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3 years ago

An object elongates from a length of 45 cm to a length of 55 cm. The percent strain is

Engineering

1 answer:

MrRissso [65]3 years ago

3 0

Answer:

Percentage Elongate = (final gage length – initial gage length)/ initial gage length) x 100.

=55-45/45 *100

=10/45 *100

=22.2%

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3 years ago

A closed, 0.4-m-diameter cylindrical tank is completely filled with oil (SG 0.9)and rotates about its vertical longitudinal axis

egoroff_w [7]

Answer: $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa

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where:

$p_{fluid}$ is the liquid's density

w is the angular velocity

r is the radius

γfluid.z is the pressure variation due to centrifugal force.

For this question, the difference between a point on the circumference and a point on the axis will be:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}.r_{B} ^{2} }{2}$ - γfluid. $z_{B}$ - ( $\frac{p_{fluid}.w^{2}.r_{A} ^{2} }{2}$ - γfluid. $z_{A}$ )

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$ - γfluid( $z_{B}$ - $z_{A}$ )

Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$

$p_{B} - p_{A} = \frac{0.9.10^3.40^2}{2}(0.2^2 - 0)$

$p_{B} - p_{A}$ = 28800

The difference in pressure between two points, one on the circumference and the other on the axis is $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa

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3 years ago

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Answer:

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4 years ago

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mars1129 [50]

Answer:

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3 years ago

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