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Lorico [155]
3 years ago
11

A 5cm diameter copper sphere (of density = 8954 kg/m3, specific heat capacity = 0.3831 kJ/kg K) is initially at a uniform temper

ature of 10oC. The sphere is placed inside an environment having a temperature of 200oC. The surface heat transfer coefficient is h = 10 W/(m2K). Under these conditions, we can assume uniform internal temperature for the sphere. What is the temperature inside the sphere after 10 minutes in this environment?

Engineering
1 answer:
Korolek [52]3 years ago
6 0

Answer:

Temperature inside sphere after 10 minutes = 19924.33K

Explanation:

Detailed explanation and calculation is shown in the image below

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Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1
steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

90.8777353 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

3 0
2 years ago
A right triangle has a base of 12 inches and a height of 30 inches, what is the centroid of the triangle?​
aliina [53]

Answer:

the correct answer is 42

4 0
2 years ago
In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water
Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0
2 years ago
The AGC control voltage: ___________
lyudmila [28]

Answer:

The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.

Explanation:

3 0
2 years ago
In Lab 7, we worked through a program that displayed the homeless shelter occupancy over time. The same approach can be used for
Bezzdna [24]

Answer:

Explanation:

The python code to generate this is quite simple to run.

i hope you understand everything written here, you can as well try out other problems to understand better.

First to begin, we import the package;

Code:

import pandas as pd

import matplotlib.pyplot as plt

name = input('Enter name of the file: ')

op = input('Enter name of output file: ')

df = pd.read_csv(name)

df['Date'] = pd.to_datetime(df["Date"].apply(str))

plt.plot(df['Date'],df['Absent']/(df['Present']+df['Absent']+df['Released']),label="% Absent")

plt.legend(loc="upper right")

plt.xticks(rotation=20)

plt.savefig(op)

plt.show()

This should generate the data(plot) as seen in the uploaded screenshot.

thanks i hope this helps!!!

6 0
2 years ago
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