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4 years ago

7

The magnetic flux that passes through one turn of a 18-turn coil of wire changes to 2.67 from 8.19 Wb in a time of 0.0386 s. The

average induced current in the coil is 275 A. What is the resistance of the wire?

1 answer:

Alik [6]4 years ago

3 0

I believe that the resistance of the wire is 12

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Problems with solar energy include _____.

german

First choice: the inability of current technology to capture
large amounts of the Sun's energy

Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low. But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.

Second choice: the inability of current technology to store
captured solar energy

No. We're pretty good at building batteries to store small amounts, or raising water to store large amounts. Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.

Third choice: inconsistencies in the availability of the resource

I think this is it. If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
of luck during long periods of overcast skies.

Fourth choice: lack of demand for solar energy

If there is a lack of demand, it's purely a result of willful manipulation
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3 years ago

Read 2 more answers

Importance of choke coil?<br><br>

Verdich [7]

Answer:The choke coil works because it can act as an inductor. When the current pass through will change as AC currents creates a magnetic field in the coil that works against that current. This is known as inductance and blocks most of the AC current from passing through.

Explanation:

8 0

2 years ago

PLEASE HELP ME I HAVE BEEN DOING THIS FOR HOURS!! How do some carnivorous plants avoid beginning the digestion process from an a

konstantin123 [22]

Answer:

Once a carnivorous plant has procured an item for dinner, it has to have some way to turn it into fertilizer. What carnivorous plants do is very similar to what humans do with their dinner after they have eaten it. Most carnivorous plants have glands that secrete acids and enzymes to dissolve proteins and other compounds. The plants may also enlist other organisms to help with digestion. The plants then absorb the nutrients made available from the prey.

Drosera releases digestive juices through the glands at the tip of its tentacles and absorbs the nutrients through the tentacles, leaf surface, and sessile glands. In order to do this it bends its tentacles and rolls or bends the leaf to get as many tentacles as possible into contact with the prey for digestion and to make as much leaf surface available for absorption. Its relative Drosophyllum has differently structured, non moving tentacles and doesn't use them directly for digestion. Instead it has specialized glands on the surface of the leaf that release the digestive enzymes (see Carniv. Pl. Newslett. 11(3):66-73 ( PDF ) for drawings and discussion).

The sealed trap of Dionaea does digestion in a way similar to the leaf surface digestion carnivores—upon capture of a prey, digestive enzymes in mucous are released. The advantage of the sealed trap of Dionaea is rain won't wash away the nutrients as digestion proceeds.

The sealed trap carnivores Aldrovanda and Utricularia already have water in their traps so they only need to release enzymes. Utricularia appears to release the enzymes continuously into its traps.

The other carnivorous plants use either a mixed mode of digestive enzymes and partner organisms (Genlisea, Sarracenia, most Nepenthes, Cephalotus, some Heliamphora, Roridula) or other organisms exclusively for digestion (most Heliamphora, some Nepenthes, Darlingtonia). Part of the reason for partnering with other organisms is that the plants actually have little choice in the matter. This could also be a factor for the leaf surface and sealed trap digesters as well. The prey will have gut flora that are quite capable of digesting their host when it dies. In addition, insect larvae, frog tadpoles, and predacious protozoans will or will attempt to take up residence in water-filled traps. The plant releasing digestive enzymes and acids into the traps will help tip the nutrition balance to themselves, but there are limits.

Explanation:

6 0

3 years ago

A student is standing 8 m from a roaring truck engine that is measured at 20 dB. The student moves 4 m closer to

Ipatiy [6.2K]

Answer:80 dB

Explanation:

5 0

3 years ago

Two people stand facing each other at roller skating rink then push off each other

9966 [12]

a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

a)

The momentum of an object is a vector quantity given by:

$p=mv$

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

$p=p_1 + p_2$

Which can be rewritten as

$p=m_1 u_1 + m_2 u_2$

where $m_1,m_2$ are the masses of the two people and $u_1,u_2$ their initial velocities.

However, the two people are initially at rest, so

$u_1 = 0\\u_2 = 0$

Therefore the total momentum is

$p=0+0=0$

b)

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

$p_i = p_f$

where

$p_i$ is the total momentum of the system before

$p_f$ is the total momentum of the system after

In this problem,

$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

$p_f = p_i = 0$

So the total momentum is zero also after they push off each other.

c)

The total momentum of the girl and the boy after they push off each other can be written as:

$p_f = m_1 v_1 + m_2 v_2$ (1)

where:

$m_1 = 30 kg$ is the mass of the girl

$v_1 = -5 m/s$ is her velocity (she moves backward, so the negative sign)

$m_2 = 50 kg$ is the mass of the boy

$v_2$ is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

$p_f = 0$ (2)

By combining eq(1) and eq(2) we get

$0=m_1 v_1 + m_2 v_2$

And solving for v2 we find the velocity of the boy:

$v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s$

and the positive sign means he is moving forward.

d)

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

$\Delta p = F\Delta t$

where

$\Delta p$ is the change in momentum

F is the force

$\Delta t$ is the time during which the force is applied

In this problem:

$\Delta t = 2.5 s$

For the boy, the change in momentum is:

$\Delta p = m_2 (v_2 - u_2)$

And since

$m_2 = 50 kg\\u_2 = 0 m/s\\v_2 = 3 m/s$

We have

$\Delta p = (50)(3-0)=150 kg m/s$

So, the force exerted between the boy and the girl is:

$F=\frac{\Delta p}{\Delta t}=\frac{150}{2.5}=60 N$

8 0

3 years ago