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QveST [7]
3 years ago
13

Tech a says the higher the numarical gear ratio (4:1), the more torque that will be applied to the wheels. Tech b says that the

lower the numarical gear ratio (2:1), the more torque that will be applied to the wheels. Who is correct?
Engineering
2 answers:
nadya68 [22]3 years ago
6 0

Answer:

gear rattlo

Explanation:

Zepler [3.9K]3 years ago
3 0

Answer:

Tech A

Explanation:

The amount of energy required to apply the same force with a 1:1 ratio is divided into 4, so you can apply 4 times as much force than a 1:1 ratio. efficiency and speed come into play here, but assuming the machine powering the gear can run at a unlimited RPM, 4:1 will have more force and a slower output speed than a 2:1 ratio.

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The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:
Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1

Here;

G_{s} is specific gravity of soil solids

y_{w} is unit weight of water = 998 kg/m^3

w is the moisture content = 0.17

e is the void ratio

y is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

w*G_{s} = S*e  ..... Eq2

S is saturation rate

Substitute Eq 2 into Eq 1

y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w}  }{1+e}

14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233

Specific gravity of soil solids

G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765

Saturated Unit Weight

y_{s} = \frac{(G_{s} + e)*y_{w}  }{1+e} \\=\frac{(4.878811765 + 1.38233)*998  }{1+1.38233}\\\\= 2622.902571 N/m^3

7 0
3 years ago
Incremental software development could be very effectively used for customers who do not have a clear idea about the systems nee
avanturin [10]

<u>Software Development and Client Needs</u>

In Incremental method of software development customers who do not have a basic idea of the development process are being carried along on like other methods that will relegate them to the background until a product is ready.

With this model and structure in place, when softwares/ products are built from several stages e.g prototype, testing, and when new features are added customers are always carried along with their valuable feedback and suggested greatly considered to achieve the customers satisfactions

This model will work well for the customers/clients who does not have a clear idea on the systems needed for their operations.

In summary the incremental model combines features from the waterfall and prototyping model.

For more information on soft ware development process kindly visit

brainly.com/question/20369682

5 0
2 years ago
3.
Andreyy89

Answer:

7

Explanation:

5 + 2 = 7

4 0
2 years ago
A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen
Lerok [7]

Answer:

Q = 424523.22 kw

Explanation:

\rho =7833 kg/m

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

\alpha = 3.91*10^{-6} m^2/s

T_s = 285 degree celcius

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15

\dot m = 14764.85 kg/s

A_s = \pi DL = \pi 0.4*2 = 2.513 m^2

Q = \dot m C \Delta T = h A_s \Delta T

\dot m C \Delta T = h A_s \Delta T

solving for h

h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}

h = 675.6 kw/m^2K

Q = h A_s\Delta T

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

7 0
3 years ago
Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
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