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ANEK [815]
4 years ago
10

What driving behavior can increase risk

Engineering
1 answer:
mario62 [17]4 years ago
6 0

Answer:

instances of hazardous practices are not utilizing headlights appropriately, driving unreasonably quick for conditions, closely following, dangerous passing, or path evolving, and so on. Inability to wear a safety belt is a high-chance conduct that frequently makes the results of an impact more terrible.

Explanation:

hope this helps :) !!

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A submarine submerges by admitting seawater (S = 1.03) into its ballast tanks. The amount of water admitted is controlled by air
ser-zykov [4K]

Answer:

Explanation:

Fw= y looking symbol sea water (2010.62-168171.2727/h+32.935)ft^3

6 0
3 years ago
If a 9V battery produces a current of 3 A through a load, what is the resistance of the load
Elden [556K]

3 ohms hope this helps :D ❤

7 0
3 years ago
Read 2 more answers
Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
3 years ago
PLZ ASAP WILL GIVE BRAINLIST
gavmur [86]

Answer: A, B, C & F (interacting w computers, making decisions & solving problems, evaluating information & getting information).

Explanation: Those are the correct & verified answers.

7 0
3 years ago
Read 2 more answers
The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint
snow_lady [41]

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = \frac{100}{100 - P} M kg  per day

put her value

mass of sludge produced = \frac{100}{100 - 78} 7240 kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge =  \frac{\rho sludge}{\rho water }

and as we know that

\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}

\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}

S sludge = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = \frac{mass sludge produce}{\rho sludge}

volume of biological sludge = \frac{32909.09}{1152}

volume of biological sludge = 28.566 m³ per day

6 0
3 years ago
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