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Posted by Shrusti Kushwah 2 years, 10 months ago (13471971)
- 3 answers
Posted by Rakhi Bagri 2 years, 10 months ago (14190667)
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Posted by Abhishek Kamath 2 years, 10 months ago (12886676)
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Anaaya Maurya 2 years, 10 months ago (13632115)
Rajju Sahu 2 years, 10 months ago (10814880)
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Amazing Techs By Nakul 2 years, 10 months ago (8235779)
Posted by Swathi Ojha 2 years, 10 months ago (12536455)
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Posted by S S 2 years, 10 months ago (13761536)
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Preeti Dabral 2 years, 10 months ago (2983787)
Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy.
To prove Bernoulli's theorem, consider a fluid of negligible viscosity moving with laminar flow. as shown in Figure.
Let the velocity, pressure and area of the fluid cloumn be p1, v1 and A1 at Q and p2, v2 and A2 at R. Let the volume bounded by Q and R move to S and T where QS = L1, and RT = L2
If the fluid is incompressible:
The work done by the pressure difference per unit volume = gain in kinetic energy per unit volume + gain in potential energy per unit volume. Now:
A1L1 = A2L2
Work done is given by:
W = F {tex}\times{/tex} d = p {tex}\times{/tex} volume
{tex}\Rightarrow{/tex} Wnet = p1 - p2
{tex}\Rightarrow{/tex} K.E = {tex}\frac{1}{2}{/tex}mv2 = {tex}\frac{1}{2}{/tex}V {tex}\rho{/tex}v2 = {tex}\frac{1}{2}{/tex}{tex}\rho{/tex}v2 ({tex}\because{/tex} V = 1)
{tex}\Rightarrow{/tex} K.Egained = {tex}\frac{1}{2} \rho\left(v_{2}^{2}-v_{1}^{2}\right){/tex}
P1 + {tex}\frac{1}{2} \rho v_{1}^{2}{/tex} + {tex}\rho{/tex}gh1 = P2 + {tex}\frac{1}{2} \rho v_{2}^{2}{/tex} + {tex}\rho{/tex}gh2
{tex}\therefore{/tex} P + {tex}\frac{1}{2} \rho v^{2}{/tex} + {tex}\rho{/tex}gh = const.
For a horizontal tube
{tex}\because{/tex} h1 = h2
{tex}\therefore{/tex} P + {tex}\frac{1}{2} \rho v^{2}{/tex} = const.
Therefore, this proves Bernoulli's theorem. Here we can see that if there is an increase in velocity there must be a decrease in pressure and vice versa.
Posted by S S 2 years, 10 months ago (13761536)
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Posted by Yashfa Khan 2 years, 9 months ago (14202784)
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Posted by Account Deleted 2 years, 10 months ago (11593177)
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Swayam Jyoti Goswami 2 years, 10 months ago (14207929)
Few rules to keep homework help section safe, clean and informative.
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Swayam Jyoti Goswami 2 years, 10 months ago (14207929)
Few rules to keep homework help section safe, clean and informative.
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Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.
Posted by Sakshi Singh 2 years, 10 months ago (3448479)
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Shreyansh Keshri 2 years, 10 months ago (12989785)
Siddharth Kumar 2 years, 10 months ago (13314010)
Posted by Vivek Raj 2 years, 10 months ago (14203902)
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Mishti Choudhary 2 years, 7 months ago (11261794)
Naman Bansal 2 years, 9 months ago (10944931)
Lavi Aggarwal 2 years, 10 months ago (13244820)
Posted by Granth Gond 2 years, 10 months ago (12256422)
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Preeti Dabral 2 years, 10 months ago (2983787)
Bal Vidya Public School, Bhopal
NOTICE
Date:- 18th October, 20XX
Title:- Annual Day Function.
Our school is going to celebrate the annual day on 26th October, 20XX and the timings would be 9:00 am to 4:30 pm. Students from VI to X are invited to participate in the events. The programme is being held in our school auditorium. All the students of our school are invited to attend this programme. For more details, contact the undersigned.
