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The elevator descends with an acceleration of 0.9 m/s². Find how much a 95 kg person weighs in it at the moment of descent. ​

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Answer 1
Refer to the photo attached.
The Elevator Descends With An Acceleration Of 0.9 M/s. Find How Much A 95 Kg Person Weighs In It At The

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For a particular nonlinear spring, the relationship betweem the magnitude of the applied force F and the resultant displacement x from equilibrium is given by the equation F = k x^2 What is the amount of work done by stretching the spring a distace x0? A) kx0^3 B) (1/2)kx0 C) (1/2)kx0^3 D) (1/3)kx0^2 E) (1/3)kx0^3

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To get the work, you have to integrate the force as a function of [tex]$x$[/tex] from zero displacement to Xo

[tex](Integral of) $\mathrm{k} \mathrm{x}^{\wedge} 2 \mathrm{dx}$ from 0 to $\mathrm{Xo}_{\mathrm{o}}=(1 / 3) \mathrm{k}\left(\mathrm{Xo}^{\wedge}\right)^{\wedge} 3$[/tex]

The work done by stretching the spring to the given distance is [tex]W=\frac{k x_0}{3}[/tex]

The given parameters:

- Applied force on the spring [tex]$=F$[/tex]

- Extension of the spring [tex]$=x_0$[/tex]

The work done by stretching the spring to the given distance is calculated as follows;

[tex]W=\frac{k x_0}{3}[/tex]

[tex]$$\begin{aligned}& W=\int_{x_a}^{x_b} F d x \\& W=\int_{x_a}^{x_b} k x^2 d x \\& W=k \int_{x_a}^{x_b} x^2 d x \\& W=k\left[\frac{x^3}{3}\right] \\& W=k\left[\frac{x_b-x_a}{3}\right] \\& W=k\left[\frac{x_0-0}{3}\right] \\& W=\frac{k x_0}{3}\end{aligned}[/tex]

Thus, the work done by stretching the spring to the given distance is

[tex]W=\frac{k x_0}{3}[/tex]

measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.

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for the circuit below all values are rms with a source frequency of 60 hertz. the generator impedance is negligible (0 w). calculate the value of ib and and determine the value of vb-n. now, calculate the rms values of apparent, real and reactive power of zb. (show units for all)

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The reactive power of the branch (Qb) can be calculated from its voltage and current 676.16 VA.

What is reactive power?

Reactive power is the power in an AC circuit that is required to establish and maintain a voltage across a load. It is associated with the storage and release of energy in the form of electric and magnetic fields. Reactive power does not contribute to the actual work output of a system and is measured in Volt-amperes reactive (VARs). Power factor is a measure of reactive power relative to the total power in a system.

Source frequency = 60 Hz
Generator impedance = 0 W
Circuit:
Vg = 170 V
R1 = 10 Ω
R2 = 20 Ω
Zb = 20 + j10 Ω

The current flowing through the generator (Ig) and the branch (Ib) can be calculated from Ohm's Law:
Ig = Vg/R1 + Vg/R2 = 170/10 + 170/20 = 17 A
Ib = Ig - Vg/Zb = 17 - 170/(20 + j10) = 17 - 16.4 + j4.4 = 0.6 + j4.4 A
Since Ib is a complex number, we can find its magnitude (|Ib|) and angle (θ):
|Ib| = √(0.6² + 4.4²) = 4.46 A
θ = tan⁻¹(4.4/0.6) = 80.16°
The voltage across the branch (Vb-n) can be calculated using Ohm's Law:
Vb-n = Ib × Zb = (0.6 + j4.4) × (20 + j10) = -8.4 + j74.4 V
The apparent power of the branch (Sb) can be calculated from its voltage and current:
Sb = Vb-n × Ib* = (-8.4 + j74.4) × (0.6 - j4.4) = -45.48 + j367.04 VA
The real power of the branch (Pb) can be calculated from its voltage and current:
Pb = Vb-n × Ib = (-8.4 + j74.4) × (0.6 + j4.4) = -45.48 - j367.04 W
The reactive power of the branch (Qb) can be calculated from its voltage and current:
Qb = Vb-n × Ib* = (-8.4 + j74.4) × (0.6 - j4.4) = 676.16 VA

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