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3 years ago

11

When driving during the daytime in reduced visibility, such as rain, smoke, or fog,

1 answer:

riadik2000 [5.3K]3 years ago

3 0

Answer:A

Explanation:

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Thank you very much this helped a lot

Len [333]

Answer:

Your welcome i guess

Explanation:

7 0

2 years ago

Read 2 more answers

A geothermal pump is used to pump brine whose density is 1050 kg/m3 at a rate of 0.3 m3/s from a depth of 200 m. For a pump effi

nasty-shy [4]

Answer:

Input power of the geothermal power will be 686000 J

Explanation:

We have given density of brine $\rho =1050kg/m^3$

Rate at which brine is pumped $V=0.3m^3/sec$

So mass of the pumped per second

Mass = volume × density = $1050\times 0.3=315$ kg/sec

Acceleration due to gravity $g=9.8m/sec^2$

Depth h = 200 m

So work done $W=mgh=315\times 9.8\times 200=617400J$

Efficiency is given $\eta =0.9$

We have to fond the input power

So input power $=\frac{617400}{0.9}=686000J$

So input power of the geothermal power will be 686000 J

5 0

3 years ago

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

Viktor [21]

Given:

Pressure, $P_{1}$ = 1300 kPa

Temperature, $T_{1}$ = $500^{\circ}$

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

$v_{2}$ = $1.15\frac{m^{3}}{kg}$

a) Mass flow rate is given by:

$m' = \frac{Av}{v_{1}}$

Now,

$m = \frac{0.2\times 40}{0.172}$ = 46.51 kg/s

b) for the power produced by turbine, $P = m'(h_{1} - h_{2})$

$P = 46.51\times(793 - 401)$ = 18.231 MW

5 0

3 years ago

2. Can you make adjustments on a saw while the trigger is in the on position.

Liula [17]

Umm... Heck no, that is very dangerous and could cause you to loose a finger... OR A HAND. To be safe I would always turn it off but if the saw has an on and off switch and a rev switch I think it would be ok.

3 0

3 years ago

Read 2 more answers

If you install special sound-reflecting windows that reduce the sound intensity level by 33.0 dBdB , by what factor have you red

Sever21 [200]

Answer:

The sound intensity is reduced by a factor of 1995.26

Explanation:

When comparing two sound intensities, the intensity level is measured in the unit of decibel or dB. The intensity of the threshold of hearing for a human being is 10^−12 W/m^2 . When the intensity level is zero, it means that the sound intensity is the same as the threshold of hearing.

The reduced sound intensity level is given as 33 dB, so

10 log (I/Io) = - 33

I : intensity of the sound

Io ( = 10^−12 W/m^2): threshold of hearing

So, the intensity ratio is

I/Io = 10^-3.3

= 5.01 x 10^-4

1/ 5.01 x 10^-4 = 1995.26

3 0

3 years ago