You walk 60m east and then 20m at 20 degrees south of west. What is your displacement?

1 answer:

5 0

Let north and east be posetive
Y=20sin20
=6.840402867 south
X=20cos20
18.79385243 west
Net horizontal components=60+(-18.79)
=41.20614758m
Net vertical=-6.84040267
X^2=41.20614758^2+(-6.684040267)^2
√x^2 =√1744.73771m
x=41.77m

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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?

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The force needed to stretch the steel wire by 1% is 25,140 N.

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diameter of the steel, d = 4 mm

the radius of the wire, r = 2mm = 0.002 m

original length of the wire, L₁

final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)

extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁

the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

$A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2$

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}$

$F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N$

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

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True or false? If the speed of an object doubles, its kinetic energy will also double.

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Concept map for kinetic energy, work and power

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Kinetic energy: the energy of motion

Work: the change in kinetic energy

Power: the rate of work done

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m is the mass of the object

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