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2 years ago

7

how are the construction steps for bisecting an angle similar to the steps you would have taken to construct and bisect a line s

egment? how are they different?

1 answer:

GREYUIT [131]2 years ago

5 0

Answer:

does the question entail anything else?

Explanation:

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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 much to build a barber clipper?

goldenfox [79]

Like the price to manufacture?

8 0

3 years ago

Your local hospital is considering the following solution options to address the issues of congestion and equipment failures at

kiruha [24]

Jsjhjrhwjdbwjwjrueiworuuwud

4 0

2 years ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0

3 years ago

An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric

lawyer [7]

Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

Temperature= 15°C

RH=80%

At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

$h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}$

At temperature= 50°C

$h_2=73 \frac{KJ}{kg}$

$So\ the\ heater\ load =h_2-h_1$

The heater load = 73 - 38 KJ/kg

The heater load =35 KJ/kg

3 0

3 years ago