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2 years ago

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When an external magnetic flux through a conducting loop decreases in magnitude, a current is induced in the loop that creates i

ts own magnetic flux through the loop. How does that induced magnetic flux affect the total magnetic flux through the loop

1 answer:

Shkiper50 [21]2 years ago

7 0

Answer:

Len's law

Explanation:

We can explain this exercise using Len's law

when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.

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A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.

konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

$W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J$

Learn more about work and kinetic energy:

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2 years ago

Why are all machines not 100% efficient<br> A. Mass<br> B. Gravity <br> C. Friction <br> D. Distance

Anon25 [30]

All machines are not 100% efficient because of <span>C. Friction</span>

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3 years ago

Read 2 more answers

A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children pl

nordsb [41]

Answer:

Please find the complete solution in the attached file.

Explanation:

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2 years ago

Un avión vuela a 10000m de altura y otro a 33300 pies, si un pie equivale a 30.48cm ¿Cuál vuela a mayor altura?

user100 [1]

Answer:

Avion A (10000 meters).

Explanation:

Deje que la altura de los aviones sea A y B respectivamente.

Dados los siguientes datos;

Altura A = 10000 metros

Altura B = 33300 pies

Para encontrar el avión que voló más alto, tendríamos que hacer alguna conversión de unidades.

Conversión:

Metros a centímetros;

1 metro = 100 cm

10000 metros = 100 * 10000 = 1.000.000 centímetros.

Por lo tanto, la altura A en cm = 1,000,000 centímetros

Pies a centímetros;

1 pie = 30,48 centímetros

33300 pies = 33300 * 30,48 = 1014984 centímetros.

Por lo tanto, la altura B en cm = 1014984 centímetros.

De los cálculos anteriores, podemos deducir que el avión A voló más alto.

6 0

2 years ago

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

kakasveta [241]

Answer:

$W=4037.36\ J$

Explanation:

Given:

mass of ice melted, $m=3.3\times10^{-2}\ kg$

time taken by the ice to melt, $t=5\ min=300\ s$

latent heat of the ice, $L=3.34\times 10^5\ J$

Now the heat rejected by the Carnot engine:

$Q_R=m.L$

$Q_R=0.033\times 3.34\times 10^5$

$Q_R=11022\ J$

Since we have boiling water as hot reservoir so:

$T_H=373\ K$

The cold reservoir is ice, so:

$T_L=273\ K$

Now the efficiency:

$\eta=1-\frac{T_L}{T_H}$

$\eta=1-\frac{273}{373}$

$\eta=26.81\%$

Now form the law of energy conservation:

Heat supplied:

$Q_S-W=Q_R$

where:

$Q_S=$ heat supplied to the engine

$Q_S-\eta\times Q_S=Q_R$

$Q_S(1-\eta)=Q_R$

$Q_S=\frac{11022}{1-0.2681}$

$Q_S=15059.36\ J$

Now the work done:

$W=Q_S-Q_R$

$W=15059.36-11022$

$W=4037.36\ J$

8 0

2 years ago