What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 14 kg dog off the floor?

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Answer 1

Answer: The diameter would be 267km speed suction i think

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A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.113V. Assuming that errors are due to random processes, how many of the readings are expected to be greater than 2.70V?

Answers

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, $\mu = 2.501 V$

standard deviation, $\sigma = 0.113 V$

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, $P(X\leq 2.70)$ :

$z = (x - \mu)/(\sigma) = (2.70 - 2.501)/(0.113) = 1.761$

Now, from the Probability table of standard normal distribution:

$P(z\leq 1.761) = 0.9608$

Now,

$P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%$

Now, for the expected no. of readings greater than 2.70 V:

$P(X\geq 2.70) = (No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V)/(Total\ no.\ of\ readings)$

No. of readings expected to be greater than 2.70 V = $P(X\geq 2.70)* Total\ no.\ of\ readings$

No. of readings expected to be greater than 2.70 V = $0.0392* 60 = 2.352$ ≈ 2

A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa

Answers

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto (1)/(V)$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 150 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v L

$V_2$ = final volume of gas = $(v)/(3)L$

$150* v=P_2* (v)/(3)$

$P_2=450kPa$

Therefore, the new pressure of the gas will be 450 kPa.

A chemical process converts molten iron (III) oxide into molten iron and carbon dioxide by using a reducing agent of carbon monoxide. The process allows 10.08 kg of iron to be produced from every 16.00 kg of iron (III) oxide in an excess of carbon monoxide. Calculate the percentage yield of iron produced in this process.

Answers

Answer:

percentage yield = 63%

Explanation:

The yield efficiency or percentage yield measure the amount of products that are formed from a given amount of reactant. For a percentage yield of 100, all the reactants are completely converted to product. Mathematically, the percentage yield is given by:

$percentage\ yield = (Actual\ yield)/(expected\ yield) * 100\nActual\ yield = 10.08kg\nExpected\ yield= 16.00kg\n\n\therefore percentage\ yield = (10.08)/(16.00) * 100 = 63 \%$

Webster is giving a speech on the benefits of moving toward the use of windmill energy instead of having to rely on fossil fuels. Most likely, he will select which method of arrangement?

Answers

Answer:

the netuon arrangment

Explanation:

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.

Answers

Answer:

The power input, in kW is -86.396 kW

Explanation:

Given;

initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

Step 1:

Determine the specific volume:

p₁v₁ = RT₁

$v_1 = (R'T_1)/(p_1) = ((0.28699.(kJ)/(kg.K) )(300 k))/((1.05 bar *\ (10^5 N/m^2)/(1 bar) *(1kJ)/(1000N.m) )) \n\nv_1 = 0.81997 \ m^3/kg$

Step 2:

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

Step 3:

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

step 4:

determine work input;

W = Q - mΔh

Where;

Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

W = -86.396 kW

Therefore, the power input, in kW is -86.396 kW

An automobile engine consumes fuel at a rate of 27.4 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.73 g/cm3, deter- mine the efficiency of this engine in percentage(

Answers

Answer:

The efficiency of the engine is 22.5%.

Explanation:

Efficiency = power output ÷ power input

power output = 55 kW

power input = specific energy×volumetric flow rate×density

specific energy = 44,000 kJ/kg

volumetric flow rate = 27.4 L/h = 27.4 L/h × 1000 cm^3/1 L × 1 h/3600 s = 7.61 cm^3/s

density = 0.73 g/cm^3 = 0.73 g/cm^3 × 1 kg/1000g = 7.3×10^-4 kg/cm^3

power input = 44,000 kJ/kg × 7.61 cm^3/s × 7.3×10^-4 kg/cm^3 = 244.4332 kJ/s = 244.4332 kW

Efficiency = 55 ÷ 244.4332 = 0.225 × 100 = 22.5%

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