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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

You might be interested in

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operat

Gwar [14]

Answer:

$\eta _{max} = 0.2413 = 24.13%$

$\eta' _{max} = 0.5061 = 50.61%$

Given:

$T_{1max} = 100^{\circ} = 273 + 100 = 373 K$

operating temperature of heat engine, $T_{2} = 10^{\circ} = 273 + 10 = 283 K$

$T_{3max} = 300^{\circ} = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _{max}$ is given by:

$\eta _{max} = 1 - \frac{T_{2}}{T_{1max}}$

$\eta _{max} = 1 - \frac{283}{373} = 0.24$

$\eta _{max} = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_{3max}$ changes to $300^{\circ}$ , so, the new maximum efficiency, $\eta' _{max}$ is given by:

$\eta' _{max} = 1 - \frac{T_{2}}{T_{3max}}$

$\eta _{max} = 1 - \frac{283}{573} = 0.5061$

$\eta _{max} = 0.5061 = 50.61%$

4 0

4 years ago

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main

andreev551 [17]

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C = 273 K

$\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5$

R = 287 J/kg K

$C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s$

$C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s$

$V_2 = \dfrac{V_1}{C_1}\timesC_2$

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 × 147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁ )

= 165.49 - 147.5

= 17.5 m/s

6 0

4 years ago

A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine

Anuta_ua [19.1K]

Answer:

force R = 846.11 N

lifting force L = 110.36 N

if cable fail complete both R and L will be zero

Explanation:

given data

mass woman mw = 60 kg

mass package mp = 9 kg

accelerates rate a = g/4

to find out

force R and lifting force L and if cable fail than what values would R and L acquire

solution

we calculate here first reaction R force

we know elevator which accelerates upward

so now by direction of motion , balance the force that is express as

R - ( mw + mp ) × g = ( mw + mp ) × a

here put all these value and a = g/4 and use g = 9.81 m/s²

R - ( 60 + 9 ) × 9.81 = ( 60 + 9 ) × g/4

R = ( 69 ) × 9.81/4 + ( 69 ) 9.81

R = 69 ( 9.81 + 2.4525 )

force R = 846.11 N

and

lifting force is express as here

lifting force = mp ( g + a)

put here value

lifting force = 9 ( 9.81 + 9.81/4)

lifting force L = 110.36 N

and

we know if cable completely fail than body move free fall and experience no force

so both R and L will be zero

5 0

3 years ago

Of the core elements of successful safety and health programs,management leadership,worker participation and what else directly

ch4aika [34]

I think it would definitely be dinvi and the ghosts just because I really relate to that story

7 0

3 years ago

Driving test. Need NOW

Flauer [41]

Answer:

4 - Double Parking

6- Perpendicular Parking

5- Hand over hand method

1 - Manual Transmission

3 - Push pull method

2 - Parking on a hill

Explanation:

Manual Transmission - requires a stick shift

Perpendicular Parking - type of parking in a parking lot.

Double Parking - Parking in the roadway

Hand-over-hand - Grab the far side of the wheel.

Push-pull Method - Slide your hand along the wheel.

Parking on a hill - Turn wheels toward curb.

3 0

3 years ago