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RideAnS [48]
3 years ago
14

Name two types of battery chargers that are used in mechanics

Engineering
1 answer:
tekilochka [14]3 years ago
7 0
Primary batteries
Secondary batteries
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A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of
Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0
3 years ago
Many jacks use hydraulic power. True or False
34kurt

Answer:true

Explanation:

7 0
3 years ago
Freeee Poinntssss 100!!!!!!<br> Hi how you doin?
Andreas93 [3]
Thank u very much , im doin good wby :)
6 0
3 years ago
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The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air
Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0
4 years ago
The pilot of an airplane reads the altitude 6400 m and the absolute pressure 45 kPa when flying over the city. Calculate the loc
garri49 [273]

Answer:

1) The absolute pressure equals = 96.98 kPa

2) Absolute pressure in terms of column of mercury equals 727 mmHg.

Explanation:

using the equation of pressure statics we have

P(h)=P_{surface}-\rho gh...............(i)

Now since it is given that at 6400 meters pressure equals 45 kPa absolute hence from equation 'i' we obtain

P_{surface}=P(h)+\rho gh

Applying values we get

P_{surface}=45\times 10^{3}+0.828\times 9.81\times 6400

P_{surface}=96.98\times 10^{3}Pa=96.98kPa

Now the pressure in terms of head of mercury is given by

h_{Hg}=\frac{P}{\rho _{Hg}\times g}

Applying values we get

h=\frac{96.98\times 10^{3}}{13600\times 9.81}=0.727m=727mmHg

7 0
3 years ago
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