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2 years ago

B. The silica cylinder of a radiant wall heater is 0.6 m long

1 answer:

6 0

So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.

<h3>Power radiated by the radiant wall heater</h3>

The power radiated by the radiant wall heater is given by P = εσAT⁴ where

ε = emissivity = 1 (since we are not given),
σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
A = surface area of cylindrical wall heater = 2πrh where
r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
h = length of heater = 0.6 m, and
T = temperature of heater

Since P = εσAT⁴

P = εσ(2πrh)T⁴

Making T subject of the formula, we have

<h3>Temperature of heater</h3>

T = ⁴√[P/εσ(2πrh)]

Since P = 1.5 kW = 1.5 × 10³ W

Substituting the values of the variables into the equation, we have

T = ⁴√[P/εσ(2πrh)]

T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]

T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]

T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]

T = ⁴√[0.01105 × 10¹⁴ K⁴)]

T = ⁴√[1.105 × 10¹² K⁴)]

T = 1.0253 × 10³ K

T = 1025.3 K

So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

Learn more about temperature of radiant wall heater here:

brainly.com/question/14548124

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3 years ago

If a car's engine gives off 65% thermal energy, what is the maximum efficiency?

stich3 [128]

Answer:

35%

Explanation:

The car's engine gives off 65% thermal energy

So only 35 % is converted into mechanical energy .

input heat = Q₁ = 100

output heat = Q₂ = 65

Work output = Q₁ - Q₂ = W

W = 100 - 65 = 35

Efficiency = W / Q₁ X 100

= (35/ 100) X 100

= 35%.

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3 years ago

The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency? Hz(b) What is its wavelength in glas

vekshin1

<h2>Answers:</h2>

The speed of a wave is given by:

$v=f.\lambda$ (1)

Where $f$ is the frequency and $\lambda$ the wavelength.

In the case of light, its speed is:

$c=f.\lambda$ (2)

On the other hand, the described situation is known as Refraction, a phenomenon in which the light changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (3)

In addition, as the light changes its direction, its wavelength changes as well:

$n=\frac{\lambda_{air}}{\lambda_{glass}}$ (4)

Knowing this, let's begin with the answers:

<h2>a) Frequency</h2>

From equation (2) we can find $f$ :

$f=\frac{c}{\lambda}$ (5)

Knowing that $1nm=(10)^{-9}m$ :

$f=\frac{3(10)^{8}m/s}{632.8(10)^{-9}m}$

$f=4.74(10)^{14}Hz}$ (6) >>>Frequency of the helium-neon laser light

<h2>b) Wavelength in glass</h2>

We already know the wavelength of the light in air $\lambda_{air}$ and the index of refraction of the glass.

So, we only have to find the wavelength in glass $\lambda_{glass}$ from equation (4):

$\lambda_{glass}=\frac{\lambda_{air}}{n}$

$\lambda_{glass}=\frac{632.8(10)^{-9}m}{1.48}$

$\lambda_{glass}=427(10)^{-9}m=427nm$ (7) >>>Wavelength of the helium-neon laser light in glass

<h2>c) Speed in glass</h2>

From equation (3) we can find the speed $v$ of this light in glass:

$v=\frac{c}{n}$

$v=\frac{3(10)^{8}m/s}{1.48}$

$v=2.027(10)^{8}m/s$ (8) >>>Speed of the helium-neon laser light in glass

7 0

2 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

3 0

3 years ago

In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in (inclu

Darya [45]

Answer:

Explanation:

When the number of slits increases, the intensity of fringes increases.

So, the fringes appear to be more bright.

As we know that the fringe width is inversely proportional to the number of slits, so as the number of slits increases, the fringe width decreases, hence the fringes are narrower, bright and close together.

7 0

3 years ago