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erma4kov [3.2K]
2 years ago
13

The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the n

umber of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter dd. Express your answer in terms of some or all of the variables rrr, RRR, ddd, and TTT. nn

Engineering
1 answer:
AleksAgata [21]2 years ago
8 0

Answer:

n = 2r³/Rd²

Explanation:

See the attached file for the derivation.

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The organic acid, ACOOH, reacts reversibly with the alcohol BOH, to form the ester ACOOB according to the stoichiometric equatio
kompoz [17]

Answer:

The time taken to achieve the 0.9 fractional conversion, t = 28.125 hours

Explanation:

The first image attached is the picture showing the question clearly.

The 2nd attached image is page 1 of the solution and the 3rd attached image is the page 3 of the solution.

4 0
3 years ago
A sprinter reaches his maximum speed in 2.5sec from rest with constant acceleration. He then maintains that speed and finishes t
liubo4ka [24]

The maximum Speed of the Sprinter from the velocity time graph of his motion is; 11.98 m/s

<h3>How to find the maximum speed?</h3>

We are given;

Initial Speed; u = 2.5 s

Total distance; d = 100 m

Total time; T = 9.6 s

The total distance is;

d = ¹/₂(9.6 + (9.6 - 2.5) * v

where v is maximum speed.

Thus;

¹/₂(9.6 + (9.6 - 2.5) * v = 100

16.7v = 200

v = 200/16.7

v = 11.98 m/s

Read more about Maximum Speed at; brainly.com/question/4931057

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3 0
2 years ago
Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output
Vera_Pavlovna [14]

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel (R_T) is gotten by using the formula:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm

The current flowing through the amplifier (I) is:

I = V / R_T = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / R_T = I²R_T. Therefore:

P = V² / R_T = 15² / R_T = 112.5 watts

6 0
2 years ago
An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th
photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

{W_{u}}= \eta  {W_{pump}}  .............1

put here value

{W_{u}}  = (0.80)(20)

{W_{u}} = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

\Delta{E_{mech}} = {m} \Delta pe   ..............2

\Delta {E_{mech}} = {m} g \Delta z  

and that can be express as

\Delta {E_{mech}} = \rho Q g \Delta z     ..................3

so

\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]      ......4

solve it we get

\Delta {E_{mech}} = 13.614 hp

so here

due to frictional effects, mechanical power lost in piping

we get here

{W_{frict}} = {W_{u}}-\Delta {E_{mech}}  

put here value

{W_{frict}} = 16 -13.614

{W_{frict}} = 2.37  hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0
2 years ago
An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205
lara [203]

Answer:

0.845\ \text{N}

Explanation:

g = Acceleration due to gravity at sea level = 9.81\ \text{m/s}^2

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2

Weight at sea level

W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}

Weight at the given height

W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}

Change in weight W_h-W=636.805-637.65=-0.845\ \text{N}

Her weight reduces by 0.845\ \text{N}.

8 0
3 years ago
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