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2 years ago

Thermal energy measured by?

Engineering

1 answer:

Harlamova29_29 [7]2 years ago

3 0

Answer:

Thermal energy is measured using a thermometer denominated in Fahrenheit, Celsius and Kelvin

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Answer: preventive maintenance

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3 years ago

) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p

andrey2020 [161]

Answer:

number of pulses produced = 162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = $\frac{linear\ displacement}{circumference}$ × No of pulses per revolution .................2

number of pulses produced = $\frac{200}{314.16}$ × 256

number of pulses produced = 162 pulses

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3 years ago

Find the velocity and acceleration of box B when point A moves vertically 1 m/s and it is 5 m

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Answer:

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4 years ago

How many seconds do you need to stop a car going 60 miles per hour, if the pavement is dry?

Anna71 [15]

Answer:

Roughly 4.6 seconds

Explanation:

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3 years ago

a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What

Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, $dq/dt = 100 W$

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, $T_0 = 20^0 C = 273 + 20 = 293 K$

Temperature of the inside of the box, $T_{in} = ?$

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

$dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C$

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4 years ago