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2 years ago

The p-type or N-type semiconductor is not of practical use. Why?

Physics

1 answer:

Savatey [412]2 years ago

6 0

Explanation:

The p-type semiconductor doesn't have greater energy than the n-type semiconductor. This is because the semiconductor itself doesn't really have an energy. Charge carriers (electrons/holes) have energies. The relative positioning of the bands tell you how much energy an electron or a hole must have for it to exist.

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For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long)

Lesechka [4]

The question is missing a diagram of the ray reflection. I attached a diagram which comes from a similar question in the answer section. The full question should be as follows:

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cmfrom their point of intersection, as shown in the figure. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long) after reflecting from the first mirror?

Answer:

34.6°

Explanation:

To strike the midpoint of the second mirror, the ray light will have to travel half of the distance vertically

i.e. 29/2 = 14.5

We can solve this through trigonometry.

Let the angle between the ray and the vertical plane mirror is known as α

tan α = 10/14.5

α = $tan^{-1} (10/14.5)$ = 34.6°

The angle of incidence is the angle between the ray and the normal line of the mirror.

Let angle of incidence of first mirror be β

β = α = 34.6

6 0

3 years ago

What are some possible materials you could use to make your battery? DON’T FORGET TO include information about why it is importa

Dima020 [189]

Explanation:

Commercially available batteries use a variety of metals and electrolytes. Anodes can be made of zinc, aluminum, lithium, cadmium, iron, metallic lead, lanthanide, or graphite. Cathodes can be made of manganese dioxide, mercuric oxide, nickel oxyhydroxide, lead dioxide or lithium oxide. Potassium hydroxide is the electrolyte used in most battery types, but some batteries use ammonium or zinc chloride, thionyl chloride, sulfuric acid or lithiated metal oxides. The exact combination varies by battery type. For example, common single-use alkaline batteries use a zinc anode, a manganese dioxide cathode, and potassium hydroxide as the electrolyt

4 0

3 years ago

A rope pulls a 82.5 kg skier at a constant speed up a 18.7Â° slope with Î¼k = 0.150. How much force does the rope exert?

Artist 52 [7]

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f = μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

6 0

3 years ago

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

Veseljchak [2.6K]

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

4 0

4 years ago

A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are

Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, $\Sigma F_y$ = 0

The vertical component of the tension, $T_y$ , in each half of the rope is given as follows;

$T_y$ = T × sin(θ)

∴ $\Sigma F_y$ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0

3 years ago

The p-type or N-type semiconductor is not of practical use. Why?​

The p-type or N-type semiconductor is not of practical use. Why?