To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
Answer:
Electric potential = 0.00054 V
Explanation:
We are given;
Charge; q = 3 pC = 3 × 10^(-12) C
Radius; r = 2 cm = 0.02 m
Formula for the electric potential of this surface will be;
V = kqr
Where;
K is a constant = 9 × 10^(9) N⋅m²/C².
Thus;
V = 9 × 10^(9) × 3 × 10^(-12) × 0.02
V = 0.00054 V
Answer:
300 m/s
Explanation:
The difference in time between the two bangs is 1 s.
Thus;
t2 - t1 = 1
We know that distance/time = speed.
Thus;
d2/v - d1/v = 1
Multiply through by v to get;
d2 - d1 = v
Where v is speed of sound in air.
d1 = 350 m
d2 = (150 × 2) + 350 = 650 m
Thus;
v = d2 - d1 = 650 - 350 = 300 m/s
Flame of fire could get put out with water
