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3 years ago

5

Compressor clutch controls are being discussed. Technician A says a clutch may be cycled on and off based on system pressure. Te

chnician B says some clutches will not be cycled on-off unless the driver operates the manual air conditioning control device. Who is correct?

1 answer:

kogti [31]3 years ago

3 0

Answer:

Technician B is correct.

Explanation:

Technician B says some clutches will not be cycled on-off unless the driver operates the manual air conditioning control device.

the role of the clutch is the transmitted power from the engine to the compressor. and to provide a means of engaging and disengaging the refrigeration system from engine operation. the clutch is driven by one or more of the drive belts fitted to the engine.

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4 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

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3 years ago

Question 26 suppose that a constant force is applied to an object. newton's second law of motion states that the acceleration of

ololo11 [35]

<span>The answer is 6 kg the mass of the second object. By using Inversely proportional formula it means that (14 kg) (3 m/s</span>²<span>) = M (7 m/s</span>²<span>). Where M is the mass of the second object. For the Newton’s second law of motion formula which is: Force = mass x acceleration, we have:</span>

<span>F = (14 kg) (3 m/s</span>²<span>) = 42 N</span>

Therefore:

<span>42 N = M (7 m/s</span>²)

<span>M = (42 N) / (7 m/s</span>²<span>)</span>

M = 6 kg mass of the second object

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