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3 years ago

10

Could anyone help with this? :)

2 answers:

bonufazy [111]3 years ago

4 0

I think the answer might be b

dangina [55]3 years ago

3 0

C 5,1,2,4,3 C is correct

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A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The w

miss Akunina [59]

The question is incomplete. The complete question is :

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).

Solution :

From flow over sphere, the mass transfer equation can be written as :

$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$

where, Sherood number, $Sh = \frac{K_L d}{D_{eff}}$

Reynolds number, $Re=\frac{Vd\rho}{\mu}$

Schmid number, $Sc= \frac{\mu}{\rho D_{eff}}$

So,

$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$

Diameter, d = 1 cm = $1 \times 10^{-2}$ m

V = 1 m/s

$\rho = 1000 \ kg/m^3$

$\mu = 10^{-3} \ kg/m/s$

$D_{eff} = 2 \times 10^{-9} \ m^2/s$

$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$

$K_L \times 5 \times 10^6=478.22$

$K_L=9.5644 \times 10^{-5}$ m/s

So the mass transfer coefficient is 9.5644 $\times 10^{-5}$ m/s. It is given solubility,

$\Delta C = 2 \ kg/m^3$

$N = Md^2 \times \Delta C \times K_L$

$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$

$N= 6 \times 10^{-8}$ kg/s (dissolution rate)

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3 years ago

What is the velocity of a car that travels from mile marker 32 on I-10 to mile marker 312 on I-10 in a time of 2 hours and 45 mi

Dimas [21]

Answer:

101.81818 MPH

Explanation:

8 0

3 years ago

PLEASE ANSWER THESE I NEED HELP!!!

Gnesinka [82]

4. Intermediate conductivity and a low melting point.

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3 years ago

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When plants and animals die, their bodies are decomposed by organisms such as bacteria or fungi. Which statement is true regardi

IceJOKER [234]

B because the the organism is changing into another chemical form

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4 years ago

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A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0

n200080 [17]

Answer:

a

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

b

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

Work done = $m*g*h$

where m is mass

g is acceleration due to gravity

and h is the height

c

Work done in terms of force and distance is = mg

where m is mass of bucket and

g is acceleration due to gravity

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = $\Delta P.E$

W = m*g*h

where g = 9.9 m/ $s^2$

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0

3 years ago