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2 years ago

What do you mean by overflow and underflow error in array?.

Engineering

1 answer:

Misha Larkins [42]2 years ago

5 0

Answer:

Overflow and underflow are both errors resulting from a shortage of space. On the most basic level, they manifest in data types like integers and floating points. Unlike the physical world, the number stored in a computer exists in a discrete number of digits.

Explanation:

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. Determine the state of stress at point A on the cross-section at section a-a of the cantilever beam. Show the results in a dif

bezimeni [28]

The answer to this problem is the results of the point A

5 0

2 years ago

A flywheel performs each of these functions except: A. Contains a gear used for engine starting B. Smoothes engine operation C.

torisob [31]

Answer:

C. Provides lubrication to parts

Explanation:

Flywheel :

Flywheel is a device which stored the mechanical energy.This energy can be use when more energy required during any operation.Due to high moment of inertia of the flywheel it resist the change in the speed .The flywheel is attached to the crank shaft at the rear side of the engine.

The flywheel perform following function:

1. It connects the crankshaft and the transmission system.

2.It makes the engine operation smooth.

3.It contains gear and other parts of the engine.

But it can not provide lubrication.

C. Provides lubrication to parts

4 0

2 years ago

We would like to measure the density (p) of an ideal gas. We know the ideal gas law provides p= , where P represents pressure, R

Nostrana [21]

Answer: =

Explanation:

= P / (R * T) P- Pressure, R=287.058, T- temperature

From the given that

Sample mean(pressure) = 120300 Pa

Standard deviation (pressure) = 6600 Pa

Sample mean(temperature) = 340K

Standard deviation(temperature) = 8K

To calculate the Density;

Maximum pressure = Sample mean(pressure) + standard deviation (pressure) = 120300+6600 = 126900 Pa

Minimum pressure = Sample mean (pressure) - standard deviation (pressure)= 120300-6600 = 113700 Pa

Maximum temperature = Sample mean (temperature) + standard deviation (temperature) = 340+8 = 348K

Minimum temperature = Sample mean (temperature) - standerd deviation (temperature) = 340-8 = 332K

So now to calculate the density:

Maximum Density= Pressure (max)/(R*Temperature (min))= 126900/(287.058*332)= 1.331

Minimum density=Pressure(min)/(R*Temperature (max))= 113700/(287.058*348)= 1.138

Average density= (density (max)+ density (min))/2= (1.331+1.138)/2= 1.2345

cheers i hope this helps

5 0

3 years ago

A beam has been fixed to the floor by the pin at B and the roller at A as shown in figure 1 below.

ahrayia [7]

What figure below???

3 0

2 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

2 years ago