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Olenka
2 months ago
12

The molecular weight of a 10g rubber band

Engineering
1 answer:
Viktor [391]2 months ago
8 0
The response to this query is 1 * 10 g/mole = 10.
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Suppose we include the lead resistance in the calculation of temperature for a class A RTD. If R3 = 1000 ohms, Ra = 18 ohms, V0
alex41 [359]

Respuesta:

La temperatura máxima que se puede medir (en °C) es 14170.27°C

Explicación:

Los RTDs son termómetros compuestos de metales cuya resistencia aumenta con la temperatura.

Para un RTD de Clase A, l, Alpha = 0.00385.

La fórmula para el RTD es

Rt = Ro ( 1 + alpha x t)

Donde

Rt es la resistencia a la temperatura t°C,

Ro es la resistencia a 0°C

Alpha es un coeficiente de temperatura constante para un RTD de clase A.

Aquí, Rt = 1000ohms,

Ro se considera como Ra = 18Ohms

Por lo tanto,

1000 = 18 ( 1 + 0.00385t)

Dividiendo ambos lados por 18

1 + 0.00385t = 1000/18

0.00385t = 55.55 - 1

0.00385t = 54.55

t = 54.55/0.00385

t = 14170.27°C

5 0
1 month ago
Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routin
Daniel [329]
For Deterministic Quicksort, which operates by selecting the first element as the pivot, consider a scenario where the pivot consistently divides the array into segments of 1/3 and 2/3 for all recursive calls. (a) The runtime recurrence for this case needs to be determined. (b) Use a recursion tree to justify that this recurrence resolves to Theta(n log n). (c) Provide distinct sequences of 4 and 13 numbers that prompt this behavior.
3 0
2 months ago
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,
Mrrafil [318]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = -\frac{dp}{dV/V}................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = \frac{\pi }{4} *d^2*l(i)................3

Final volume v2 = \frac{\pi }{4} *d^2*l(f)................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

Substituting all values yields

0.0130 = \frac{\pi }{4} *2^2*dl

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0
2 months ago
1. You do not need to remove the lead weights inside tires before recycling them.
Kisachek [356]
That's correct -.-.-.-.-.-.-.-.-.-.-.-.-.- Easy.
7 0
3 months ago
A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k = 70 W/m·K) with a uniform cl
Kisachek [356]

Answer:

(a) the rate of heat transfer to the coolant is Q = 139.71W

(b) the surface temperature of the shaft T = 40.97°C

(c) the mechanical power wasted by the viscous dissipation in oil 22.2kW

Explanation:

See explanation in the attached files

5 0
1 month ago
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