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4 years ago

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Five loops are formed of copper wire of the same gauge (cross-sectional area). Loops 1-4 are identical; Loop 5 has the same heig

ht as others but is longer. At the instant shown, all the loops are moving at the same speed in the directions indicated.
There is a uniform magnetic field pointing out of the page in region I; in Region II there is no magnetic field. Ignore any interactions between the loops.

A) For any loop that has an induced current, indicate the direction of the current.

B) Rank the magnitude of the emfs around the loops. Explain your reasoning.

C) Rank the magnitudes of the currents in the loops. Explain your reasoning.

1 answer:

nata0808 [166]4 years ago

3 0

The answer to this question is A

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Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de

ahrayia [7]

Answer:

a) El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

$\Delta z = \frac{v_{out}^{2}}{2\cdot g}$

Donde:

$\Delta z$ - Diferencia de altura, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

$v_{out} = \sqrt{2\cdot g \cdot \Delta z}$

Si $g = 9.807\,\frac{m}{s^{2}}$ y $\Delta z = 0.3\,m$ , entonces la rapidez de salida del chorro es:

$v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}$

$v_{out} \approx 2.426\,\frac{m}{s}$

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

$\dot V_{out} = v_{out}\cdot A_{t}$

Donde:

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

$A_{t}$ - Área transversal del orificio, medido en metros cuadrados.

$\dot V_{out}$ - Caudal de salida del chorro, medido en metros cúbicos por segundo.

Dado que $v_{out} = 2.426\,\frac{m}{s}$ y $A_{t} = 5\,cm^{2}$ , el caudal de salida del chorro es:

$\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)$

$\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}$

El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

5 0

3 years ago

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart

Ber [7]

Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
$\sum F = ma$
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force $F_f$ acting in the opposite direction. So Newton's second law can be rewritten as
$F-F_a = ma$
or
$F=ma + F_f$

since the frictional force is 15 N and we want to achieve an acceleration of $a=1.50 m/s^2$ , we can substitute these values to find what is the force the man needs:
$F=(30 kg)(1.5 m/s^2)+15 N=60 N$

8 0

4 years ago

Answer all 3 parts of this essay question: A) (5pts) If you have a Periodic Table that is NOT color coded, describe where to loo

cestrela7 [59]

Answer:

A) If you look at the vertical rows you can tell how many electrons there are on the outer shell eg. Group 7 (eg. nitrogen) has 7 electrons on its outer shell

B) They are in the same group

C) Helium,Argon and Organessom are all Nobel gasses

8 0

4 years ago

Is oxygen a reactant or product

dezoksy [38]

Answer:

Methane and oxygen (oxygen is a diatomic — two-atom — element) are the reactants, while carbon dioxide and water are the products

Explanation:

3 0

3 years ago

Read 2 more answers

A capacitor is charged to a potential difference of 12v it delivers 40% of its stored energy to a lamp what is the final potenti

Korolek [52]

Call the capacitance C.
<span>Note the energy in a capacitor with voltage V is E =½CV². </span>

<span>Initial energy = ½C(12)² = 72C </span>

<span>40% of energy is delivered, so 60% remains.in the capacitor. </span>
<span>Remaining energy = (60/100) x 72C =43.2C </span>

<span>If the final potential difference is X, the energy stored is ½CX² </span>
<span>½CX² = 43.2C </span>
<span>X² = 2 x 43.2 = 86.4 </span>
<span>X = 9.3V</span>

6 0

3 years ago