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weqwewe [10]
3 years ago
14

You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun us

es a plunger which is operated by pulling back on a handle. You must squeeze the handle very hard to get the caulk to come out of the narrow opening because
Physics
1 answer:
ZanzabumX [31]3 years ago
3 0

Answer:

The answer is

A. Pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

Explanation:

 The question is incomplete, here is a complete question with full options

You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun uses a plunger which is operated by pulling back on a handle. You must squeeze the handle very hard to get the caulk to come out of the narrow opening because:_________.

A. pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

B. viscous drag between the walls of the tip and the caulk causes the caulk to swirl around chaotically.

C. Newton’s third law requires most of the energy in the caulk to be used to push back on the plunger rather than moving it through the tip.

D. the high density of the caulk impedes its flow through the small opening.

Since the caulk is thick and the exit nozzle is small, the pressure needed to deliver the caulk will be very high as pressure is uniformly distributed at the plunger side at every part of the caulk, hence very high pressure is needed to deliver the caulk which is why the handle needed the very hard squeeze

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Explanation:

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A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
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Answer:

851.33 N

Explanation:

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v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

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Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

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Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

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3 years ago
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