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1 year ago

Why a brass lid ring on a glass canning jar will loosen when heated.

Engineering

1 answer:

Zolol [24]1 year ago

4 0

hi here is your answer hope it helps or then sry

Explanation:

A brass lid on a glass canning jar will loosen when heated because brasshas the greater coefficient of thermal expansion

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An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30

sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

$F = LwY_{avg}$

where;

w is the width of the annealed copper

$Y_{avg}$ is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

$L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm$

Now, determine the average true stress, $Y_{avg}$ , for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

$F = LwY_{avg}$

$F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN$

The power required in this operation is given by;

$P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW$

5 0

2 years ago

5. Which of these materials in a shop contain metals and toxins and can pollute the environment? A) Antifreeze B) Solvents C) Ba

Elza [17]

As a multiple choice the only correct answer is D

3 0

2 years ago

Read 2 more answers

A person puts a few apples into the freezer at -15oC to cool them quickly for guests who are about to arrive. Initially, the app

frosja888 [35]

Answer:

Temperature at center of apples = 11.2⁰C

Temperature at surface of apples = 2.7⁰C

Amount of Heat transferred = 17.2kJ

Explanation:

The properties of apple are given as:

k = 0.418 W/m.°C

ρ = 840 kg/m³

Cр = 3.81 kJ/kg.°C

α = 1.3*10 ⁻⁷ m²/s

h = 8 W/m².°C

d = 0.09m

r = 0.045m

t = 1 hour = 3600s

<h2>Solution</h2>

Biot number is given as:

$Bi = \frac{hr}{k}= \frac{8\cdot0.045}{0.418}=0.861$

The constants λ₁ and A₁ corresponding to Biot number (from the table) are:

λ₁ = 1.476

A₁ = 1.239

Fourier Number is:

$T = \frac{a\cdot{t}}{r^2} = \frac{(1.3\cdot10^{-7})(3600)}{0.045^2}= 0.231> 0.2$

As Fourier Number > 0.2 , one term approximates solutions are applicable

The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.

8 0

2 years ago

Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

Yuki888 [10]

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

$60.1cm*\frac{1m}{100cm}=0.601m$

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

$V_{water}=\pi r^{2}h$

$V_{water}=\pi (\frac{0.601m}{2})^{2}*120m$

$V_{water}=113.28m^{3}$

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

$d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}$

$d_{water}=1000\frac{Kg}{m^{3}}$

Now, water density is given by the equation $d=\frac{m}{V}$ , where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

$m_{water}=d_{water}.V_{water}$

$m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}$

$m_{water}=113280Kg$

With the water mass we can find the weight of water:

$w_{water}=m_{water} *g$

$w_{water}=113280kg*9.8\frac{m}{s^{2}}$

$w_{water}=1110144N$

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

$w_{total}=w_{water}+w_{pipe}$

$w_{total}=1110144N+2500N$

$w_{total}=1112644N$

Converting this total weight to kN, we have:

$1112644N*\frac{0.001kN}{1N}=1113kN$

7 0

2 years ago

Technician A that shielding gas nozzles may have different shapes. Technician B says that gelding gas nozzles is attached to the

Lilit [14]

B is correct! in this senecio

5 0

2 years ago