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QveST [7]
3 years ago
7

An aluminium bar 600mm long1 with diameter 40mm, has a hole drilled in the centre of the bar. The hole is 30mm in diameter and i

s 100mm long. If the modulus of elasticity for the aluminium is 85GN/M^2 Calculate the total contraction on the bar due to a compressive load of 180KN

Physics
1 answer:
nexus9112 [7]3 years ago
6 0

Answer:

Total contraction = 1.2277 mm

Explanation:

the solution is given in the picture below and it is more explanatory

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A cyclist overcomes a resistive force of 30 N in order to cycle 30 m. It takes her 6 seconds to cycle this distance. Calculate t
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Answer:

Power = 15[W]

Explanation:

This problem can be solved using the work definition.

Work is equal to the product of the force by the distance, for this problem we have:

F = force = 30 [N]

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w = work = F * d = 30*30 = 900 [J], "units in joules"

The power is defined as the work done in an interval of time.

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therefore

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The frequency of which type of electromagnetic wave is just higher than that of visible light?
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Choice E.

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The emission spectrum of iodine is shown below.<br> Which is the absorption spectrum?
garik1379 [7]
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6 0
3 years ago
Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

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The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

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0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

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The momentum goes to the wall
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