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stepan [7]
3 years ago
11

Technician A says that thrust angle refers to the direction the front wheels are pointing. Technician B says that scrub radius r

efers to the vertical centreline of the tire in relation to an imaginary line through the steering knuckle pivots. Who is correct?
Engineering
1 answer:
Novay_Z [31]3 years ago
8 0

Answer:

both technicians are correct

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I need this asap thank you :) plzzzzz When the spring on a mousetrap car is fully unwound, the force acting on the car is _____.
Usimov [2.4K]

Answer:  (only friction) the friction lets it keep its speed and not slow down and it creats volocity between the serface of where the mousecar is running and the wheels on the ground

sorry if im wrong i tried my best

6 0
3 years ago
A steady‐flow gas furnace supplies hot air at a rate of 850 cfm and conditions of 120F and 1.00 atm. The air splits into two bra
V125BC [204]

Answer:

if I am not wrong the volumetric flow rate into the finance if the year inter 868 1.00 pm

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3 years ago
Use the map to answer the question.
djverab [1.8K]

Answer:

Captain falcon

Explanation:

7 0
2 years ago
A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th
Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ    

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

 cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    

b)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

x_A and x_B are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, x_A = 248 /(248 + 42.8) = 0.83

x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

5 0
3 years ago
The autorotation spin characteristics of a straight-wing aircraft are induced by Group of answer choices
NemiM [27]

Answer:

More Drag on the down going wing and More Lift on the up going wing

Explanation:

The autorotation spins of blades used in airborne wind energy technology sectors help drive and move the winds and water propeller-type turbines or shafts of generators to produce electricity at altitude and transmit the electricity to earth through conductive tethers.

Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.

As a result, more Drag usually induces the autorotation spin characteristics of a straight-wing aircraft on the downgoing wing and More Lift on the up-going wing.

7 0
3 years ago
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