In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.
a). In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.
b). In series, the total current is V / (3R) .
In parallel, the total current is 3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.
c). In series, the power dissipated by the circuit is
(V) · V/3R = V² / 3R .
In parallel, the power dissipated by the circuit is
(V) · 3V/R = 3V² / R .
Changing from series to parallel, the power dissipated by
the circuit (also the power delivered by the battery) increases
to 9 times its original value.
Answer:
1989.6Kg
Explanation:
The computation of the mass of the other body is given below:
As we know that
F = G × m1 × m2 ÷ r²
Here the G would have the constant value i.e. 6.67 × 10^-11Nm² / kg².
Now
6.5 × 10^-7N = 6.67 × 10^-11Nm² / kg² × 60Kg × m2 / (3.5m) ²
m2 = (F × r²) / (G × m1)
m2 = (6.5 × 10^-7N × (3.5m) ²) ÷ (6.67 × 10^-11Nm² / kg² × 60Kg)
= 1989.6Kg
Answer:
d = 1.27m
Explanation:
Given m = 8.2×10-26kg, v = 1.1×10⁶m/s, q = +5e = 5×1.6×10‐¹⁹ C.
E = 49kN/C = 49000N/C
The displacement is given by
d = 1/2× mv²/qE = 1/2 × 8.2×10-²⁶ × (1.1×10⁶)²/(5×1.6×10-¹⁹ ×49000) = 1.27m
In paint industry ..and also in steel industry
