All categories

3 years ago

The use of alcohol can cause

Engineering

2 answers:

Ann [662]3 years ago

7 0

Answer:mental changes

Explanation:

makkiz [27]3 years ago

6 0

Answer:

Depression, and health problems

You might be interested in

What information is usually gathered during vehicle crash tests?

aleksley [76]

Time, distance, damage

7 0

3 years ago

Explain 3 types of lubrication occurs in bearings?

Angelina_Jolie [31]

Answer:

The three types of lubrication occurred in the bearing are:

When the lubrication occur in the bearing it play an important role in the rolling life of bearing element. Basically, lubricant task is to reduced the friction.

1) Grease lubricant: Grease contain various type of additives which help in enhance the performance. It consist of oil where thickness are added as, this thickness improved the characteristics of grease.

2) Oil lubricant: This is used to reduced or minimized the friction and the lubricant oil is used to in those vehicles when they are motorized. Bearing lubricant oil is used to higher the speed capability.

3) Solid films: These are non fluid coating surface that are applied in the friction surface for prevention. It is used in very extreme situation where oil and grease types of lubricant does not work.

8 0

3 years ago

A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$

r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Now by putting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

Ф=18.03°

So the shear angle is 18.03°.

4 0

4 years ago

What is the air change rate (ACH) for a 100 ft^2 (9.3 m^2) space with a 10 ft (3.0 m) ceiling and an airflow rate of 200 cfm (95

kakasveta [241]

Answer:

The ACH is 12/h

Solution:

As per the question:

Area of the space, $A_{s} = 100 ft^{2}$

Height of the given space, h = 10 ft

Air flow rate, $Q_{a} = 200 cfm$

Now, to find the Air Change Rate (ACH):

We calculate the Volume of the given space:

$V_{s} = A_{s}\times h = 100\times 10 = 1000 ft^{3}$

Now, the ACH per min:

= $\frac{V_{s}}{Q_{a}} = \frac{1000}{200} = 5/min$

Now, ACH per hour:

= $\frac{60}{5} = 12/h$

3 0

4 years ago

13–27. The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-k

Feliz [49]

Answer:

See explanation for step by step procedure to get answer.

Explanation:

Given that:

The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.

See the attachments for complete steps to get answer.

4 0

3 years ago