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SCORPION-xisa [38]
2 years ago
9

Increase vehicle control through lane position and speed adjustments is not necessary risk management, just common sense.

Engineering
1 answer:
Vesnalui [34]2 years ago
6 0
True

If it was wrong I’m sorry
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A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in
Ierofanga [76]

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, \sigma=\frac {F}{A}

where F is applied force and A is original area

Also, engineering strain, \epsilon=\frac {\triangle l}{l} where l is original area and \triangle l is elongation

We also know that Hooke's law states that E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

\triangle l= 0.1 mm= 0.1\times 10^{-3} m

By substitution we obtain

E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa

5 0
3 years ago
Drag each item to show if it is an element or not an element.
Flauer [41]

Answer:

CARBON

Explanation:

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5 0
3 years ago
Input Energy ---> Output Energy
uranmaximum [27]

Answer:

motion ------> electrical. winds push the turbines which generate a magnetic fields which in turn, generates electricity

4 0
2 years ago
What is a radio wave made up of? Molecules? Electrons? Other?
mafiozo [28]

Radio waves are radiated by charged particles when they are accelerated. They are produced artificially by time-varying electric currents, consisting of electrons flowing back and forth in a specially-shaped metal conductor called an antenna. ... Radio waves are received by another antenna attached to a radio receiver.

4 0
2 years ago
Read 2 more answers
A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can
Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a)   ln(T2/T1)=μβ         β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π  radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π  

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb     T1=80 lb   μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

4 0
3 years ago
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