A fountain shoots a jetof water straight up. The nozzle is 1 cm in diameter and the speed of the water exiting the nozzle is 30
m/s. What is the force exerted by the water jet
1 answer:
Answer:
Explanation:
mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water
A = 3.14 x .005²
= 785 x 10⁻⁷ m²
mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg
momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .
Rate of change of momentum = 706.5 x 10⁻⁴
Let force be F
F - mg = 706.5 x 10⁻⁴
F = mg + 706.5 x 10⁻⁴
F = 23.55 x 10⁻⁴ x 9.8 + 706.5 x 10⁻⁴
= 937.3 x 10⁻⁴ N .
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