Ask question

Ask question

All categories

2 years ago

6

. In order to prevent injury from inflating air bags, it is recommended that vehicle occupants sit at least __________ inches aw

ay from the air bag.

2 answers:

scZoUnD [109]2 years ago

8 0

10 inches is the answer

Umnica [9.8K]2 years ago

8 0

10 inches is the answer to that question sir

You might be interested in

Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flo

deff fn [24]

Answer:

Z = 3 + 0.23t

The water level is rising

Explanation:

Please see attachment for the equation

8 0

2 years ago

Read 2 more answers

Technician A ay that acid rain doe the mot harm when it firt fall on a finih. Technician B ay that hard water potting i uually j

Tamiku [17]

Technician B is right say that hard water potting i usually jut a Surface problem that can be wahed off.

What do you mean by Hard water?

The amount of dissolved calcium and magnesium in the water determines its hardness. Calcium and magnesium are the main dissolved minerals in hard water. The last time you washed your hands, you might have actually felt the effects of hard water.

What do you mean by acid rain?

Any type of precipitation that contains acidic elements, such as sulfuric or nitric acid, that falls to the ground from the atmosphere in wet or dry forms is referred to as acid rain, also known as acid deposition. Rain, snow, fog, hail, and even acidic dust can fall under this category.

Some plants are sensitive to excessive moisture around their root zone, so it may be necessary to increase drainage when growing plants in pots. Additionally, standing water at the bottom of the pot can cause root rot.

Many university agriculture extension agencies have thoroughly debunked the old garden myth that adding rocks to the bottom of a pot will increase drainage.

Learn more about hard water click here:

brainly.com/question/28178305

#SPJ4

6 0

1 year ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

2 years ago

The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi

Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,

1 gallon = 0.133 cubic feet (cf)

Therefore,

Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133

= 4.28 x 10˄8

Applying Mass balance i.e

Accumulation = Mass In - Mass out (Eq. 01)

Here

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation = 0.5 - 11

= - 10.5 cfs

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

= 37,800

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

= 907,200

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200

= 471 Days.

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0

2 years ago

Asolid rectangular rodhas a length of 90mm, made of steel material (E =207,000 MPa, Syield= 300 MPa), the cross section of the r

erastova [34]

Answer:

13.6mm

Explanation:

We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.

See attachment for the step by step solution of the problem

3 0

3 years ago