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4 years ago

A Scalar can only be a positive quantity that has a magnitude but no direction ? a)-True b)-False

Engineering

1 answer:

Bogdan [553]4 years ago

8 0

false

im not sure

but they haven't a direction

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What year did Vidal Sassoon come to America ?

tekilochka [14]

Answer:

1970

Explanation:

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3 years ago

Read 2 more answers

What is the governing ratio for thin walled cylinders?

Ann [662]

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

if $\frac{radius}{thickness} >10$ then you have a thin walled cylinder

or using the diameter:

if $\frac{diameter}{thickness} >20$ then you have a thin walled cylinder

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3 years ago

A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque

Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

τ = 16*T / pi*d^3

T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

T_c = τ_c*pi*d_c^3 / 16

T_c = 12*1000*pi*1^3 / 16

T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

T_s = τ_s*pi*d_s^3 / 16

T_s = 15*1000*pi*0.25^3 / 16

T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

T = min ( 2356.2 , 46.02 )

T = 46.02 lb-in

6 0

3 years ago

A large building will need several different types of workmen to install and repair pipes for water, heating,

siniylev [52]

Answer:

Plumber and pipefitters

Explanation:

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3 years ago

An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of

uranmaximum [27]

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

$\Delta L=L\alpha\Delta T$

Now by putting the values

$\Delta L=750\times \25.5\times 10^{-6}\times 200$

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

3 0

4 years ago