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1 year ago

a low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a

1 answer:

5 0

A low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a powder-actuated tool.

<h3>What is a fastening system?</h3>

It should be noted that a fastening system is important to fixing rails to railroad ties.

In this case, a low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a powder-actuated tool.

Learn more about fasteners on:

brainly.com/question/16350719

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Water flows through a multisection pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s a

Alex_Xolod [135]

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b. 2.3 kPa.

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This situation can be modelled by Bernoulli's Principle, as there are no energy interaction throughout the multisection pipe and current lines exists between both ends. Likewise, this system have no significant change in gravitational potential energy since it is placed horizontally on the ground and is described by the following model:

$P_{1} + \rho \cdot \frac{v_{1}^{2}}{2} = P_{2} + \rho \cdot \frac{v_{2}^{2}}{2}$

Where:

$P_{1}$ , $P_{2}$ - Pressures at the beginning and at the end of the current line, measured in kilopascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Fluid velocity at the beginning and at the end of the current line, measured in meters per second.

Now, the pressure difference between these two points is:

$P_{1} - P_{2} = \rho \cdot \frac{v_{2}^{2}-v_{1}^{2}}{2}$

If $\rho = 1000\,\frac{kg}{m^{3}}$ , $v_{1} = 3\,\frac{m}{s}$ and $v_{2} = 2.1\,\frac{m}{s}$ , then:

$P_{1} - P_{2} = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \frac{\left(2.1\,\frac{m}{s} \right)^{2}-\left(3\,\frac{m}{s} \right)^{2}}{2}$

$P_{1} - P_{2} = -2295\,Pa$

$P_{1} - P_{2} = -2.295\,kPa$ (1 kPa is equivalent to 1000 Pa)

Hence, the right answer is B.

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2 years ago