The post Comparison of Stopping Distance and Time for Two Vehicles appeared first on MyRank.
]]>When the body is moving with a certain velocity and suddenly brakes are applied, then the body stops completely after covering a certain distance, this is called as Stopping Distance. It is the distance travelled between the time when the body decides to stop a moving vehicle and the time when the vehicle stops completely. Two vehicles of masses \({{m}_{1}}\) and \({{m}_{2}}\) are moving with velocities \({{v}_{1}}\) and\({{v}_{2}}\), when they are stopped by the same reading force (F).
The ratio of their stopping distances, \(\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{1}}v_{1}^{2}}{{{m}_{2}}v_{2}^{2}}\),
The ratio of their stopping time, \(\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}{{v}_{2}}}\),
(i) If vehicles possess same velocities, \({{v}_{1}}={{v}_{2}}\), \(\frac{{{x}{1}}}{{{x}{2}}}=\frac{{{m}{1}}}{{{m}{2}}}\); \(\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\),
(ii) If vehicles possess same kinetic momentum, \({{P}_{1}}={{P}_{2}}\),
\(\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\left( \frac{P_{1}^{2}}{2{{m}_{1}}} \right)\left( \frac{2{{m}_{2}}}{P_{2}^{2}} \right)=\frac{{{m}_{2}}}{{{m}_{1}}}\) \(\Rightarrow \frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=1\),
(iii) If vehicle possess same kinetic energy:
\(\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=1\) \(\Rightarrow \frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{\sqrt{2{{m}_{1}}{{E}_{1}}}}{\sqrt{2{{m}_{2}}{{E}_{2}}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\),
If vehicle is stopped by friction then,
Stopping distance \(\left( x \right)=\frac{\frac{1}{2}m{{v}^{2}}}{F}=\frac{\frac{1}{2}m{{v}^{2}}}{ma}=\frac{{{v}^{2}}}{2\mu g}\) \(\left[ as,\,\,a=\mu g \right]\),
Stopping time \(\left( t \right)=\frac{mv}{F}=\frac{mv}{m\mu g}=\frac{v}{\mu g}\).
The post Comparison of Stopping Distance and Time for Two Vehicles appeared first on MyRank.
]]>The post Sine of the Difference and Sum of Two Angles appeared first on MyRank.
]]>1. sin (A-B) = sinA cosB + cosA sinB
Proof: L.H.S: sin (A-B)
= sinA cosB + cosA sinB
L.H.S sin (A – B) = cos (90 – (A – B))
(since cos (90 – θ) = sin θ)
= cos (90 – (A – B))
= cos [(90 – A) + B]
= cos (90° – A) cos B – sin (90° – A) sin B
(since cos (90 – θ) = sin θ, sin (90 – θ) = cos θ)
= sin A cos B – cos A sin B
sin(A-B) = sinA cosB + cosA sinB
Hence proved
2. sin(A+B) = sinA cosB + cosA sinB
Proof: L.H.S: Sin (A + B)
= sin (A – (-B))
= sinA cos(-B) – cosA sin(-B)
(since cos(-θ) = cosθ
and sin(-θ) = -sinθ)
= sinA cosB + cosA sinB
sin(A+B) = sinA cosB + cosA sinB
Hence proved
3. sin (A + B) sin (A – B) = sin²A – sin²B
Proof: L.H.S: Sin (A + B) sin (A – B)
= (sinA cosB + cosA sinB) (sinA cosB – cosA sinB)
= sinA cosB x sinA cosB – sinA cosB x cosA sinB + cosA sinB x sinA cosB – cosA sinB x cosA sinB
= sinA cosB x sinA cosB – cosA sinB x cosA sinB
= sin²A cos²B – cos²A sin²B
= sin²A (1-sin²B) – (1-sin²A) sin²B
= sin²A – sin²A x sin²B – sin²B + sin²A x sin²B
= sin²A – sin²B
Sin (A + B) sin(A – B) = sin²A – sin²B
Hence proved.
