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Physics Homework Help
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About Us:
At Homework1.com we offer authentic and 100% accurate
online homework help and study assistance to students from
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students only statistics assignment help service to complete
their study project; rather we offer our best effort to teach our
student-clients about the assignment we have solved. Our
tutors are not only subject matter experts, they are avid
student-mentors and are ready to walk extra miles to make
them understand the fundamentals of the assignment done,
and help them to learn the solution by heart.
We are available online by 24×7 and we can be reached via email, live chat, as well as
by direct phone calls. Our USP is quick turnaround time with stringent quality assurance
about the assignment we undertake. We offer assistance is writing dissertations,
academic and project related essay writing, and in writing and reviewing research
papers. These research papers are done by best subject matter experts available and we
offer 100% plagiarism free content by following proper and prescribed house style.
We work on simple methodology and we maintain best quality. Our domain for physics
homework help assistance is quite extensive. Starting from complicated numerical
problems of rotational motion, simple harmonic motion, velocity, we offer our services
for Neutron’s law, electric current, and many more. Send us your assignment’s soft copy
and we will get back to you immediately with our proposal. All our works are quality
submission and we are able to take your physics assignments even at the most last hour
before deadline.
Sample of Physics Homework Illustrations and Answers:
Illustration:1
(c) force.
Deduce the dimensional formula of : (a) Velocity, (b) acceleration, and
Solution :
(a) Dimension of velocity
= Dimension of length/ Dimension of time = [L]/[T] = [LT-1]
Hence the dimensional formulae for velocity will be [M0 LT-1].
(b) Dimension of acceleration
= Dimension of velocity/Dimension of time = [πΏπ −1 ]/[T] = [LT-2]
= [M0 LT-2]
(c) [Dimension of force] = [Dimension of mass] × [Dimension of acceleration]
= [M] × [LT-2] = [MLT-2]
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Illustration:2
Deduce the dimension of (a) Pressure, (b) Work and (c) Power.
Solution :
(a) [Dimension of pressure]
= [Dimension of force]/[Dimension of area ] = [MLT-2] / [L2] = [ML-1T-2]
(b) Dimension of work.
= [Dimension of force] × [Dimension of distance]
= [MLT-2] × [L] = [ML2T-2]
(c) Dimension of power
= [Dimension of work] / [Dimension of time] = [ML2T-2]/[T]
= [ML2T-3].
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Illustration:3 Deduce the dimensional formula for (a) modulus of elasticity, and (b)
coefficient of viscosity.
Solution :
(a) Modulus of elasticity.
Deduce the dimensional formula for (a) modulus of elasticity, and (b) coefficient of
viscosity.
Solution : (a) Modulus of elasticity.
Y = Stress/Strain = Force/Area/Change in length/Original length
∴ Dimension of Y
= (Dimension of force) × (Dimension of length)/ (Dimension of area) × (Dimension of
length)
= [MLT-2] × [L]/ [L2] × [L] = [ML-1T-2]
(b) the coefficient of viscosity (n) of a liquid is defined as tangential force required per
unit area to maintain unit velocity gradient between the two layers of the liquid unit
distance apart or
n = F/A 1/(dv/dx)
= Force/ Area × (Velocity change/Distance)
Dimension of n =
(Dimension of force) × (Dimension of distance)/ (Dimension of area) × (Dimension of
velocity)
= [MLT-2] × [L]/[L2] × [LT-1] = [ML2T-2]/ [L3T-1].
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Illustration: 4
Convert one Newton into Poundal (unit of force in F.P.S. system).
Solution :
The dimensional formula of force is MLT-2. One Poundal = 1 Ib-ft/sec2 and one Newton =
1 Kg- m/sec2 1 Ib = 0.4536 kg and I ft = 0.3048 metre
∴ M units in F.P.S. system
= 0.4536 M units in M.K.S. system.
L units in F.P.S. system.
= T units in M.K.S. system.
∴ I Newton = 1 Poundal (1/0.4536) × (1/0.3048) × (1/1)-2
or 1 N = 7.233 Poundal.
Illustration: 5 If the value of G in C.G.S. system is 6.67 × 10-8 dynes cm2/g2, what
will be its value in M.K.S. system ?
Solution : The dimensional formula for G is [M-1 L3 T-2] metre = 100 cm, 1 kg = 1000 g
∴ M unit of M.K.S. system.
= 1000 M unit of C.G.S. system.
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L unit of M.K.S. system
= 100 L units of C.G.S. system.
T units of M.K.S. system,
= 100 L units of C.G.S. system
= T units of C.G.S. system.
If x is the value of G in M.K.S. system then we have
X = 6.67 × (1/1000)-1 × (1/100)3 × (1/1)-2
= 6.67 × 10-8 × 103 ×10-6 × 1
= 6.67 × 10-11 Nm2/kg2.
