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1 month ago

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A suspension system is being designed for a 1300 kg vehicle. When the vehicle is empty, its static deflection is measured as 2.5

0 mm. It is estimated that the largest cargo carried by the vehicle will be 1000 kg. What is the minimum value of the damping coefficient such that the vehicle will be subject to no more than 5% overshoot, whether it is empty or filled?

1 answer:

Keith_Richards [3.2K]1 month ago

3 0

Answer: damping coefficient = 1.5×10^5Ns/m

Explanation:

Refer to the attached file for the solution

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Response:

D. You accelerate

Clarification:

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1 month ago

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

serg [3582]

Answer:

$v = 3369.2 m/s$

Explanation:

The beacon is rotating at an angular speed of

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so we have

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$\omega = 1.047 rad/s$

We know that

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2 months ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

serg [3582]

Response:

(a) $W=-19.25J$

(b) $W=-52.8J$

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

$W=nRTln(\frac{V_1}{V_2} )$

The moles are calculated based on the provided mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Consequently, the work amounts to:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Best regards.

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1 month ago

An electron is projected with an initial speed of 3.9 × 105 m/s directly toward a proton that is fixed in place. If the electron

Keith_Richards [3271]

Explanation:

Attached is a document that contains the solution.

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1 month ago

If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t

Sav [3153]

Para calcular el peso utilizamos la fórmula:
$F=m*g=882N$

Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
$F=G* \frac{ M_{p}*M_{E} }{ r^{2} }$
Donde:
G = constante de gravedad = $6.67* 10^{-11} m^{3} kg^{-1} s^{-2}$
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra = $5.97* 10^{24} kg$
r = distancia entre la Tierra y la persona

A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
$m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }$

Resolviendo esta ecuación para r:
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r=6371116m = 6371.116km

Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.

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2 months ago