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baherus [9]
3 years ago
13

A ship tows a submerged cylinder, which is 1.5 m in diameter and 22 m long, at 5 m/s in fresh water at 208C. Estimate the towing

power, in kW, required if the cylinder is (a) parallel and (b) normal to the tow direction.

Physics
2 answers:
Scilla [17]3 years ago
8 0

Answer: a) P = 120kw

b) P = 800kw

Explanation: Please find the attached file for the solution

den301095 [7]3 years ago
5 0

Answer:

(a) P = 121kW (b) P = 823kW

Explanation:

Given

U = 5m/s, D = 1.5m, L = 22m, ρ = 998kg/m³

(a) Parallel, L/D = 15,

Re = 998×5×22/0.001 = 1.1×10⁸,

Area = π ×D²/4 = π × 1.5²/4 = 1.77m²

From Table estimate Cd,frontal = 1.1

Fd = 1/2 × Cd,frontal × ρ × U² × Area

Fd = 1/2 × 1.1 × 998 × 5² × 1.77 = 24289N

P = F×U = 24289 × 5 = 121445W ≈ 121kW

(b) Normal, Re = 998×5×1.5/0.001 = 7.5×10⁶,

Area = DL = 1.5 × 22 = 33m²

From Table estimate Cd,frontal = 0.4

Fd = 1/2 × Cd,frontal × ρ × U² × Area

Fd = 1/2 × 0.4 × 998 × 5² × 33 = 164670N

P = F×U = 164670 × 5 = 121445W ≈ 823kW

Fd is the drag force exerted by the fluid (water) on the submerged body. This force tends to oppose the motion of the submerged cylinder through it. This force increases with area and also with velocity. So for large area the force of resistance of the water would be large and for a small area it will also be small. The same is true for velocity.

In order for the ship to tow this cylinder successfully, it must provide enough power to overcome the resisting force of the water (the drag force as the name implies.

This force is given by the relation,

Fd = 1/2 × Cd,frontal × ρ × U² × Area

Where Fd = Drag force

Cd , frontal = coefficient of drag

ρ = density of fluid (water,)

U = velocity other body in the fluid

A = surface area of the object made available to the fluid in the direction of travel.

Power = F × U

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Answer:

x_total = 4662.5 m

Explanation:

This is a one-dimensional math problem, we must find the displacement in each section and then add them together.

1) let's use the equation

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where it indicates that part of rest for which v₀ = 0

       x₁ = ½ a₁ t²

       x₁ = ½ 2.60 11²

       x₁ = 157.3 m

2) The second displacement is at constant speed,

   let's find the final speed of the previous displacement

       v = v₀ + a₁ t₁

       v = a₁ t₁

       v = 2.60 11

       v = 28.6 m / s

now we use the uniform speed

      v = x₂ / t₂

      x₂ = v t₂

let's reduce the time to SI units

      t₂ = 2.60 min (60 s / 1min) = 156 s

      x₂ = 28.6  156

      x₂ = 4461.6 m

3) it is braking

       x₃ = v t₃ - ½ a₃ t₃²

       x₃ = 28.6 3.05 - ½ 9.38 3.05²

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4 years ago
A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?
klasskru [66]

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

where a is the acceleration, F is the net force and m is the mass of the object.

Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

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4 years ago
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You are in a car traveling at 20 m/s. An ambulance is behind you traveling 35 m/s in the same direction. What frequency do you h
arlik [135]

Answer:

The frequency heard is 576.78 Hz

Explanation:

The Doppler effect is defined as the apparent frequency change of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

This is what happens in the first part of this problem, where the sender is train A and the receiver is train B. They are both moving in opposite directions. In this case, where both are in motion, the frequency perceived by the receiver will increase when receiver and transmitter increase their separation distance and will decrease whenever the separation distance between them is reduced. The following expression is considered the general case of the Doppler effect:

f'=f*\frac{v+-vR}{v+-vE}

Where:

f ', f: Frequency perceived by the receiver and frequency emitted by the issuer respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s-1)

v: Velocity of propagation of the wave in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 343 m / s

vR, vE: Speed ​​of the receiver and the emitter respectively. Its unit of measure in the S.I. is the m / s

±, ∓:

We will use the + sign:

  • In the numerator if the receiver approaches the emitter
  • In the denominator if the emitter moves away from the receiver

We will use the sign -:

  • In the numerator if the receiver moves away from the emitter
  • In the denominator if the emitter approaches the receiver

In this case you are in a car traveling at 20 m/s and an ambulance is behind you traveling 35 m/s in the same direction.

In this case the receiver, you in the car, moves away from the emitter, while the emitter, the ambulance, approaches the receiver behind you in the same direction. So the frequency is calculated by the expression:

f'=f*\frac{v-vR}{v-vE}

Being:

  • f= 550 Hz
  • v=343 m/s
  • vR= 20 m/s
  • vE= 35 m/s

and replacing:

f'=550 Hz*\frac{343 m/s-20 m/s}{343 m/s-35 m/s}

you get:

f'= 576.78 Hz

The frequency heard is 576.78 Hz

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