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Gnoma [55]
2 years ago
5

At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c

oefficient of only about 0.090. a. What is the drag force acting on such a dolphin swimming at 7.5 m/s? b. Using the dolphin’s diameter as its characteristic length, what is the Reynolds number as it swims at this speed in 20° C water?
Physics
1 answer:
fiasKO [112]2 years ago
8 0

Answer:

497.00977 N

3742514.97005

Explanation:

\rho = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N

The drag force on the dolphin's nose is 497.00977 N

at 20°C

\mu = Dynamic viscosity = 1.002\times 10^{-3}\ Pas

Reynold's Number

Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005

The Reynolds number is 3742514.97005

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Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com
Lubov Fominskaja [6]

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm   ==>   area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

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Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

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1 year ago
Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg
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Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
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2 years ago
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A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point
denis-greek [22]

Answer:

V0=27.4 m/s; t=0.8 s

Explanation:

Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:

y_f=v_0t-0.5*g*t^2 We solve for initial speed

v_0=\frac{y_f+0.5gt^2}{t}=\frac{37+4.9*2.3^2}{2.3}=27.4m/s

Now, using the same expression we estimated time to first reach 18.5 m :

18.5=27.4t-4.9t^2 Second order equation with solutions

t1=0.8 s and t2=4.8 s

The first time corresponds to the first reach.

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The velocity of a 10.0 kg object that has 720 J of kinetic energy is m/s. (Report the answer to two significant figures.)
nikdorinn [45]
Kinetic energy<span> is the </span>energy<span> of motion. An object that has motion - whether it is vertical or horizontal motion - has </span>kinetic energy<span>. It is expressed as:

KE = mv^2 /2

720 = 10.0v^2 /2
v = 12 m/s

Hope this answers the question. Have a nice day.</span>
4 0
2 years ago
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Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿
Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.  

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0
2 years ago
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