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ryzh [129]
4 years ago
15

Two disadvantages of spreadsheet

Physics
2 answers:
vodka [1.7K]4 years ago
7 0
It require so much time for maintenance, substantial startup reliant on the database
liraira [26]4 years ago
3 0
One disadvantage of spreadsheet is Vulnerable to fraud. Many company losses billion-dollars due to Fraudulent manipulation in company excel.  The reason behind this spreadsheet vulnerability is lack of controls, which makes it so easy to alter formulas , values or dependencies without being detected. Another disadvantage is Susceptible to trivial human errors. Sometimes you may missed negative signs and misaligned rows may sound harmless.
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Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?
ozzi

Answer: The forces acting on both of them will increase in magnitude.

Explanation:

According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.

6 0
3 years ago
Which list places the layers of the sun in the correct order from outermost to innermost?
Valentin [98]
Answer - corona, chromosphere, photosphere
4 0
3 years ago
How many Joules of potential
s2008m [1.1K]

Answer:

40 j, 80j.

Explanation:

P.E= mgh. G=10 m/s².

For 4m, P.E=1*10*4=40 joules.

For 8m, P.E=1*10*8=80 joules.

4 0
4 years ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
TiliK225 [7]

Answer:

E = 6.77 *10^{3} N/C

Explanation:

THE GIVEN sheet can be taken as two horizontal force  with surface charge density is

at one surface is ∈_1 = 95*10^ {-9} C/mm2

at oher surface is ∈_2= -25*10^{-9} C/mm2

the magnitude of electric field due to surface charge is given as\frac{∈}{2∈_O}

So, electric field at P (2 CM below from surface is) = E_1 +E_2

E = \frac{∈_1}{2∈_O} +\frac{∈_2}{2∈_O}

E = \frac{95*10^{-9} +25*10^{-9}}{2*8.85*10^{-12}}

E = 6.77 *10^{3} N/C

8 0
4 years ago
The resistanceless inductor is connected across the ac source whose voltage amplitude is 24.5 V and angular frequency is 850 rad
timurjin [86]

Answer:

Part A 2.88 A, PART B 28.82 mA PART C  0.288 mA

Explanation:

We have given angular frequency \omega =850rad/sec

Voltage source has an amplitude of 24.5 V, So V= 24.5 volt

Part A

We have given inductance L=10^{-2}H

So inductive reactance X_L=\omega L=850\times 10^{-2}=8.5ohm

So current i=\frac{V}{X_L}=\frac{24.5}{8.5}=2.8823A

Part B

We have given inductance L=1 H

So inductive reactance X_L=\omega L=850\times 1=850ohm

So current i=\frac{V}{X_L}=\frac{24.5}{850}=28.82mA

Part C

We have given inductance L=100 H

So inductive reactance X_L=\omega L=850\times 100=85000ohm

So current i=\frac{V}{X_L}=\frac{24.5}{85000}=0.288mA

8 0
3 years ago
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