In an RL parallel circuit, VT = 240 V, R = 330 Ω, and XL = 420 Ω. What is the Apparent Power (VA)?

Engineering

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer:

that answer is correct

Explanation:

This answer is correct because they explained everything they needed.

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Airbags will deploy in a head on collision but not in a collision that occurs from angle

Aneli [31]

Answer:

Airbags will deploy in almost any angle.

Explanation:

Cars have sensors around them, so when the car gets hit, the sensors detect a crash and deploy the airbags to keep you safe.

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3 years ago

Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the approp

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Answer:

The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.

Explanation:

The detailed steps and appropriate calculation is as shown in the attached file.

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3 years ago

Someone should help me and explain stress strain curve

dimulka [17.4K]

Answer:

In engineering and materials science, a stress–strain curve for a material gives the relationship between stress and strain. It is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined (see tensile testing).

Explanation:

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3 years ago

A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2

Furkat [3]

Answer:

$P_{1} = 403,708\,kPa\,(58.553\,psi)$

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

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

The initial pressure is:

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

The speed at outlet is:

$v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}$

$v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }$