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Mashutka [201]
3 years ago
13

Are the lines in the diagram perpendicular, parallel, skew, or none of these?

Mathematics
2 answers:
ivolga24 [154]3 years ago
5 0
<span>Are the lines in the diagram perpendicular, parallel, skew, or none of these?

l and m: neither
l and m intersect, but are not perpendicular. 

l and n: skew
l and n are not parallel, but they do not intersect because they are not on the same plane. 

m and n: perpendicular.
m and n intersect at a right angle</span>
MissTica3 years ago
4 0

1. None Of These

2. Skew

3. Perpendicular

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Answer:

a) 0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The spring must be 5.6 centimeters far from its natural length.

Step-by-step explanation:

a) The work done to stretch the ideal spring from its natural length is defined by the following definition:

W = \frac{1}{2}\cdot k\cdot (x_{f}-x_{o})^{2} (1)

Where:

k - Spring constant, measured in newtons.

x_{o}, x_{f} - Initial and final lengths of the spring, measured in meters.

W - Work, measured in joules.

The spring constant is: (W = 5\,J, x_{o} = 0.36\,m, x_{f} = 0.51\,m)

k = \frac{2\cdot W}{(x_{f}-x_{o})^{2}}

k = \frac{2\cdot (5\,J)}{(0.51\,m-0.36\,m)^{2}}

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If we know that k = 444.44\,\frac{N}{m}, x_{o} = 0.41\,m and x_{f} = 0.46\,m, then the work needed is:

W = \frac{1}{2}\cdot \left(444.44\,\frac{N}{m} \right)\cdot (0.46\,m-0.41\,m)^{2}

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0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The elastic force of the ideal spring (F), measured in newtons, is defined by the following formula:

F = k\cdot \Delta x (2)

Where \Delta x is the linear difference from natural length, measured in meters.

If we know that k = 444.44\,\frac{N}{m} and F = 25\,N, then the linear difference is:

\Delta x = \frac{F}{k}

\Delta x = \frac{25\,N}{444.44\,\frac{N}{m} }

\Delta x = 0.056\,m

The spring must be 5.6 centimeters far from its natural length.

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