(b) To check the correctness of a derived relationship between physical
quantities. By making use of the fact that the dimension of the quantities on the left
hand side must be same as that of the dimensions of the quantities of the right hand
side of the dimensional equation (this principle is known as the dimensional equation
(this principle is known as the principal of homogeneity of dimension), we can check the
correctness of any relation connecting various physical quantities. Let us consider the
following examples :
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Physics Homework Help
Physics Homework Help Service
Contact Us
Homework1
3422 SW 15 Street
Suite #8924
Deerfield Beach, FL, US 33442
Tel: +1-626-472-1732
Web: https://homework1.com/
Email: [email protected]
Facebook: https://www.facebook.com/homework1com
Linkedin: https://www.linkedin.com/in/homework1
Twitter: https://twitter.com/homework1_com
Google Plus: https://plus.google.com/118210863993786098250/
Pinterest: https://www.pinterest.com/homeworkone/
Copyright © 2014-2015 Homework1.com, All rights reserved
Homework1
About Us:
At Homework1.com we offer authentic and 100% accurate
online homework help and study assistance to students from
USA, UK, Australia, and Canada. However, we don’t offer
students only statistics assignment help service to complete
their study project; rather we offer our best effort to teach our
student-clients about the assignment we have solved. Our
tutors are not only subject matter experts, they are avid
student-mentors and are ready to walk extra miles to make
them understand the fundamentals of the assignment done,
and help them to learn the solution by heart.
We are available online by 24×7 and we can be reached via email, live chat, as well as
by direct phone calls. Our USP is quick turnaround time with stringent quality assurance
about the assignment we undertake. We offer assistance is writing dissertations,
academic and project related essay writing, and in writing and reviewing research
papers. These research papers are done by best subject matter experts available and we
offer 100% plagiarism free content by following proper and prescribed house style.
We work on simple methodology and we maintain best quality. Our domain for physics
homework help assistance is quite extensive. Starting from complicated numerical
problems of rotational motion, simple harmonic motion, velocity, we offer our services
for Neutron’s law, electric current, and many more. Send us your assignment’s soft copy
and we will get back to you immediately with our proposal. All our works are quality
submission and we are able to take your physics assignments even at the most last hour
before deadline.
Sample of Physics Homework Illustrations and Answers:
Illustration:1
(c) force.
Deduce the dimensional formula of : (a) Velocity, (b) acceleration, and
Solution :
(a) Dimension of velocity
= Dimension of length/ Dimension of time = [L]/[T] = [LT-1]
Hence the dimensional formulae for velocity will be [M0 LT-1].
(b) Dimension of acceleration
= Dimension of velocity/Dimension of time = [πΏπ −1 ]/[T] = [LT-2]
= [M0 LT-2]
(c) [Dimension of force] = [Dimension of mass] × [Dimension of acceleration]
= [M] × [LT-2] = [MLT-2]
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Illustration:2
Deduce the dimension of (a) Pressure, (b) Work and (c) Power.
Solution :
(a) [Dimension of pressure]
= [Dimension of force]/[Dimension of area ] = [MLT-2] / [L2] = [ML-1T-2]
(b) Dimension of work.
= [Dimension of force] × [Dimension of distance]
= [MLT-2] × [L] = [ML2T-2]
(c) Dimension of power
= [Dimension of work] / [Dimension of time] = [ML2T-2]/[T]
= [ML2T-3].
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Illustration:3 Deduce the dimensional formula for (a) modulus of elasticity, and (b)
coefficient of viscosity.
Solution :
(a) Modulus of elasticity.
Deduce the dimensional formula for (a) modulus of elasticity, and (b) coefficient of
viscosity.
Solution : (a) Modulus of elasticity.
Y = Stress/Strain = Force/Area/Change in length/Original length
∴ Dimension of Y
= (Dimension of force) × (Dimension of length)/ (Dimension of area) × (Dimension of
length)
= [MLT-2] × [L]/ [L2] × [L] = [ML-1T-2]
(b) the coefficient of viscosity (n) of a liquid is defined as tangential force required per
unit area to maintain unit velocity gradient between the two layers of the liquid unit
distance apart or
n = F/A 1/(dv/dx)
= Force/ Area × (Velocity change/Distance)
Dimension of n =
(Dimension of force) × (Dimension of distance)/ (Dimension of area) × (Dimension of
velocity)
= [MLT-2] × [L]/[L2] × [LT-1] = [ML2T-2]/ [L3T-1].
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Illustration: 4
Convert one Newton into Poundal (unit of force in F.P.S. system).
Solution :
The dimensional formula of force is MLT-2. One Poundal = 1 Ib-ft/sec2 and one Newton =
1 Kg- m/sec2 1 Ib = 0.4536 kg and I ft = 0.3048 metre
∴ M units in F.P.S. system
= 0.4536 M units in M.K.S. system.
L units in F.P.S. system.
= T units in M.K.S. system.
∴ I Newton = 1 Poundal (1/0.4536) × (1/0.3048) × (1/1)-2
or 1 N = 7.233 Poundal.
Illustration: 5 If the value of G in C.G.S. system is 6.67 × 10-8 dynes cm2/g2, what
will be its value in M.K.S. system ?
Solution : The dimensional formula for G is [M-1 L3 T-2] metre = 100 cm, 1 kg = 1000 g
∴ M unit of M.K.S. system.
= 1000 M unit of C.G.S. system.
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L unit of M.K.S. system
= 100 L units of C.G.S. system.
T units of M.K.S. system,
= 100 L units of C.G.S. system
= T units of C.G.S. system.
If x is the value of G in M.K.S. system then we have
X = 6.67 × (1/1000)-1 × (1/100)3 × (1/1)-2
= 6.67 × 10-8 × 103 ×10-6 × 1
= 6.67 × 10-11 Nm2/kg2.
(b) To check the correctness of a derived relationship between physical
quantities. By making use of the fact that the dimension of the quantities on the left
hand side must be same as that of the dimensions of the quantities of the right hand
side of the dimensional equation (this principle is known as the dimensional equation
(this principle is known as the principal of homogeneity of dimension), we can check the
correctness of any relation connecting various physical quantities. Let us consider the
following examples :
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