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3 years ago

Why data structure is important

Engineering

1 answer:

Readme [11.4K]3 years ago

8 0

Answer:

Data structures are important as it allows the user to insert, update, arrange, rearrange, delete, and retrieve data in an efficient manner, from the database. And to accomplish the said tasks algorithms are used. It is used to manipulate the stored data within the database in the required manner. …

Explanation:

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A jet impinges directly on to a plate that is oriented normal to the axis of the jet. The mass flow rate of the jet is 50 kg/min

PilotLPTM [1.2K]

Answer:

166.67 N

Explanation:

Given:

Mass flow rate = 50 kg/min = 50 kg / 60 seconds = 0.833 kg/s

Initial velocity = 200 m/s

after striking the normal board the water will flow in the normal direction, thus the final velocity in the direction of the initial flow will be zero

therefore,

Force = (change in momentum)

Force = Initial momentum - final momentum

Force = 0.833 × 200 - 0.833 × 0

Force = 166.67 N

3 0

3 years ago

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

8 0

4 years ago

Technician A says that squeeze-type resistance spot welding (STRSW) may be used on open butt joints. Technician B says that repl

Mamont248 [21]

Answer:

B only

Explanation:

Squeeze-type resistance spot welding (STRSW)is a type of electric resistance welding that brings about the weld on interfacing sheet metal pieces through which heat generated from electric resistance bring about fusion and welding of the two pieces together

Therefore, it is not meant for opening but joints but it can be used for making replacement spot welds adjacent to the original spot weld due to the smaller heat affected zone (HAZ) created by the STRSW process.

6 0

3 years ago

Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe

NemiM [27]

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075