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Ask QuestionPosted by Lakshya Pratap Singh 4 years, 3 months ago
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Posted by Tejal Singhal 4 years, 3 months ago
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Posted by Jhilik S 4 years, 3 months ago
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Yogita Ingle 4 years, 3 months ago
The dimensional formula for capacitance is {tex}M^{-1} L^{-2} T^{4} I^2.{/tex}
Capacitance can be defined as the ratio of the change in an electric charge to the corresponding change in its electric potential in a system.
Posted by ?Ritesh Gupta☺️ 4 years, 3 months ago
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Posted by Mayank Vekariya 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
The dimensional formula of gravitational constant is given by,
M-1 L3 T-2
Where,
- M = Mass
- L = Length
- T = Time
Derivation
From Newton’s law of gravitation,
Force (F) = [GmM] × r-2
Gravitational Constant (G) = F × r2 × [Mm]-1 . . . . (1)
Since, Force (F) = Mass × Acceleration = M × [LT-2]
∴ The dimensional formula of force = M1 L1 T-2 . . . . (2)
On substituting equation (2) in equation (1) we get,
Gravitational Constant (G) = F × r2 × [Mm]-1
Or, G = [M1 L1 T-2] × [L]2 × [M]-2 = [M-1 L3 T-2].
Therefore, the gravitational constant is dimensionally represented as M-1 L3 T-2.
Posted by Saksham Kaushik 4 years, 3 months ago
- 1 answers
Gaurav Seth 4 years, 3 months ago
The dimensional formula of coefficient of friction is given by,
[M0 L0 T0]
Where,
- M = Mass
- L = Length
- T = Time
Derivation
Coefficient of friction (μ) = Frictional Force × [Normal Force]-1 . . . . (1)
Since, Force (F) = Mass × acceleration = Mass × velocity × [Time]-1
And, velocity = Displacement × [Time]-1
∴ The dimensions of Force = [M] × [LT-1] × [T]-1 = [M1 L1 T-2] . . . . (2)
On substituting equation (2) in equation (1) we get,
Coefficient of friction (μ) = Frictional Force × [Normal Force]-1
Or, μ = [M1 L1 T-2] × [M1 L1 T-2]-1 = [M0 L0 T0].
Therefore, the coefficient of friction is dimensionally represented as [M0 L0 T0].
Posted by Payal Roy 4 years, 3 months ago
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Himanshu Kumar 4 years, 3 months ago
Himanshu Kumar 4 years, 3 months ago
Posted by Khushi ??? 4 years, 3 months ago
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Posted by Adi=1 Sikarwar 4 years, 3 months ago
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Gautami Patil 4 years, 3 months ago
Posted by Akash Singh 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Kinetic Theory and Gas Pressure
The pressure of a gas is the result of continuous bombardment of the gas molecules against the walls of the container. According to the kinetic theory, the pressure P exerted by an ideal gas is given by
• Boyle’s Law
According to this law, the volume (V) of a fixed mass of a gas is inversely proportional to the pressure (P) of the gas, provided temperature of the gas is kept constant.
• Charle’s Law
According to this law, the volume (V) of a given mass of a gas is directly proportional to the
temperature of the gas, provided pressure of the gas remains constant.
• Gay Lussac’s Law (or Pressure Law)
According to this law, the pressure P of a given mass of a gas is directly proportional to its absolute temperature T, provided the volume V of the gas remains constant.
Posted by Muskan Kumari Upadhyay 4 years, 3 months ago
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Posted by Rahul Byadagihal 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Instantaneous velocity describes how fast an object is moving at different instants of time in a given time interval. It is also defined as average velocity for an infinitely small time interval.
Here lim is taking operation of taking limit with time tending towards 0 or infinitely small.
dx/dt is differential coefficient – Rate of change of position with respect to time at an instant.
Posted by Anchal Singh 4 years, 3 months ago
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Posted by Aryan Kumar 4 years, 3 months ago
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Siya Sharma 4 years, 3 months ago
Posted by Aryan Kumar 4 years, 3 months ago
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Siya Sharma 4 years, 3 months ago
Posted by Prabhat Singh 4 years, 3 months ago
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Siya Sharma 4 years, 3 months ago
Posted by Vanshika Xxxx 4 years, 3 months ago
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Posted by Vanshika Xxxx 4 years, 3 months ago
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Akash Pandey 4 years, 3 months ago
Posted by Vanshika Xxxx 4 years, 3 months ago
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Yogita Ingle 4 years, 3 months ago
Case I:
let initial velocity = u m/s
distance=s=12m
Time =t=2sec
S=u+a(2t-1)/2
12=u+(2x2-1)/2
12=u+3a/2
2u+3a=24--------------(1)
Case II:
Time=t=4sec
distance =s=20m
20=u+a(2x4-1)/2
20=u+7a/2
2u+7a=40--------(2)
from 1 and 2:
2u+3a=24
2u+7a=40
- - -
***********
-4a=-16
a=4m/s2
substitute the value of a in equation 1 we get :
u=6m/s
Case iii:
Now, distance covered in 4sec after 5second=distance covered in 9th sec-distance covered in 5 sec
=(6x9+1/24x9x9)-(6x5+1/2*4*5*5)
=54+162-(30+50)
=136m
Posted by Komalpreet Kaur 4 years, 3 months ago
- 1 answers
Yogita Ingle 4 years, 3 months ago
Instantaneous velocity can be equal to average velocity when the acceleration is zero or velocity is constant because in this condition all the instantaneous velocities will be equal to each other and also equal to the average velocity. Or distance -time graph will be a straight line whose slope (velocity) is same at every point .
Posted by Navtej Singh 4 years, 3 months ago
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Hemant Kumar Bahrod 4 years, 3 months ago
Gaurav Seth 4 years, 3 months ago
Every object in the universe attracts every other object with a force which is called the force of gravitation.
Gravitation is one of the four classes of interactions found in nature.
These are
(i) the gravitational force
(ii) the electromagnetic force
(iii) the strong nuclear force (also called the hadronic force).
(iv) the weak nuclear forces.
Although, of negligible importance in the interactions of elementary particles, gravity is of primary importance in the interactions of objects. It is gravity that holds the universe together.
Posted by Avni Sharma 4 years, 3 months ago
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Ã.P.W.B.Đ ⚡ 4 years, 3 months ago
Posted by Adi Bhau . 4 years, 3 months ago
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Ã.P.W.B.Đ ⚡ 4 years, 3 months ago
Posted by Nehal Yadav 4 years, 3 months ago
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Gaurav Seth 4 years, 3 months ago
Let Time period =T
Mass of the bob = m
Acceleration due to gravity = g
Length of string = L
Let
⇒a=0 ⇒ Time period of oscillation is independent of mass of the bob
-2b=1
⇒b=-
b+c = 0
- + c =0
c=
Giving values to a,b and c in first equation
The real expression for Time period is
Posted by Prince Arora 4 years, 3 months ago
- 1 answers
Gaurav Seth 4 years, 3 months ago
Given,
Hight of Hall( H) = 25 m
Let ball is thrown with speed 40m/s at an angle ∅ with horizontal .
We know,
H = u²sin²∅/2g
25= (40)² × sin²∅/2× 10
25 = 80 × sin²∅
Sin²∅ = 25/80
Sin∅ = 5/4√5
Cos∅ = √55/4√5
Now, horizontal range = u²sin2∅/g
= (40)²× 2sin∅×cos∅/g
= 160 × 2 × 5/4√5 × √55/4√5
= 20√55 m
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