Answer:
1. Effect of air pressure
2. air- powered wheel chair
3. Pneumatic valves
Explanation:
1. In any pneumatic device, the mipact of air pressure to produce the moving effect on an heavy object is unexpected.
2. pneumatice demultiplexer when air in comprressed tank is allowed released to cause movement of the chair.
3. In industries, a pneumatic valve operates by force of air when actuated. A signal causes actuation of coil. When coil is energized, compressed high pressure air is allowe to enter in a small cylinder and cause operation of valve
Answer:
n this question, we are asked to find the probability that
R1 is normally distributed with mean 65 and standard deviation 10
R2 is normally distributed with mean 75 and standard deviation 5
Both resistor are connected in series.
We need to find P(R2>R1)
the we can re write as,
P(R2>R1) = P(R2-R1>R1-R1)
P(R2>R1) = P(R2-R1>0)
P(R2>R1) = P(R>0)
Where;
R = R2 - R1
Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.
u = u2-u1
u = 75 - 65 = 10ohm
sd = √sd1² + sd2²
sd = √10²+5²
sd = √100+25 = 11.18ohm
Now we will calculate the z-score, to find P( R>0 )
Z = ( X -u)/sd
the z score of 0 is
z = 0 - 10/11.18
z= - 0.89
Answer:
A. True
Explanation:
When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, 
Number of passes necessary for this reduction, 
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
minimum time required to reduce the depth of the plate by 20 mm:
number of passes * Time/pass
n * Time/pass
40 * 40
1600 = 26 mins 40 secs
Answer:
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