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3 years ago

WANT POINTS? JUST ANSWER ME:)

Engineering

2 answers:

bulgar [2K]3 years ago

7 0

Answer:

mcfndvkernfnkvfvjnfdkvfv

Explanation:

Lapatulllka [165]3 years ago

6 0

Answer:

yes why thank you

snarfs :)

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Nearlyof all serious occupational injuries and illnesses stem from overexertion of repetitive motion.

Ostrovityanka [42]

Answer:

a) 1/2

Explanation:

Overexertion accounted for more than half of all events that resulted in a disabling condition.

Furthermore, 30% of all overexertion cases were reported in the services industry, on the other hand, 25% of injuries resulting from contact with objects and equipment occurred in the manufacturing industry.

The above piece of information is taken from the bureau of labor statistics, Survey of Occupational Injuries and Illnesses

"LOST-WORKTIME INJURIES AND ILLNESSES: CHARACTERISTICS AND RESULTING DAYS AWAY FROM WORK, 2002"

8 0

3 years ago

A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?

stiks02 [169]

Answer:

The tax on this property is $3750$ dollars

Explanation:

Given

Tax on per $100 is $2.50

Tax on every $1 is $\frac{2.5}{100} = 0.025$ dollars

Tax on property of value $150,000 is

$150,000 * 0.025 = 3750$ dollars

The tax on this property is $3750$ dollars

7 0

2 years ago

Yall know what this is called?

aliya0001 [1]

Answer:

oof no bro

Explanation:

5 0

2 years ago

Please help me fast, I don’t have time

Anna71 [15]

Answer: precision

Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer

5 0

3 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

2 years ago