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3 years ago

In the layout of a printed circuit board for an electronic product, 12 different locations can accommodate chips.

Engineering

1 answer:

iren [92.7K]3 years ago

4 0

Answer:

a) 244,140,625 different ways

b) 390,625 different ways

Explanation:

a) If there are 5 ways to place a chip on each location, and there are 12 locations overall, we have:

5^12 ways of placing them

This would mean a total of 244,140,625 different ways

b) If five chips are of the same type, we can first find how many ways we can place chips on the remaining 7 locations:

5^7 = 78,125

Next we can multiply this by the number of ways the next 5 chips could be the same:

78,125 * 5 = 390,625 different ways

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All of the following have the same units except: Group of answer choices resistance capacitive reactance. inductance. impedance.

Brilliant_brown [7]

Answer:

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Explanation:

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2 years ago

A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.

Monica [59]

Answer: (a). E = 3.1656×10³⁴ √k/m

(b). f = 9.246 × 10¹² Hz

(c). Infrared region.

Explanation:

From Quantum Theory,

The energy of a proton is proportional to the frequency, from the equation;

E = hf

where E = energy in joules

h = planck's constant i.e. 6.626*10³⁴ Js

f = frequency

(a). from E = hf = 1 quanta

f = ω/2π

where ω = √k/m

consider 3 quanta of energy is lost;

E = 3hf = 3h/2π × √k/m

E = (3×6.626×10³⁴ / 2π) × √k/m

E = 3.1656×10³⁴ √k/m

(b). given from the question that K = 15 N/m

and mass M = 4 × 10⁻²⁶ kg

To get the frequency of the emitted photon,

Ephoton =hf = 3h/2π × √k/m (h cancels out)

f = 3h/2π × √k/m

f = 3h/2π × (√15 / 4 × 10⁻²⁶ )

f = 9.246 × 10¹² Hz

(c). The region of electromagnetic spectrum, the photon belongs to is the Infrared Spectrum because the frequency ranges from about 3 GHz to 400 THz in the electromagnetic spectrum.

6 0

3 years ago

How many power station do we have

loris [4]

Answer: 9,719

Explanation:

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3 years ago

The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter

Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

( $p_t$ )₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ $[$ α₂/( $p_t$ )₂ $]^{1/2$ ------- let this be equation 2

where ( $p_t$ )₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ from equation for (σf)₂ in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ = 2(6( σ₀ )₁) $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

divide both sides by 2(σ₀)₁

$[$ α₁/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]$ = 36 $[$ 0.84/( $p_t$ )₂ $]$

1 / ( $p_t$ )₁ = 30.24 / ( $p_t$ )₂

( $p_t$ )₂ = 30.24( $p_t$ )₁

( $p_t$ )₂/( $p_t$ )₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

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3 years ago

A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai

lbvjy [14]

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

See that attached pictures for explanation.

3 0

3 years ago