Compare and contrast mechanical properties of plastics, metals and ceramics.

3. (9 points) A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 10

Bezzdna [24]

Answer:

a) 49.95 watts

b) The self locking condition is satisfied

Explanation:

Given data

weight of the square-thread power screw ( w ) = 100 kg = 1000 N

diameter (d) = 20 mm ,

pitch (p) = 2 mm

friction coefficient of steel parts ( f ) = 0.1

Gravity constant ( g ) = 10 N/kg

Rotation of electric power screwdrivers = 300 rpm

A ) Determine the power needed to raise to the basket board

first we have to calculate T

T = Wtan (∝ + Ф ) * $\frac{Dm}{2}$ ------------- equation 1

Dm = d - 0.5 ( 2) = 19mm

Tan ∝ = $\frac{L}{\pi Dm}$ where L = 2*2 = 4

hence ∝ = 3.83⁰

given f = 0.1 , Tan Ф = 0.1. hence Ф = 5.71⁰

insert all the values into equation 1

T = 1.59 Nm

Determine the power needed using this equation

$\frac{2\pi NT }{60}$ = $\frac{2\pi * 300 * 1.59}{60}$

= 49.95 watts

B) checking if the self-locking condition of the power screw is satisfied

Ф > ∝ hence it is self locking condition is satisfied

3 0

3 years ago

Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter ???

Andrej [43]

We need to define the variables,

So,

$F_x (x) = 1-e^{-\lambda x}\\F_x (x) = 1-e^{-0.5x}$

Therefore, the probability that the repair time is more than 4 horus can be calculate as,

$P(x>4)=1-P(x4)= 1-F_x(4)\\P(x>4) = 1-e^{-0.5*4}\\P(x>4) = 1-0.98\\P(x>4) = 0.018$

The probability that the repair time is more than 4 hours is 0.136

b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,

$P(x\geq 12|x>7)=P(X\geq7+5|x>7)\\P(x\geq12|x>7)=P(X\geq5)\\P(x\geq12|x>7)=1-P(x\leq 5)\\P(x\geq12|x>7)=1-e^{-0.5(2)}$

$P(x\geq 12|x>7)=0.6321$

The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63

3 0

3 years ago

A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

lidiya [134]

Answer:

Q = 62 ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

$n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol$

$T_1 = 27^0 \ C = ( 27+273)K = 300 K$

$P_1 = 200 \ kPa$

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if $V_1 = x\\V_2 = 2V_1$

Using Charles Law; since pressure is constant

$V \alpha T$

$\frac{V_2}{V_1} =\frac{T_2}{T_1}$

$\frac{2V_1}{V_1} =\frac{T_2}{300}$

$T_2 = 300*2\\T_2 = 600$

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg × 40

mass of He = 280 kg

Now; the amount of Heat Q transferred = $m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T$

From gas table

$Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K$

∴ Q = $12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)$

Q = $62.389 *10^6$

Q = 62 MJ

Q = 62 ( since we are instructed not to include the units in the answer)

5 0

3 years ago

Read 2 more answers

true or false incident reports, such as situation reports and status reports enhance situational awareness and ensure that perso

Degger [83]

True becauseidbajqkfkmxzmbakwlgof

4 0

3 years ago

A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 54 percent. Determine the rate of hea

Gennadij [26K]

Answer:

$\dot Q _{L} = 511.111 MW$ . Heat transfer can be higher if themal efficiency is lower.

Explanation:

The heat transfer rate to the river water is calculated by this expression:

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = (\frac{1}{\eta_{th}}-1 )\cdot \dot W\\\dot Q_{L} = (\frac{1}{0.54}-1)\cdot (600 MW)\\\dot Q _{L} = 511.111 MW$

The actual heat transfer can be higher if the steam power plant reports an thermal efficiency lower than expected.

8 0

3 years ago