Question

A firefighter holds a water pipe of cross-section 1 mm” The water comes out at a rate 200 cc's The reactionary force exerted on the hand of the fire fighter is ​

Answer 1

Answer:

answer is 10 N

Explanation:

Mark me at the brainliest answer

Answer 2

Answer:

40 N

Explanation:

A=1mm² = 102cm²

Discharge rate = 200 cc / s

So, v = 200 / A m/s

= 20* 10⁴ m/s

New change in momentum in one second

delta(M)= PAV

delta(P) = delta(M)*V = ρAV xv = ρAv²

F =P. (AV) v

= 1× 200 × 2 × 10* dyne

=40×105 dyne

Favy = 40 newton