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IrinaK [193]
4 years ago
14

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

cm in diameter and the outlet flow is 50 L/s.
Calculate the propulsion force at the moment of departure, that is, with the boat at rest. The pressure at the inlets and outlets is assumed to be practically atmospheric (water density = 1000 Kg /m3).
Physics
1 answer:
Phoenix [80]4 years ago
7 0

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

propulsion force, F = 1000 x 0.005 x 9.8

propulsion force, F = 49 N      

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

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Three moons can fit inside the volume of the sun.

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Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

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6 0
2 years ago
A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

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3 years ago
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Answer:

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A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must
umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

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Hence the student must push at a distance of 6.58m

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