The post Sine of the Difference and Sum of Two Angles appeared first on MyRank.
]]>The post Kalinga Institute of Industrial Technology Entrance Exam (KIITEE) 2020 Notification Released appeared first on MyRank.
]]>Sl. No | Subject | No. of Questions | No. of Marks |
---|---|---|---|
1 | Physics | 40 | 160 |
2 | Chemistry | 40 | 160 |
3 | Mathematics/ Biology | 40 | 160 |
Duration: 3 Hrs (180 Minutes) | |||
1 mark will be deducted for every wrong answer |
Submission of Online Application | 16-11-2019 |
Last Date for Submission of Online Application | 31-03-2020 |
Last date of hosting Admit Card in the website | 05-04-2020 |
Date of Entrance Examination (Online) | 14-04-2020 to 23-04-2020 |
Declaration of Result | 26-04-2020 |
Counseling starts from | 12-05-2020 |
The post Kalinga Institute of Industrial Technology Entrance Exam (KIITEE) 2020 Notification Released appeared first on MyRank.
]]>The post Conditional Identities (Tanθ) – Problems appeared first on MyRank.
]]>1. If A + B + C = nπ, n Є Z that TanA + TanB + TanC = TanA TanB TanC.
Proof: We have
\(\operatorname{Tan}\left( A+B+C \right)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan A\tan C}\),
A + B + C = nπ
Tan (A + B + C) = tan(nπ)
= tan(π) = 0
\(0=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan A\tan C}\),
TanA + TanB + TanC – TanA TanB TanC = 0
TanA + TanB + TanC = TanA TanB TanC
Hence proved.
Note: in ΔABC, A + B + C = π then TanA + TanB + TanC = TanA TanB TanC
2. If A + B + C = π, then Tan(A/2) tan(B/2) + tan(C/2) tan(B/2) + tan(C/2) tan(A/2) = 1
Proof: A + B + C = π
(A/2) + (B/2) + (C/2) = π/2
(A/2) + (B/2) = π/2 – (C/2)
\(\tan \left( \frac{A}{2}+\frac{B}{2} \right)=\tan \left( \frac{\pi }{2}-\frac{C}{2} \right)\) \(\left( \because \ \tan \left( \frac{\pi }{2}-\frac{C}{2} \right)=\cot \left( \frac{C}{2} \right) \right)\),
= cot(C/2)
\(\frac{\tan \left( \frac{A}{2} \right)+\tan \left( \frac{B}{2} \right)}{1-\tan \left( \frac{A}{2} \right)\tan \left( \frac{B}{2} \right)}=\cot \left( \frac{C}{2} \right)\),
\(\frac{\tan \left( \frac{A}{2} \right)+\tan \left( \frac{B}{2} \right)}{1-\tan \left( \frac{A}{2} \right)\tan \left( \frac{B}{2} \right)}=\frac{1}{\tan \left( \frac{C}{2} \right)}\),
\(\tan \left( \frac{C}{2} \right)\left( \tan \left( \frac{A}{2} \right)+\tan \left( \frac{B}{2} \right) \right)=1-\tan \left( \frac{A}{2} \right)\tan \left( \frac{B}{2} \right)\),
\(\tan \left( \frac{A}{2} \right)\tan \left( \frac{C}{2} \right)+\tan \left( \frac{B}{2} \right)\tan \left( \frac{C}{2} \right)=1-\tan \left( \frac{A}{2} \right)\tan \left( \frac{B}{2} \right)\),
\(\tan \left( \frac{A}{2} \right)\tan \left( \frac{C}{2} \right)+\tan \left( \frac{B}{2} \right)\tan \left( \frac{C}{2} \right)+\tan \left( \frac{A}{2} \right)\tan \left( \frac{B}{2} \right)=1\),
Tan(A/2) tan(B/2) + tan(C/2) tan(B/2) + tan(C/2) tan(A/2) = 1
Hence proved.
The post Conditional Identities (Tanθ) – Problems appeared first on MyRank.
