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2 years ago

What is the difference between Drift current & Diffusion current ?

1 answer:

dsp732 years ago

8 0

The main difference between the drift current and diffusion current is the corresponding reaction of the semiconductor materials in the electric fields. In electric fields, when semiconductor material's movement go along with the direction of the electric field it is called drift current, but when it tends to go from high concentrated area to lower concentrated area then it is diffusion current.

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Two astronauts are floating together with zero speed in a gravity-free region of space. The mass of astronaut A is 110 kg and th

Kobotan [32]

Answer:

The recoil speed of Astronaut A is 0.26 m/s.

Explanation:

Given that,

Mass of astronaut A, $m_A=110\ kg$

Mass of astronaut B, $m_B=74\ kg$

Astronaut A pushes B away, with B attaining a final speed of 0.4, $v_B=0.4\ m/s$

We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :

$m_Av_A=m_Bv_B\\\\v_A=\dfrac{m_Bv_B}{m_A}\\\\v_A=\dfrac{74\times 0.4}{110}\\\\v_A=0.26\ m/s$

So, the recoil speed of Astronaut A is 0.26 m/s.

4 0

2 years ago

What is the relationship between the frequency and the pitch of a sound?

jarptica [38.1K]

The pitch of a sound come from the frequency of the soundwave, soundwaves that have bigger frequencies sound with a high pitch and soundwaves with less frequencie have low pitch.

6 0

2 years ago

A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of th

Tamiku [17]

Answer:

the angle is about 67.79 degrees

Explanation:

We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)

We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

$half\,\,max-height = \frac{v_{yi}^2}{4\,g}$

we can use this information to find the y component of the velocity at that height via the formula:

$v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}$

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

$1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\$

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

$tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o$

3 0

2 years ago

What other structures is it near hypothalamus?

Colt1911 [192]

The other structures near the hypothalamus are thalamus and the pituitary gland.
Hypothalamus is located below the thalamus and is part of the limbic system. It is an integral part of the brain, it is a small cone-shaped structure that projects down ward from the brain, ending in the pituitary stalk, a tubular connection to the pituitary gland.

3 0

3 years ago

Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it

MakcuM [25]

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r = r-h / r

2h = r-h

3h = r

h = r / 3

5 0

2 years ago