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Slav-nsk [51]
3 years ago
15

Calculate the Reynolds number for a person swimming through maple syrup. The density of syrup is about 1400 kg/m^3 and the visco

sity is about 0.5 Pa's. A person is about 2m in length and can swim about 1 m/s.
Physics
1 answer:
Tema [17]3 years ago
5 0

Answer:

The Reynolds number is 5600.

Explanation:

Given that,

Density = 1400 kg/m³

Viscosity = 0.5 Pa's

Length = 2 m

Speed = 1 m/s

We need to calculate the Reynolds number

Using formula of Reynolds number

R_{e}=\dfrac{\rho V\times L}{\mu}

Where, \rho = density of fluid

v = speed of syrup

l = length of a person

\mu=Viscosity

Put the all value into the formula

R_{e}=\dfrac{1400\times1\times2}{0.5}

R_{e}=5600

Hence, The Reynolds number is 5600.

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How much energy is needed to melt 5 g of ice? The specific latent heat of melting for water is 334000 J/kg.
Katen [24]

Answer:

The needed energy to melt of ice is 1670 J.

Explanation:

Given that,

Mass of ice = 5 g

Specific latent heat = 334000 J/kg

We need to calculate the energy

Using formula of energy

Q=mL

Where, m = mass

L = latent heat

Put the value into the formula

Q=5\times10^{-3}\times334000

Q=1670\ J

Hence, The needed energy to melt of ice is 1670 J.

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The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k
Brrunno [24]

Answer:

The gravitational force changing velocity is

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Explanation:

The expression for the gravitational force is

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Differentiate the above equation

\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}

The velocity is the distance in at time so

V=\frac{dr}{dt}=0.4 \frac{km}{s}

\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}

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Read 2 more answers
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
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