]]>The post Equation of a Plane Progressive Wave appeared first on MyRank.
]]>1) If during propagation of a progressive wave, the particles of the medium perform Simple Harmonic Motion about their mean position, then the waves is known as a Harmonic progressive wave.
2) Suppose a plane simple harmonic wave travels from the origin along the positive direction of x – axis from left to the right as shown in below figure.
The displacement y of a particle (1) at O from its mean position at any time t is given by: y = a sinωt. The wave reaches the particle 2 after time, \(t=\frac{x}{v}\). Hence, displacement y of a particle (2) is given by:
\(y=a\sin \left( t-\frac{x}{v} \right)=a\sin \left( \omega t-kx \right)\) \(\left( \because \,\,k=\frac{\omega }{v} \right)\),
3) The general equation of a plane progressive wave with initial phase is:
y (x, t) = a sin (ωt ± kx ± φ₀)
Where, y = Displacement; a = Amplitude; ω = Angular Frequency; k = Propagation Constant; φ₀ = Initial Phase and x = Position
4) Particle velocity: The rate of change of displacement y with respect to time t is known as particle velocity. Hence, from y = a sin (ωt – kx).
Particle Velocity \(({{v}_{p}})=\frac{\partial y}{\partial t}=a\omega \cos \left( \omega t-kx \right)\); Maximum particle velocity (v_{p})_{max} = aω
Also, \(\frac{\partial y}{\partial t}=-\frac{\omega }{k}\times \frac{\partial y}{\partial x}\Rightarrow {{v}_{p}}=-v\times \) Slope of wave at that point.
The post Equation of a Plane Progressive Wave appeared first on MyRank.
]]>The post Conditional Identities appeared first on MyRank.
]]>Some standard Identities in Triangle:
1. sin2A + sin2B + sin2C = 4 sinA sinB sinC
Proof: L.H.S sin2A + sin2B + sin2C
(since sinC + sinD = 2sin (C + D)/2 cos (C – D)/2),
\(\sin 2A+\sin 2B=2\sin \left( \frac{2A+2B}{2} \right)\cos \left( \frac{2A-2B}{2} \right)\),
\(\sin 2A+\sin 2B=2\sin \left( A+B \right)\cos \left( A-B \right)\),
\(\sin 2A+\sin 2B+\sin 2C=2\sin \left( A+B \right)\cos \left( A-B \right)+\sin 2C\) \(\left( \because \ \sin 2x=2\sin x\cos x \right)\),
\(=2\sin \left( A+B \right)\cos \left( A-B \right)+2\sin C\cos C\) \(\left( \begin{align} & \because A+B+C=\pi \\ & \Rightarrow A+B=\pi -C \\\end{align} \right)\),
\(=2\sin \left( \pi -C \right)\cos \left( A-B \right)+2\sin C\cos C\),
\(=2\sin \left( C \right)\cos \left( A-B \right)+2\sin C\cos C\),
\(=2\sin C\left( \cos \left( A-B \right)+\cos C \right)\),
\(=2\sin C\left( \cos \left( A-B \right)+\cos \left( \pi -\left( A+B \right) \right) \right)\),
\(=2\sin C\left( \cos \left( A-B \right)-\cos \left( A+B \right) \right)\) \(\left( \because \ \cos \left( A+B \right)-\cos \left( A-B \right)=2\sin A\sin B \right)\),
\(=2\sin C\times 2\sin A\sin B\),
\(=4\sin A\sin B\sin C\),
Hence proved
2. cosA + cosB + cosC = 1 + 4sin(A/2) sin(B/2) sin(C/2)
Proof: L.H.S cosA + cosB + cosC – 1
\(=2\cos \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right)+\cos C-1\) \(\left( \because \frac{A+B+C}{2}=\frac{\pi }{2} \right)\),
\(=2\cos \left( \frac{\pi }{2}-\frac{C}{2} \right)\cos \left( \frac{A-B}{2} \right)+\cos C-1\),
\(=2\sin \left( \frac{C}{2} \right)\cos \left( \frac{A-B}{2} \right)+\cos C-1\) \(\left( \because \ {{\sin }^{2}}x=\frac{1-\cos 2x}{2} \right)\),
\(=2\sin \left( \frac{C}{2} \right)\cos \left( \frac{A-B}{2} \right)+1-2{{\sin }^{2}}\left( \frac{C}{2} \right)-1\),
