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ozzi
3 years ago
13

Answer?...................

Engineering
1 answer:
torisob [31]3 years ago
5 0

Answer:

The correct option is;

c. Leaving the chuck key in the drill chuck

Explanation:

A Common safety issues with a drill press leaving the chuck key in the drill chuck

It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.

It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable

Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.

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A motorist is driving his car at 60km/hr when he observes that a traffic light 250m ahead turns red. The traffic light is
Alecsey [184]

Explanation:

Okay soo-

Given-

u = 60 km/hr = 60×1000/3600=50/3 m/s

t = 20 s

s = 250 m

a = ?

v = ?

Solution -

Here, acceleration is uniform.

(a) According to 2nd kinematics equation,

s = ut + ½at^2

250 = 50/3 ×20 + 0.5×a×20×20

250-1000/3=200a

(750-1000)/3=200a

a = -250/(3×200)

a = -5/12

a = 0.4167 m/s^2

The required uniform acceleration of the car is 0.4167 m/s^2.

(b) According to 1st kinematics equation

v = u + at

v = 50/3 + (-5/12)×20

v = 50/3-25/3

v = 25/3

v = 8.33 m/s

The speed of the car as it passes the traffic light is 8.33 m/s.

Good luck!

5 0
3 years ago
Consider an ideal cogeneration steam plant to generate power and process heat. Steam enters the turbine from the boiler at 7 Mpa
igomit [66]

Answer:

1. The diagram T-s or H-s is attached to this answer.

2. The fraction of the steam extracted is 4.088Kg/s

3. The net Power produced per kg of steam exiting the boiler is 1089.5KJ/Kg.

4. The mass flow rate of steam supplied by the boiler is 16.352Kg/s

5. the net power produced by the plant is 11016.2KJ/s.

6. The utilization factor is 0.218.

Explanation:

To analyze this problem we need to find all the thermodynamic coordinates of the system. In the second image attached to this answer, we can see the entire ideal cogeneration steam plant system.

From a water thermodynamic properties chart, we can obtain the information for each point.

+ Steam enters the turbine from the boiler at 7 Mpa and 500 degrees C:

h₆=3410.56KJ/Kg

s₆=6.7993 KJ/Kg

This is an ideal cogeneration steam system, therefore: s₆=s₇=s₈

+One-fourth of the steam is extracted from the turbine at 600-kPa:

h₇(s₇) = 2773.74 KJ/Kg (overheated steam)

+The remainder of the steam continues to expand and exhausts to the condenser at 10 kPa.

h₈(s₈)=2153.58 KJ/Kg (this is wet steam with title X=0.8198)

h₁(P=10Kpa)= 191.83 KJ/Kg (condensed water) s₁=0.64925KJ/Kg

-This flow is pumped to 600KPa, so:

s₂=s₁

h₂(s₂)=192.585KJ/Kg

+The steam extracted for the process heater is condensed in the heater:

h₃(P=600KPa)=670.42KJ/Kg (condensed water)

+The steam extracted for the process heater is condensed in the heater and mixed with the feed-water at 600 kPa:

The mixing process of the flow of point 2 and 3 is an adiabatic process, therefore:

\dot{Q}_4=\dot{Q}_2+\dot{Q}_3=\dot{m}_2 h_2+\dot{m}_3h_3\\\dot{m}_4 h_4=\dot{m}_2 h_2+\dot{m}_3h_3=\dot{m}_2 0.75h_4+\dot{m}_30.25h_4\\h_4=0.75h_2+0.25h_3=312.043KJ/Kg

s₄=1.02252

+the mixture is pumped to the boiler pressure of 7 Mpa:

s₅=s₄

h₅(s₅)=323.685KJ/Kg

1)Now we have all the thermodynamic coordinates and we can draw the diagram of the system.

2) To determine the fraction of steam, the mass flow that is extracted from the turbine at state 7, we use the information that this flow is used to generate 8600KJ/s in a process of heat. Therefore:

P=8600KJ/s=\dot{m}_{3-7}(h_7-h_3)\\\dot{m}_{3-7}=8600KJ/s/(h_7-h_3)=4.088Kg/s

3)The net power produced per kg of steam exiting the boiler can be obtained as the rest between all the power obtained in the turbine less the power used in the pumps:

P_{turb}/Kg=(h_6-h_7)+0.75(h_7-h_8)=1101.94KJ/Kg\\P_{pump1}/Kg=h_2-h_1=0.755KJ/kg\\P_{pump2}/kg=h_5-h_4=11.642KJ/Kg\\P_{net}/kg=P_{turb}-P_{pump1}-P_{pump2}=1089.543KJ/Kg

4) To determine the mass flow rate of steam that must be supplied by the boiler, we only have to remember that the flow used in point 2) is a fourth of the total flow. therefore:

0.25\dot{m}_{tot}=\dot{m}_{3-7}\\\dot{m}_{tot}=4\dot{m}_{3-7}=16.352Kg/s

5)The net power supplied by the plant is the net power calculated in point 3) less the power used in the heat process 7-3:

P_{net sys}=P_{net}/kg \cdot \dot{m}_t-6800Kj/s=11016.2KJ/s

6) The utilization factor is obtained as the division between net power supplied by the plant and the power used to heat the steam. In this case:

UF=P_{net sys}/P_{boil}=P_{net sys}/[\dot{m}_t(h_6-h_5)]=0.21

3 0
3 years ago
Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave f
fenix001 [56]

Answer:

hello your question has a missing part below is the missing part

Consider the string length equal to \pi

answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)

Explanation:

Given string length = \pi

distorted function f(x) = 2sin(2x) - 10sin(10x)

Determine the wave formed in the string

attached below is a detailed solution of the problem

8 0
3 years ago
The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diam
timama [110]

Answer:

See explanations for completed answers

Explanation:

Given that; The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diameter of 50 mu m has a strength of 140 MPa. (a) Determine the constants for the Hall-Patch equation for this alloy, (b) How much more should you reduce the grain size if you desired a strength of 220 MPa

See attachment for completed solvings

6 0
3 years ago
The velocity of a fluid flowing between 2 parallel plates is given as a function of the distance between the plates (y) by the e
daser333 [38]

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4 0
3 years ago
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