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2 years ago

13

which of the following tools is used for measuring small diameter holes which a telescoping gauge cannot fit into? A. telescopin

g gauges B. Feeler Guages C. Dial indicator D. Small hold gauges

1 answer:

Alex2 years ago

7 0

Answer:

C. Dial indicator

Explanation:

This meassers small diameters

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A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to

poizon [28]

Answer:

i) The torque required to raise the load is 15.85 N*m

ii) The torque required to lower the load is 6.91 N*m

iii) The minimum coefficient of friction is -0.016

Explanation:

Given:

dm = mean diameter = 0.03 m

p = pitch = 0.004 m

n = number of starts = 1

The lead is:

L = n * p = 1 * 0.004 = 0.004 m

F = load = 7000 N

dc = collar diameter = 0.035 m

u = 0.05

i) The helix angle is:

$tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43$

The torque is:

$T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm$

ii) The torque to lowering the load is:

$T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm$

iii)

$T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\ 0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}$

Clearing u:

u = -0.016

5 0

3 years ago

python Write a program that takes a date as input and outputs the date's season. The input is a string to represent the month an

kupik [55]

Answer:

month = input("Input the month (e.g. January, February etc.): ")

day = int(input("Input the day: "))

if month in ('January', 'February', 'March'):

season = 'winter'

elif month in ('April', 'May', 'June'):

season = 'spring'

elif month in ('July', 'August', 'September'):

season = 'summer'

else:

season = 'autumn'

if (month == 'March') and (day > 19):

season = 'spring'

elif (month == 'June') and (day > 20):

season = 'summer'

elif (month == 'September') and (day > 21):

season = 'autumn'

elif (month == 'December') and (day > 20):

season = 'winter'

print("Season is",season)

Explanation:

4 0

2 years ago

A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stre

disa [49]

Answer:

0.234

Explanation:

True stress is ratio of instantaneous load acting on instantaneous cross-sectional area

σ = k × (ε)^n

σ = true stress

ε = true strain

k = strength coefficient

n = strain hardening exponent

ε = ( σ / k) ^1/n

take log of both side

log ε = $\frac{1}{n}$ ( log σ - log k)

n = ( log σ - log k) / log ε

n = (log 578 - log 860) / log 0.20 = 0.247

the new ε = ( 600 / 860)^( 1 / 0.247) = 0.234

6 0

2 years ago

Read 2 more answers

The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24C, and a wet-bulb temperature of 17C. Using the psychrome

TEA [102]

Answer:

(a) Relative Humidity = 48%,

Specific humidity = 0.0095

(b) Enthalpy = 65 KJ/Kg of dry sir

Specific volume = 0.86 m^3/Kg of dry air

(c/d) 12.78 degree C

(e) Specific volume = 0.86 m^3/Kg of dry air

8 0

2 years ago

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in $2.25 * 10^4 BTU$ . It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - $1 BTU = 778.17 lbf*ft$

$2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft$

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.

$h_2$ is the final elevation and $h_1$ is the initial elevation.

Replacing W in the Ep equation

$Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\$

Finally, the next step is to replace the variables of the problem.

$h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft$

The elevation at the high point of the road is 12186.5 in ft.

3 0

2 years ago