Electrical Engineering - Quantities and Units - Discussion
Discussion Forum : Quantities and Units - General Questions (Q.No. 1)
1.
When these numbers are multiplied, (6 × 103) (5 × 105), the result is
Discussion:
50 comments Page 3 of 5.
Ela said:
10 years ago
6*5 = 30.
10^3*10^5 = 10^(3+5) = 10^8.
30*10^8.
10^3*10^5 = 10^(3+5) = 10^8.
30*10^8.
Raji said:
1 decade ago
Since the given numbers have a significant figure in each (6, 5), its product has two significant figures after multiplication.
= 6*10^3*5*10^5 = 30*10^8.
= 6*10^3*5*10^5 = 30*10^8.
Shailu said:
1 decade ago
6*5 = 30.
10 is common for both the values,
Powers are added.
10^3+5 = 10^8.
Total value = 30*10^8.
10 is common for both the values,
Powers are added.
10^3+5 = 10^8.
Total value = 30*10^8.
Shailu said:
1 decade ago
Is correct answer?
Santhosh kumar said:
1 decade ago
The power dissipated by a resistor of 10 ohm when a current of 2A passes through it is.
Nandu said:
1 decade ago
When we are multiplying of these numbers the powers of those numbers are added.
Ashik said:
1 decade ago
(a^m)*(a^n)=a^(m+n)
[p*(a^m)]*[q*(a^n)]=(p*q)*[a^(m+n)]
Therefore (6 x 10^3)*(5 x 10^5)=(6*5)*10^(3+5)
=30*10^8
[p*(a^m)]*[q*(a^n)]=(p*q)*[a^(m+n)]
Therefore (6 x 10^3)*(5 x 10^5)=(6*5)*10^(3+5)
=30*10^8
Rameshbabu said:
1 decade ago
Thank you for easy understandable explanation.
Murali said:
1 decade ago
Simply, the numbers 6 and 5 normal multiplication (6*5=30).
In power the numbers has been added (3+5=8).
In power the numbers has been added (3+5=8).
Kitty said:
1 decade ago
If the numbers are mulitiplied the exponents are added.
Thanks for discussion.
Thanks for discussion.
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