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3 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

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

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<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

Horizontal distance he must clear = 22 meter

The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

We have equation of motion, $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when displacement = 1.5 meter.

$1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

So minimum speed required = 22/0.553 = 39.78 m/s

Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

$-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

Horizontal speed of car = V cos 7 = 0.993V

So it has to travel 22 meter in t seconds

0.993Vt = 22, Vt = 22.155 m

Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

Speed required = 22.155/0.926 = 23.93 m/s

Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

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3 years ago