frac{1}{R'} + \frac{1}{R_{4}} = \frac{1}{3R} + \frac{1}{R} = \frac{4R}{3R^{2}}$

$R'' = \frac{3}{4}R$

Luego, esta resistencia está en serie con las resistencias R₅ y R₆, por lo tanto:

$R''' = R'' + R_{5} + R_{6} = \frac{3}{4}R + 2R = \frac{11}{4}R$

Esta resistencia está ahora en paralelo con R₇, entonces:

$\frac{1}{R''''} = \frac{1}{R'''} + \frac{1}{R_{7}} = \frac{4}{11R} + \frac{1}{R} = \frac{15R}{11R^{2}}$

$R'''' = \frac{11}{15}R$

Finalmente, esta resistencis está en serie con las resistencias R₈ y R₉, por lo tanto la resistencia total es:

$R_{t} = R'''' + R_{8} + R_{9} = \frac{11}{15}R + 2R = \frac{41}{15}R = \frac{41}{15}*125 \: \Omega = 341.7 \: \Omega$

b) Para este inciso debemos usar la Ley de Kirchhoff, pues tenemos tres mallas. Supondremos que las corriente de cada malla fluiran en sentido horario, por lo tanto las ecuaciones de para cada malla serán:

Malla 1

Segun la ley de Ohm tenemos:

$V-i_{1}R-i_{3}R=0$ (1)

Malla 2

$-i_{2}R-i_{5}R-i_{2}R+i_{3}R=0$ (2)

Malla 3

$-i_{4}R-i_{4}R-i_{4}R+i_{5}R=0$ (3)

Recordemos tambien que:

$i_{1}=i_{2}+i_{3}$ (4)

$i_{2}=i_{4}+i_{5}$ (5)

Lo que debemos hacer ahora es resolver el sistema de ecuaciones y encontrar los valore de las corrientes. Por lo tanto, los valores de las corrientes serán:

$i_{1}=3/13\: A$

$i_{2}=4/65\: A$

$i_{3}=11/65\: A$

$i_{4}=1/65\: A$

$i_{5}=3/65\: A$

Finalmente:

La corriente correspondiente a R8 y R9 es 0.23 A
La corriente correspondiente a R7 es 0.17 A.

Pudes aprender mas de mallas aquí:

https://brainly.lat/tarea/11593276

4 0

3 years ago

Pls someone I need it urgently and explain Solving and explanation so I can understand Thank you

Temka [501]

Answer:

f = 6.37 Hz, T = 0.157 s

Explanation:

The expression you have is

y = 5 sin (3x - 40t)

this is the equation of a traveling wave, the general form of the expression is

y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

w = 2π f

f = w / 2π

from the equation w = 40 rad / s

f = 40 / 2π

f = 6.37 Hz

frequency and period are related

f = 1 / T

T = 1 / f

T = 1 / 6.37

T = 0.157 s

4 0

3 years ago

Enter a value of 3/4 of a full rotation

solmaris [256]

Answer:

270°

Explanation:

A full rotation is 360° so half (2/4) of it is 180° meaning a quarter (1/4) is 90°

So you can do 90+90+90=270

or 180+90=270

6 0

3 years ago

Calculate the Reynolds number for an oil gusher that shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter

Alenkasestr [34]

Answer:

$Re=1992.24$

Explanation:

Given:

vertical height of oil coming out of pipe, $h=25\ m$

diameter of pipe, $d=0.1\ m$

length of pipe, $l=50\ m$

density of oil, $\rho = 900\ kg.m^{-3}$

viscosity of oil, $\mu=1\ Pa.s$

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

<u>Using the equation of motion:</u>

$v^2=u^2-2gh$

where:

v = final velocity

u = initial velocity

Putting the respective values:

$0^2=u^2-2\times 9.8\times 25$

$u=22.136\ m.s^{-1}$

<u>For Reynold's no. we have the relation as:</u>

$Re=\frac{\rho.u.d}{\mu}$

$Re=\frac{900\times 22.136\times 0.1}{1}$

$Re=1992.24$

6 0

3 years ago