All categories

3 years ago

Airbrakes on some trucks have how many second lag time with for the engage properly?

1 answer:

6 0

The answer that best completes the given statement above is ONE SECOND (1). There should only be one second of lag time to engage properly for some of the airbrakes of the trucks. One half to one second is the normal lag time and it should not exceed beyond this time. Hope this helps.

You might be interested in

20. Threshold braking in the vehicle's braking system

amm1812

The answer is b hope this helps

7 0

3 years ago

A general contractor drives 5 miles to the hardware store in 10 minutes. He shops for needed tools for 30 minutes, then drives t

Alex17521 [72]

Given parameters:

Distance to hardware shop = 5 miles

Time to reach hardware shop = 10 minutes

Time spent at the shop = 30 minutes

Average speed to customer home = 45 mph

Time taken for the travel = 20 minutes

Unknown:

Average speed of the contractor = ?

Solution:

Average speed is the total distance covered divided by the total time taken.

Average speed = $\frac{total distance}{total time taken}$

total distance = distance to hardware shop + distance to customer's home

We do not know the distance to customer's home but we have been given the speed and time, so we can find the distance.

Distance = speed x time

Convert the time to hrs and solve;

60 minutes = 1 hr

20 minutes = $\frac{20}{60} hr$ = $\frac{1}{3} hr$

So, Distance = 45mph x $\frac{1}{3} hr$ = 15miles

Now;

Total distance = 5 + 15 = 20miles

Total time = time to reach hardware shop + time to reach customer's house

= 10 + 20

= 30 minutes

Convert the time from minutes to hrs;

60 minutes = 1hr

30 minutes = $\frac{30}{60}$ = 0.5hr

So;

Average speed = $\frac{20}{0.5}$ = 40mph

The average speed is 40mph

3 0

4 years ago

PLS HELP

Ahat [919]

"This was the idea that non-living objects can give rise to living organisms."

7 0

3 years ago

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion

stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 240 K

= initial volume of gas =

= final volume of gas =

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0

3 years ago

How many licks dose it take to eat a lolipop

murzikaleks [220]

Answer: Around 364 to 480

3 0

3 years ago

Read 2 more answers