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Agata [3.3K]
2 years ago
11

A sailcraft is stalled on a windless day. A fan is attached to the craft and blows air into the sail which bounces backward upon

impact. The boat can what?
Physics
1 answer:
mylen [45]2 years ago
7 0

Answer:

Impulse = change in momentum w bounce

There are 2 impulses acting. Recoil of the fan going the negative direction and the impulse of the air bouncing off the sail. The greater impulse will bounce so the direction will be to the right moving the craft.

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When a guitar string is plucked, in what direction does the wave travel? In what directions does the string vibrate?
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A rocket is fired from rest from the ground (y = 0) at time t0 = 0 s. As the rocket is burning its fuel, it moves vertically upw
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3 years ago
A 75,000-watt radio station transmits at 88 MHz. Determine the number of joules transmitted per second. 88 J/s 8800 J/s 2,000,00
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2 years ago
A boat has a propulsion system that consists of a pump that sucks water at the bow and presses it on the stern. All tubes are 5
Phoenix [80]

Answer:

The propulsion force at the moment of departure, is 49 N  

Explanation:

Given;

diameter of tubes = 5 cm = 0.05 m

volumetric flow rate, V = 50 L/s = 0.005 m³/s

density of water, ρ = 1000 Kg /m³

hydraulic / propulsion force, F = ρVg

where;

ρ is the density of the fluid (water)

V is the volumetric flow rate of water

g is acceleration due to gravity

propulsion force, F = ρVg

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7 0
3 years ago
Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa
zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, \rho _o = 926 kg/m³

density of the wood, \rho _{wood} = 974 kg/m³

density of water, \rho _w = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood}  -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood}  -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0
3 years ago
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