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777dan777 [17]
1 year ago
11

Technician A says that the pressure differential switch may need to be re-centered after bleeding the brakes. Technician B says

that there are several types of pressure differential switches, and each requires a different method of re-centering. Who is correct?
a. Technician A
b. Technician B
c. Both Technician A and Technician B
d. Neither Technician A nor Technician B
Physics
1 answer:
Snezhnost [94]1 year ago
6 0

Option C

Both Technician A and Technician B is correct

<h3><u>Explanation:</u></h3>

The pressure differential valve is the equipment that warns you if you leak one of your brake circuits. After draining and refilling some brake system pressure differential switch may be actuated and an indication light may be illuminated. If the indication light sojourns sparked after draining the valve piston may demand to be re-centered. If the pressure is replaced to the brake line and your brakes appear to be running accurately over, then the reset button operated.

The common practical method to re-center the piston is to fix the failure, drain the system and then generate a pressure loss opposite what moved the piston in the first place. Each type has a unique kind of re-centering technique.

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Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
1 year ago
If a 10 meter ramp helps you move a 500 kg object up 1 meter. What was the mechanical advantage of the ramp?
sattari [20]
The mechanical advantage of an inclined plane is

(Length of the incline) / (its height)

= (10m) / (1m)

= 10 .

It's the same for any load, and doesn't depend on the mass that you're trying to move up or down the ramp.
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1 year ago
A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resis
snow_lady [41]

Answer:

The resistance will be 2×R

Explanation:

We note that the resistivity of a cylindrical wire is given by the following relation;

\rho = \frac{RA}{L}

Where:

ρ = Resistivity of the wire

R = The wire resistance

A = Cross sectional area of the wire = π·D²/4

L = Length  of the wire

Rearranging, we have;

R= \frac{\rho L}{A}

If the length and the diameter are both cut in half, we have;

L₂ = L/2

A₂ =π·D₂²/4 = \pi \cdot \left (\frac{D}{2}   \right )^{2} \times \frac{1}{4}  = \pi \cdot \frac{D^{2}}{16} = A/4

Therefore, the new resistance, R₂ can be expressed as follows;

R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times  \frac{\rho L}{A}

Hence, the new resistance R₂ =  2×R, that is the resistance will be doubled.

8 0
2 years ago
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andrew11 [14]

Answer:b)1770 kWh

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Given

volume of water V=40,000 gallons

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c=4186 J/kg-^{\circ} C

also 1 kg mass is approximately is 1 gallon

therefore 40,000 gallon is equivalent to 3.8\times 40000 kg

heat Required to raise temperature is

Q=mc\Delta T

Q=3.8\times 40000\times 4186\times 10

Q=63,627.2\times 10^5 J

Q=63,672.2\times 10^2 KJ

Q=1767.42 kWh\approx 1770 kWh

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