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serg [7]
3 years ago
14

A teacher wants to perform a classroom demonstration that illustrates both chemical and physical changes. Which would be the bes

t demonstration that she could use?
Physics
2 answers:
goblinko [34]3 years ago
4 0
Putting something in acid, as the object would chemically be broken down and physically break down.
Alex17521 [72]3 years ago
3 0
She can put chalk in vinegar as the vinegar will disintegrate the chalk chemically demonstrating chemical changes. But for physical changes she can break the chalk into small pieces by smashing it with something or her hand.
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Samples of different materials, A and B, have the same mass, but the sample
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B. The particles that make up material B have more mass than the

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Which term best describes those processes that move weathered rock materials and soils downslope?
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An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

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