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PSYCHO15rus [73]

3 years ago

15

Technician A says when diagnosing an overheating hydraulic system, be sure that the oil cooler is not plugged and the cooler’s f

an is working. Technician B says running the hydraulic system at a lower operating temperature will reduce the possibly of oil oxidation. Who is correct?

1 answer:

mel-nik [20]3 years ago

8 0

Answer:

Tech B is correct and Tech A is incorrect.

Explanation:

Here Tech A is wrong because when diagnosing an overheating hydraulic system, it is not necessary to un plugg the oil cooler rather it should be plugged to for proper diagnosis.

Technician B says running the hydraulic system at a lower operating temperature will reduce the possibly of oil oxidation is correct statement because at temperature oil's physical as well as chemical property tend to change.

Hence, Tech B is correct and Tech A is incorrect.

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A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi

Dmitriy789 [7]

Answer:

a

$A = 0.081 \ m$

b

The value is $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

The mass is $m = 0.750 \ kg$

The spring constant is $k = 17.5 \ N/m$

The instantaneous speed is $v = 39.0 \ cm/s= 0.39 \ m/s$

The position consider is x = 0.750A meters from equilibrium point

Generally from the law of energy conservation we have that

The kinetic energy induced by the hammer = The energy stored in the spring

So

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=> $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=> $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

=> $A = 0.081 \ m$

Generally from the law of energy conservation we have that

The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=> $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=> $u = 0.2569 \ m/s$

3 0

3 years ago

6 Fig. 6.1 is a full-scale diagram that represents a sound wave travelling in air

Oxana [17]

From the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

<h3>What is the frequency of a wave?</h3>

The frequency of a wave is the number of complete oscillation per second completed by a wave.

Frequency is related to wavelength and speed by the following formula:

Frequency = velocity/wavelength

Velocity of sound in air = 330 m/s

The measured wavelength = 5.0 cm = 0.05 m

Frequency = 330/0.05 = 6660 Hz

Therefore, based on the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

Learn more about frequency of sound at: https://brainly.in/question/15373132
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7 0

2 years ago

Can the boiling point of water exceed 100c

lisabon 2012 [21]

Superheated water is liquid water under pressure at temperatures between the usual boiling point, 100c (212 F ) and the critical temperature , 374 C (705F) . It is also known as “subcritical water”
Or “pressurized hot water”.

6 0

3 years ago

If a Stone with an of original velocity of zero is falling from a ledge and takes eight seconds to hit the ground what is the fi

DaniilM [7]

Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.

3 0

3 years ago

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N

7 0

3 years ago