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2 years ago

What is software certification? Discuss its importance in the changing scenario of software industry.

1 answer:

6 0

The software industry is evolving, and software certification is crucial. Software certification is a digitally signed certificate that verifies the identification of an individual or company.

<h3>What is software certification?</h3>

A certificate, sometimes known as a digital certificate, is a special document that has been digitally signed and authenticates the identity of a person or business.

Its validity may be checked using public-key cryptography to make sure that the program or website you are accessing is reliable.

Software developers must demonstrate that their program executes the XBRL conformance suites successfully in order to be certified.

They must also supply extra test cases so that XBRL employees can test the program further. The validity of certification is one year, renewable yearly.

Hence the software industry is evolving, and software certification is crucial.

To learn more about the software certification refer;

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erik [133]

Answer:

thanks hot hot

Explanation:

4 0

3 years ago

A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to

Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 = (16/3) * 10^2 V

c) E = 64/15 J

d) dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the ﬁnal potential difference across each capacitor

(d) the decrease in energy when the capacitors are connected.

Solution:

- The initial charge in the circuit is the one carried by the first charged capacitor.

Q_initial = C_1*V

Q_initial = 20*10^-6 * 800

Q_initial = 16 * 10^-3 C

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

Q_2 = Q_1*C_2/C_1

Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

Q_initial = 1.5*Q_1

Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

V_2 = V_1 = Q_1 / C_1

= 10.67*10^-3 / 20*10^-6

= (16/3) * 10^2 V

- The energy in the system is:

E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

E = 64/15 J

- The decrease in energy of the capacitors is:

dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

dE = 0.5*10^-6*20*800^2 - (64/15)

dE = 32/5 - 64/15 = 32/15 J

5 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

3 years ago

Create a series of eight successive displacements that would program a robot to move in an octagonal path that is as close as yo

Komok [63]

Answer:

bts biot bts biot jungkukkk

jungkukkkbiot

Explanation:

bts biot bts biot jungkukkk

jungkukkkbiot

5 0

3 years ago

What is software certification? Discuss its importance in the changing scenario of software industry. ​

What is software certification? Discuss its importance in the changing scenario of software industry.