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KonstantinChe [14]
3 years ago
6

Tech a says you should push the wrench when braking a fastener loose. Tech b says that you should pull the wrench when braking a

fastener loose. Who is correct?
Engineering
1 answer:
Kay [80]3 years ago
4 0

Answer:

tech b because gut feeling

Explanation:

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While discussing what affects the amount of pressure exerted by the brakes: Technician A says that the shorter the line, the mor
harina [27]

Answer:

Only Technician B is right.

Explanation:

The cylindrical braking system for a car works through the mode of pressure transmission, that is, the pressure applied to the brake pedals, is transmitted to the brake pad through the cylindrical piston.

Pressure applied on the pedal, P(pedal) = P(pad)

And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)

If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.

If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.

This whole setup does not depend on the length of the braking lines; it only depends on the applied force and cross sectional Area (size) of the piston.

5 0
2 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD
Stels [109]

Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr=<u>304,000</u>

      900

=33.77

5 0
3 years ago
Read 2 more answers
How is the energy harnessed and converted into useful energy?
garri49 [273]

Answer:

1. How energy is harnessed?  

Another way to tap solar energy is by collecting the sun's heat. Solar thermal power plants use heat from the sun to create steam, which can then be used to make electricity. On a smaller scale, solar panels that harness thermal energy can be used for heating water in homes, other buildings, and swimming pools.

2. How is solar energy converted into useful energy?

Solar panels convert the sun's light into usable solar energy using N-type and P-type semiconductor material. When sunlight is absorbed by these materials, the solar energy knocks electrons loose from their atoms, allowing the electrons to flow through the material to produce electricity.

Explanation:

hope it helps, please mark as brainliest

6 0
3 years ago
A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima
zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

\sigma_c = \frac{P}{A}

\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}

\sigma_c = 81.5MPa

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}

Where,

\sigma_a=Fatigue limit for comined alternating and mean stress

S_e =Fatigue Limit

\sigma_c=Mean stress (due to static load)

\sigma_u = Ultimate tensile stress

Fs =Security Factor

We can replace the values and assume a security factor of 1, then

\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}

Re-arrenge for \sigma_a

\sigma_a = 254.8Mpa

We know that the stress is representing as,

\sigma_a = \frac{M_c}{I}

Then,

Where M_c=Max Moment

I= Intertia

The inertia for this object is

I=\frac{\pi d^4}{64}

Then replacing and re-arrenge for M_c

M_c = \frac{\sigma_a*\pi*d^3}{32}

M_c = \frac{260.9*10^6*\pi*0.05^3}{32}

M_c = 3201.7N.m

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0
3 years ago
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