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FinnZ [79.3K]
2 years ago
13

5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its in

itial value as it flows through the pipe. The correct statement about the average velocity V is
Physics
1 answer:
prisoha [69]2 years ago
5 0

The options are;

a. V2 equals 2V1.

b. V2 equals (V1)/2.

c. V2 equals V1.

d. V2 equals (V1)/4.

e. V2 equals 4V1.

Answer:

Option A: V2 equals 2V1

Explanation:

Since the flow is steady, then we can say;

mass flow rate at input = mass flow rate at output.

Formula for mass flow rate is;

m' = ρVA

Thus;

At input;

m'1 = ρ1•V1•A1

At output;

m'2 = ρ2•V2•A2

So, m'1 = m'2

Now, we are told that the density of the fluid decreases to half its initial value.

Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

ρ1•V1•A1 = (ρ1)/2•V2•A2

Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

We now have;

ρ1•V1•A1 = (ρ1)/2•V2•A1

A1 and ρ1 will cancel out to give;

V1 = (V2)/2

Thus, V2 = 2V1

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Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

3 0
3 years ago
Your science teacher gives you three liquids to pour into a jar. After pouring, the liquids layer as seen here. What property of
Blizzard [7]
A. mass
b. color
c. density
d. texture

i believe A
6 0
3 years ago
Read 2 more answers
When using a calorimeter, the initial temperature of a metal is 70.4C. The initial temperature of the water is 23.6C. At the end
Sunny_sXe [5.5K]

1) 29.8 C

At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).

2) 6.2 C

The temperature change of the water is given by the difference between its final temperature and its initial temperature:

\Delta T = T_f - T_i

where

T_f = 29.8 C\\T_i = 23.6 C

Substituting into the formula,

\Delta T=29.8 C-23.6 C=6.2 C

And the positive sign means that the temperature of the water has increased.

3) -40.6 C

The temperature change of the metal is given by the difference between its final temperature and its initial temperature:

\Delta T = T_f - T_i

where

T_f = 29.8 C\\T_i = 70.4 C

Substituting into the formula,

\Delta T=29.8 C-70.4 C=-40.6 C

And the negative sign means the temperature of the metal has decreased.

5 0
3 years ago
Read 2 more answers
Can someone help me in this one :)
djyliett [7]
I believe it is “runs on parallel circuits”! my bad if incorrect
3 0
3 years ago
I REALLY NEED HELP WITH PHYSICS ASAP!!!<br>Vf^2 = v0^2 + 2a (xf - x0)<br><br><br>Solve for a
ELEN [110]

Answer:

a. solve for a

vf ^{2} = vo ^{2}  + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2}    -  vo ^{2}   \\ a =  \frac{vf^{2} - vo^{2}  }{2(xf - xo)}    \\ a =  \frac{vf ^{2} - vo ^{2}  }{2xf - 2xo}

I hope I helped you ^_^

7 0
3 years ago
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