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gayaneshka [121]
3 years ago
7

Define electric current and drift velocity.

Physics
2 answers:
Luba_88 [7]3 years ago
8 0

Answer:

An electric current is the stream of changed particals, such as electrons & ions, moving through an electrical conductor.

The average velocity attained bycharged partical ,such as electrons,in a material due to electric fields

Blizzard [7]3 years ago
3 0

Answer:

Current- the flow of free charges, such as electrons and ions

Drift velocity- the average speed at which these charges move

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A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
mario62 [17]

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0
3 years ago
Why is Pluto now called a dwarf planet
zvonat [6]
Before Pluto was discovered, it was predicted. Astronomers had observed that massive objects can affect the orbits of its neighbors, and, after seeing deviations in the orbits of Uranus and Neptune, assumed something substantial existed beyond their orbits.
When Pluto was spotted, it was thought to be the predicted object and was identified as a ninth planet.
A few decades later, astronomers started discovering more and more objects around other stars and didn’t know whether to call them planets or not. There appeared to be a need to define what a planet means, and that led to what some people consider Pluto’s demotion to a dwarf planet.
The International Astronomical Union decided that full-sized planets must orbit the sun, have a round shape, and have cleared their orbits of other objects. Pluto fulfills the first two criteria, but not the third.
It still goes around the sun, it’s round enough, it’s got moons, and behaves like a planet, but the idea is that Pluto did not form the same way as the rest of the planets. Pluto’s orbit is both eccentric and inclined more than the rest of the planets by about 17 degrees. That’s suggests something is different about this object.
This debate about whether to call it a planet or not is silly, because it doesn’t matter to Pluto what you call it. It is an interesting object, goes around the sun, and shows geology and an atmosphere.
There’s a tendency to define objects based on what they are now, but nothing is constant in the universe. There are some issues with the nomenclature, and a definition today may not apply to the same object tomorrow.
7 0
3 years ago
Read 2 more answers
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
3 years ago
A playground merry-go-round spins about its axis with negligible friction. A child moves from the center of the merry-go-round t
Juli2301 [7.4K]

Answer:

B

Explanation:

In this process on merry go round there is not any external torque so angular momentum will be conserve. Mass is always conserved.

4 0
2 years ago
You would convert from grams to moles by using the ___________ of the substance.
Firdavs [7]

You would convert from grams to moles by using the molar mass of the substance. The answer is letter B. for example, the molar mass Carbon dioxide is 44.01 g/mol. It means that for 1 mole of carbon dioxide, it contains 44.01 grams of Carbon dioxide.

7 0
3 years ago
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