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tatuchka [14]
3 years ago
12

How often do you rely on media for scientific information to make decisions in your life? (No weather answer please)

Physics
2 answers:
vodka [1.7K]3 years ago
6 0
The majority of the time


Nina [5.8K]3 years ago
4 0
Well, taking out the weather option ... I hardly ever do.
If I'm looking at a specific technical problem, then a scan on a source I'm not allowed to cite here, it seems, or some such may give me a hint as to whether it's going to be any help to me. If it is, i continue. If it's isn't, then i'll go away and decide whether I should pursue the problem or not (ie am I wasting my energy here ?)
I suspect, though, that I'm not representative of users.
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4 0
3 years ago
Will give brainlist.
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Answer:

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Explanation:

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How will heat flow between objects?
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3 years ago
Read 2 more answers
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

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