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2 years ago

8

Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5kW at

a 0.707 pf lagging. Determine: (a) the pf of the second load, (b) the parallel element required to correct the pf to 0.9 lagging for the two loads.

1 answer:

Anna [14]2 years ago

6 0

Answer:

a) The pf of the second load will be 0.948.

b) The parallel element required will be a capacitor of 117.5 uF

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A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6

adell [148]

Answer:

1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

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Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

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In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y =

monitta

Answer:

Ψ = 10(y^2) + c

<em><u>y = 1.067m</u></em>

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =<em>u/y</em>

<em><u>u = 20y</u></em>

<em><u>Ψ = ∫20ydy</u></em>

<em><u>Ψ = 10(y^2) + c</u></em>

<em><u>(b)</u></em>

<em><u>the flow is half below y = 1.6*(2/3)=1.067 m</u></em>

<em><u>this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m</u></em>

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Which of the following is NOT true about hydraulic valves? A. Directional control valves determine the path of a fluid in a give

Lelechka [254]

Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path

Explanation:

The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction

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A shipment of rebar that weighs 745 kg would weigh roughly how much in pounds

Andre45 [30]

Answer:

Dont no but will check

Explanation:

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