PHYSICS HELP

1 answer:

3 0

My calculator is about 1cm thick, 7cm wide, and 13cm long.

Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .

The question wants me to assume that the density of my calculator
is about the same as the density of water. That doesn't seem right
to me. I could check it easily. All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically.
I won't do that. I'll accept the assumption.

If its density is actually 1 g/cm³, then its mass is about 91 grams.

The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.

My result of 91 grams is about 100 grams ... about 10² grams.

Your results could be different.

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A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$