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1 year ago

What is a compound gear used for creating the most torque

1 answer:

8 0

In short, a lower gear produces more torque at the wheel, as well as faster acceleration and lower vehicle speed. Higher gears result in less torque at the wheel, as well as slower acceleration and higher car speed.

<h3>What is torque?</h3>

TORQUE (Terascale Open-source Resource and QUEue Manager) is a distributed resource manager that manages batch jobs and distributed compute nodes.

Torque is a force that can cause an object to rotate around an axis. In the same way that force causes an object to accelerate in linear kinematics, torque causes an object to acquire angular acceleration.

Torque is measured as a vector quantity. Torque is defined as a twisting or turning force that causes rotation around an axis, which could be a center of mass or a fixed point.

Torque can also be defined as the ability of a rotating object, such as a gear or a shaft, to overcome turning resistance.

To learn more about torque, refer to:

brainly.com/question/19104073

#SPJ1

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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

3 years ago

The electron concentration in silicon at T = 300 K is given by

puteri [66]

Answer:

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

Explanation:

From the question we are told that:

Temperature of silicon $T=300k$

Electron concentration $n(x)=10^{16}\exp (\frac{-x}{18})$

$\frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})$

Electron diffusion coefficient is $Dn = 25cm^2/s \approx 2.5*10^{-3}$

Electron mobility is $\mu n = 960 cm^2/V-s \approx0.096m/V$

Electron current density $Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2$

Generally the equation for the semiconductor is mathematically given by

$Jn=qb_n\frac{dn}{dx}+nq \mu E$

Therefore

$-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E$

$E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}$

$E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m$

7 0

3 years ago

Some Tiny College staff employees i s are information technology (IT) personnel. Some IT personnel provide technology support fo

Westkost [7]

Answer:

solution in the picture attached

Explanation:

3 0

3 years ago

Thick fluids such as asphalt and waxes and the pipes in which they flow are often heated in order to reduce the viscosity of the

Gala2k [10]

Answer:

what id the answer

Explanation:

7 0

3 years ago

The concentration of carbon monoxide (co) in an exhaust gas is 1x10^4 ppmv. what is the concentration in mg/m^3 at 25 and 1 atm

Mrrafil [7]

Answer:

1.1451 x 104 (11451.13)mg/m3

Explanation:

1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.

1ppmv = 1mL/m3

Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)

Molecular weight of CO = 12 + 16

= 28g/mol

Temperature = 273.15 + 25

= 298.15K

Pressure = 1 x 101.325kPa

= 101.325kPa

Ppmv = 1 x 10-4ppmv

Gas constant, R = 8.3144 L.kPa/mol.K

Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)

= 1.1451 x 104mg/m3

= 11451.13 mg/m3

3 0

3 years ago