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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?

Engineering

1 answer:

stiks02 [169]3 years ago

7 0

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

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Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 573 oC (this temperature is below the m

inn [45]

Answer:

$2.9\times 10^{-6}$

Explanation:

$Q_s$ = Energy for defect formation = 1.86 eV

T = Temperature = $573^{\circ}\text{C}=573+273.15=846.15\ \text{K}$

k = Boltzmann constant = $8.62\times 10^{-5}\ \text{eV/K}$

The fraction of lattice sites that are Schottky defects is given by

$\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}$

The required ratio is $2.9\times 10^{-6}$ .

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3 years ago

A single square-thread screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a diameter of 40 mm and a pitch of

Sonbull [250]

Answer:

Axial Resisting Load, F = 31.24kN

Efficiency = 16.67%

Explanation:

Given

Input Power = P,in = 3kW = 3,000W

Speed, S = 1rev/s

Pitch, p = 8mm

Thread frictional coefficient = μt = 0.18

Collar frictional coefficient = μc = 0.09

Friction radius of collar, Rc = 50mm

First, we calculate the torque while the load is being lifted in terms of 'F'.

This is calculated by

T = ½FDm[1 + πDmμt]/[πDm - μtp]

By substituton.

T = ½F(40-4)[1 + π(40-4)0.18]/[π(40-4) - 0.18 * 8]

T = 18F(1 + 6.48π)/(36π - 1.44)

T = 3.44F.Nmm

T = 3.44 * 10^-3F Nm

Then we calculate the torque due to friction from the collar

T = Fμc * Rc

T = F * 0.09 * 50

T = 4.5F. Nmm

T = 4.5 * 10^-3F Nm

Then, we calculate the axial resisting load 'F' by using the the following power input relation.

P,in = Tw

P,in = (T1 + T2) * 2πN

Substitute each value

3,000 = (3.44 + 4.5) * 10^-3 * F * 10^-3 * 2 * π * 2

F = 3000/((3.44 + 4.5) * 10^-3 * 10^-3 * 2 * π * 2

F = 31,247.69N

F = 31.24kN

Hence, the axial resisting load is

F = 31.24kN

Calculating Efficiency

Efficiency = Fp/2πP

Efficiency = 2Fp/P,in

Substitute each value

Efficiency = 2 * 31,247.69 * 8 * 10^-3/3000

Efficiency = 0.166654346666666

Efficiency = 16.67%

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4 years ago

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Nesterboy [21]

Answer:

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lukranit [14]

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4 years ago

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hram777 [196]

Answer:

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