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3 years ago

Which si unit whould be most appropriate for describing the length of a shoe?

Physics

1 answer:

aniked [119]3 years ago

4 0

Answer:

Milimeters

Explanation:

This kind of question must be checked with everyday elements, in this way we will analyze each of the options until we reach the appropriate:

kilograms can not be because this unit serves to measure the mass not the length.

kilometers (km) is a unit of length, but however it does not serve for this measurement since a km is equivalent to 1000 meters, therefore measuring the length of a shoe with a ruler measuring 1000 meters is very impractical.

milliliters is a unit of volume, and can often be found in the measurement of volume content of liquids, for example 700 milliliters of milk or 1000 milliliters of water. Therefore it is impractical to measure a shoe with a unit of volume.

In this way the way to measure a shoe correctly is using a ruler or measuring instrument graduated in millimeters.

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When the saw slices wood, the wood exerts a 104-N force on the blade, 0.128 m from the blade’s axis of rotation. If that force i

Soloha48 [4]

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

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2 years ago

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2 years ago

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000 cycles

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3 years ago