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.
Rajashree kamble said:
1 decade ago
(6*10^3) (5*0^5).
6*5 = 30 and power 3+5 = 8.
30*10^8.
6*5 = 30 and power 3+5 = 8.
30*10^8.
Lallan.k.puskar said:
1 decade ago
(6*10^3)*(5*10^5) = 6*5*10^3+5 = 30*10^8 Simple formula of use (a^n)*(a^m) = a^n+m.
Aashish jayshwal said:
1 decade ago
a^m*a^n = a^m+n.
6*5 = 30 normally multiple 10^5*10^3 = 10^8.
6*5 = 30 normally multiple 10^5*10^3 = 10^8.
Sanghamitra Samal said:
1 decade ago
(6*10^3)*(5*10^5).
10^3*10^5 = 10^8.
6*5 = 30.
10^3*10^5 = 10^8.
6*5 = 30.
Anand kumar said:
1 decade ago
{a*10^m}*{b*10^n}.
a and b multiply each other.
The power of 10 sum of each other.
{a*b}*10^(m+n).
{6*5}*10^(3+5).
30*10^8.
a and b multiply each other.
The power of 10 sum of each other.
{a*b}*10^(m+n).
{6*5}*10^(3+5).
30*10^8.
Khaja said:
1 decade 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.
Santhosh kumar said:
1 decade ago
The power dissipated by a resistor of 10 ohm when a current of 2A passes through it is.
Shailu said:
1 decade ago
Is correct answer?
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.
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.
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