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DerKrebs [107]
2 years ago
10

A minivan starts from rest on the road whose constant radius of curvature is 40 meters and whose bank angel is 10 degrees. the m

otion occurred in a horizontal plane. If the constant forward acceleration of the minivan is 1.8 m/s², determine the magnitude a of it's total acceleration 5 second after starting.
Engineering
1 answer:
zimovet [89]2 years ago
8 0

Based on the calculations, the magnitude (a) of it's total acceleration is equal to 2.71 m/s².

<u>Given the following data:</u>

  • Angle of inclination = 10°.
  • Radius of curvature, r = 40 meters.
  • Acceleration of the minivan, A = 1.8 m/s².
  • Initial velocity, u = 0 m/s (since it's starting from rest).
  • Time, t = 5 seconds.

<h3>How to determine the magnitude (a) of it's total acceleration?</h3>

First of all, we would determine the final velocity of the minivan by applying the first equation of motion as follows:

V = u + at

V = 0 + 1.8 × 5

V = 9 m/s.

Next, we would calculate the centripetal acceleration of this minivan:

Ac = V²/r

Ac = 9²/40

Ac = 2.025 m/s².

Now, we can determine the magnitude (a) of it's total acceleration:

a = √(Ac² + A²)

a = √(2.025² + 1.8²)

a = 2.71 m/s².

Read more on acceleration here: brainly.com/question/24728358

#SPJ1

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Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
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Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

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