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3 years ago

- Scrap tire management is primarily regulated at the

Engineering

2 answers:

krek1111 [17]3 years ago

5 0

Answer:D

Explanation:

Scrap tire management is primarily regulated at the state

level.

kompoz [17]3 years ago

3 0

Scrap tire management is primarily regulated at the state level.

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The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F,

Black_prince [1.1K]

Answer:

attached below

Explanation:

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3 years ago

QUESTÃO 13. Explique o uso das aspas no trecho "Darei a cada uma de vocês

lesya [120]

Answer: speaks Portuguese

Eu disse a todos a tradução para que possam te ajudar

Explanation: Y’all can help I have no idea

QUESTION 13. Explain the use of quotation marks in the excerpt "I will give each of you

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3 years ago

When changing a tire size 195/75 R15 to a 5 percent lower profile, the correct tire size would be

loris [4]

Answer:

205/70 R15

Explanation:

The change of sizes is determined a formula which changes tire to 5% lower profile with 10 mm wider cross section. This implies that the reducing aspect ratio is 5. Now, adding 10 to 195 makes it 205 but subtracting 5 from 75 making it 70. By changing the tire size of 195/75 R15 using the this formula, this will be changed to size of 205/70 R15. Therefore, the right option is 205/70 R15.

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3 years ago

The lab technician you recently hired tells you the following: Boss, an undisturbed sample of saturated clayey soil was brought

harkovskaia [24]

Answer:

The water of the saturated clayed soil is 66.67 %.

Explanation:

Given;

mass of saturated clayed soil, $M_s$ = 600 g

mass of dry soil sample, $M_d$ = 200 g

mass of water content, $M_w$ = $M_s$ - $M_d$ = 600 g - 200 g = 400 g

The water content is determined as;

$M_w(\%) = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%) = \frac{600-200}{600} *100 \% \\\\M_w(\%) = 66.67 \%$

Therefore, the water of the saturated clayed soil is 66.67 %.

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3 years ago

A pump operating at steady state receives liquid water at 20°C 100 kPa with a mass flow rate of 53 kg/min. The pressure of the w

coldgirl [10]

Answer:

The power required by pump is 6.18 KW.

Explanation:

The ideal output power of the pump is given by:

Pout = ΔP (Volume Flow Rate)

ΔP = Pf - Pi = 5 MPa - 100 KPa = 5 x 10^6 Pa - 0.1 x 10^6 Pa

ΔP = 4.9 x 10^6 Pa

Volume Flow Rate = (Mass Flow Rate)/(Density of Water)

Volume Flow Rate = (53 kg/min)(1 min / 60s)/(1000 kg/m^3)

Volume Flow Rate = 8.83 x 10^(-4) m^3/s

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Pout = 4328.33 Watt = 4.33 KW

Now, the required Power (Pin) is related with isentropic efficiency as:

Isentropic Efficiency = Pout/Pin

Pin = [Pout]/[Isentropic Efficiecy]

Pin = (4328.33 Watt)/(0.7)

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3 years ago