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2 years ago

7

Reducing waste benefits your shop by cutting down on pollution and __________. A)Lowering costs B)keeping employees healthy C)Re

ducing tripping hazards D) Improving employee efficiency

1 answer:

taurus [48]2 years ago

3 0

Answer:

I think A

Explanation:

A is the only one talking about costs

You might be interested in

: The interior wall of a furnace is maintained at a temperature of 900 0C. The wall is 60 cm thick, 1 m wide, 1.5 m broad of mat

Snowcat [4.5K]

Answer:

<em>Heat is lost at the rate of 750 J/s or W</em>

<em>The thermal resistance is 1 K/W</em>

Explanation:

interior temperature $T_{2}$ = 900 °C

wall thickness t = 60 cm = 0.6 m

width = 1 m

breadth = 1.5 m

thermal conductivity k = 0.4 W/m-K

outside temperature $T_{1}$ = 150 °C

heat through the wall = ?

The area of the wall A = w x b = 1 x 1.5 = 1.5 m^2

Temperature difference $dt$ = $T_{2}$ - $T_{1}$ = 900 - 150 = 750 °C

note that $dt$ is also equal to 750 K since to convert from °C to K we'll have to add 273 to both temperature, which will still cancel out when we subtract the two temperatures.

To get the heat that escapes through the wall, we use the equation

Q = Ak $\frac{dt}{t}$

substituting values, we have

Q = 1.5 x 0.4 x $\frac{750}{0.6}$ = <em>750 J/s or W</em>

Thermal resistance $R_{t}$ = $\frac{dt}{Q}$

$R_{t}$ = 750/750 =<em> 1 K/W</em>

7 0

3 years ago

The viscosity of the water was 2.3×10^−5lb⋅⋅s/ft^2 and the water density was 1.94 slugs/ft^3. Estimate the drag on an 88-ft diam

Lina20 [59]

Answer:

hello your question is incomplete attached below is the complete question

answer : Drag force = 1.3 Ib

Explanation:

we have to represent the dimensions of the drag force in terms of FLT

i.e : D = f( <em>d,v,p,u </em>) represented in terms of FLT

D = F , V = LT^-1, d = L, p = FL^-4 T^2

u = FL^-2 T, Number of independent terms = 5

attached below is the detailed solution

6 0

4 years ago

You are hired as the investigators to identify the root cause and describe what should have occurred based on the following info

creativ13 [48]

Answer:

The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.

Explanation:

Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.

So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.

To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.

It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg

To find the mass of the extra 4,916 L of fuel added, we have

m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³

m' = 803 kg/m³ × 4.916 m³ = 3947.548 kg

So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg

<u>Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.</u>

4 0

3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

Free_Kalibri [48]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

then $\frac{x_1^2}{t_1}=\frac{x_2^2}{t_2}$ we have given $x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm$ and we have to find $t_2$ putting all these value in equation

$\frac{2^2}{15}=\frac{6^2}{t_2}$

so $t_2=135\ hour$

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

3 years ago