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zavuch27 [327]
2 years ago
11

you spray your sister with water from a garden hose. the water is supplied to the hose at a rate of 0.649x10-3 m^3/s and the dia

meter of the nozzle you hold is 5.45x10^-3 m. at what speed does the water exit the nozzle?
Physics
1 answer:
Ivanshal [37]2 years ago
7 0

Answer:

27.82 m/s

Explanation:

The radius of the hose is half of its diameter

r = d/2 = 5.45\times10^{-3}/2 = 0.002725 m

So its area must be

A = \pi r^2 = \pi 0.002725^2 = 2.33\times10^{-5} m^2

The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose

v = \dot{V}/A =0.000649 / 2.33\times10^{-5} = 27.82 m/s

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A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. If the bullet has a mass of 22.
Bogdan [553]

Answer:

A. 1.172 metres

B. 6.82 Ns

C. 4.796 m/s

Explanation:

The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.

The 1.40 kg block is at rest so velocity is zero and has no momentum.

The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s

Momentum = 310*0.022

Momentum = 6.82 Ns.

If the bullet gets embedded they will both have common velocity v

6.82 = (0.022+1.40)v

6.82 = 1.422v

V = 6.82/1.422

V = 4.796 m/s

How high the block will rise after the bullet is embedded is given by

H = (U²Sin²tita)/2g

Where tita is 90°

H = (4.796² * sin²(90))/(2*9.81)

H =( 23.001616*1)/19.62

H = 1.172 metres

6 0
3 years ago
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