Aryan,
Head Boy
Adhishtha Agnihotri 2 years, 8 months ago (12376102)
Posted by Sakshi Singh 2 years, 10 months ago (3448479)
- 1 answers
Preeti Dabral 2 years, 10 months ago (2983787)
P = Rs. 1800
R = 8% per annum
{tex} = \frac{1}{4} \times 8\% {/tex} quarterly
= 2% quarterly
n = 1 year
= 4 quarters
{tex}\therefore A = P{\left( {1 + \frac{R}{{100}}} \right)^n}{/tex}
{tex} = 1800{\left( {1 + \frac{2}{{100}}} \right)^4}{/tex}
{tex} = 1800{\left( {1 + \frac{1}{{50}}} \right)^4}{/tex}
{tex} = 1800{\left( {\frac{{51}}{{50}}} \right)^4}{/tex}
{tex} = 1800 \times \frac{{51}}{{50}} \times \frac{{51}}{{50}} \times \frac{{51}}{{50}} \times \frac{{51}}{{50}}{/tex}
= Rs. 1948.38.
Posted by Pro Gamerz Sharma 2 years, 10 months ago (14194490)
- 0 answers
Posted by Mansi Mansi 2 years, 10 months ago (14207859)
- 1 answers
Preeti Dabral 2 years, 10 months ago (2983787)
Glucose solution is given to athletes to provide instant energy.
Posted by Surendra Pratap 2 years, 10 months ago (8825276)
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Posted by Taha Choudhari 2 years, 10 months ago (11997276)
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Ayush Pradhan 2 years, 10 months ago (13833924)
Posted by Pardhu Moram 2 years, 10 months ago (14207442)
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Posted by Adarsh Verma 2 years, 10 months ago (13956739)
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Posted by Adarsh Verma 2 years, 10 months ago (13956739)
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Posted by Adarsh Verma 2 years, 10 months ago (13956739)
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Posted by Yash Sharma 2 years, 10 months ago (14206354)
- 3 answers
Kamana Yadav 2 years, 8 months ago (14367144)
Posted by Pushkar Lohiya 2 years, 10 months ago (10053699)
- 1 answers
Preeti Dabral 2 years, 10 months ago (2983787)
Given, linear equation is 2x + 3y - 8 = 0 ...(i)
Given: 2x + 3y - 8 = 0 ..... (i)
- For intersecting lines, {tex}\frac { a _ { 1 } } { a _ { 2 } } \neq \frac { b _ { 1 } } { b _ { 2 } }{/tex}
{tex}\therefore{/tex} Any line intersecting with eq (i) may be taken as 3x + 2y - 9 = 0
or 3x + 2y - 7 = 0 - For parallel lines, {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c_ { 2 } }{/tex}
{tex}\therefore{/tex} Any line parallel with eq(i) may be taken as 6x + 9y + 7 = 0
or 2x +3y - 2 = 0 - For coincident lines, {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c_ { 2 } }{/tex}
{tex}\therefore{/tex} Any line coincident with eq (i) may be taken as 4x + 6y - 16 = 0
or 6x + 9y - 24 = 0
Posted by Shivani S Pradeep 2 years, 10 months ago (14188100)
- 1 answers
Siddharth Kumar 2 years, 10 months ago (13314010)
Posted by Ujwal Saraf 2 years, 10 months ago (14206874)
- 0 answers
Posted by Rohit Raghuwanshi 2 years, 10 months ago (13874039)
- 2 answers
Hitesh Kumar 2 years, 10 months ago (13686371)
Kanan K 2 years, 10 months ago (14053217)
Posted by Aryan Kumar 2 years, 10 months ago (12169805)
- 1 answers
Preeti Dabral 2 years, 10 months ago (2983787)
Kalpasutra is a Jain ancient text containing the biographies of the last two Jain Tirthankaras, Parshvanath and Mahavira. It contains detailed life histories with illustrations.
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Priyanshi Kushwah 2 years, 10 months ago (13981709)
1 Thank You