\(=2\sin \left( \frac{C}{2} \right)\cos \left( \frac{A-B}{2} \right)-2{{\sin }^{2}}\left( \frac{C}{2} \right)\),
\(=2\sin \left( \frac{C}{2} \right)\left( \cos \left( \frac{A-B}{2} \right)-\sin \left( \frac{C}{2} \right) \right)\),
\(=2\sin \left( \frac{C}{2} \right)\left( \cos \left( \frac{A-B}{2} \right)-\sin \left( \frac{\pi }{2}-\left( \frac{A+B}{2} \right) \right) \right)\),
\(=2\sin \left( \frac{C}{2} \right)\left( \cos \left( \frac{A-B}{2} \right)-\cos \left( \left( \frac{A+B}{2} \right) \right) \right)\),
\(=2\sin \left( \frac{C}{2} \right)\left( 2\sin \frac{A}{2}\sin \frac{B}{2} \right)\),
\(=4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\),
Hence proved.
The post Conditional Identities appeared first on MyRank.
]]>The post Water Equivalent appeared first on MyRank.
]]>Water equivalent of a body is defined as the mass of water which would absorb or evolve the same amount of heat as is done by the body in rising or falling through the same range of temperature. It is represented by W. It is the quantity of water whose thermal capacity is same as the heat capacity of the body.
ΔQ = mcΔθ … (1)
Where,
m = Mass of the body,
c = Specific heat of body,
Δθ = Rise in temperature.
If same amount of heat is given to W gram of water and its temperature also rises by Δθ. Then:
Heat given to water (Q) = W x 1 x Δθ … (ii)
From equations (i) and (ii): ΔQ = mcΔθ = W x 1 x Δθ.
Water equivalent (W) = mc gm
The product of the mass of a substance by its specific heat equal numerically to the mass of water that is equivalent in thermal capacity to the substance is known as water equivalent of that substance.
The SI unit of Water equivalent is Kg and the dimensional formula is [M¹L⁰T⁰]. Thermal capacity of the body and its water equivalent are numerically equal. If thermal capacity of a body is expressed in terms of mass of water it is called as Water Equivalent of the body.
The water equivalent of a substance is numerically equal to that heat capacity of a substance in CGS system but no dimensionally.
The post Water Equivalent appeared first on MyRank.
]]>The post Uttaranchal University 2020 Application Started appeared first on MyRank.
]]>Submission of Online Application | Started |
The post Uttaranchal University 2020 Application Started appeared first on MyRank.
]]>The post Manav Rachna University (MRNAT) 2020 Application Started appeared first on MyRank.
]]>Submission of Online Application | Started |
The post Manav Rachna University (MRNAT) 2020 Application Started appeared first on MyRank.
]]>The post REVA University Common Entrance Test (REVA CET) 2020 Notification Released appeared first on MyRank.
]]>Sl. No | Subject | No. of Questions | No. of Marks |
---|---|---|---|
1 | Physics | 40 | 40 |
2 | Chemistry | 40 | 40 |
3 | Mathematics | 40 | 40 |
Duration: 2 Hrs (120 Minutes) | |||
No Negative Marking |
Submission of Online Application | Started |
Last Date for Submission of Online Application | 30^{th} March 2020 |
Generation of Online Test Admit Card | 15^{th} April 2019 |
Entrance Examination | 24^{th} April – 27^{th} April 2020 |
Declaration of Result | 11^{th} May, 2020 |
Counselling and Admission | 23^{rd} May – 25^{th} May 2020 |
The post REVA University Common Entrance Test (REVA CET) 2020 Notification Released appeared first on MyRank.
]]>