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Ask QuestionPosted by Geetika Saini 4 years ago
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Posted by Raj Singh 4 years ago
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Posted by Himanshu Sharma 4 years ago
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Yogita Ingle 4 years ago
Full names of the units, even when they are named after a scientist should not be written with a capital letter. Eg: Newton, watt, ampere, meter.
- The unit should be written either in full or in agreed symbols only.
- Units do not take the plural form. Eg: 10 kg but not 10 kgs, 20 w but not 20 ws.
- No full stop or punctuation mark should be used within or at the end of symbols for units. Eg: 10 W but not 10 W.
Dimensions of a physical quantity are the powers to which the fundamental units are raised to obtain one unit of that quantity.
Posted by Aadya Singh 4 years ago
- 5 answers
Aseem Mahajan 4 years ago
Tsin4⁰ = ma
Tcos4⁰ = mg
tan4⁰=a/g
a= 9.8 × tan4⁰
Solve tan4 and get answer
</a>Posted by Aadya Singh 4 years ago
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Aseem Mahajan 4 years ago
Posted by Tanisha Nehra 4 years ago
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Aseem Mahajan 4 years ago
$$tan \theta = \dfrac{B cos \theta}{A + B sin \theta}$$
Posted by Aadya Singh 4 years ago
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Aseem Mahajan 4 years ago
Posted by Aadya Singh 4 years ago
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Posted by Aadya Singh 4 years ago
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Posted by Satish Bhardwaj 4 years ago
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Abirami Kamalbabu 3 years, 11 months ago
Posted by Aadya Singh 4 years ago
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Satish Bhardwaj 4 years ago
Posted by Aadya Singh 4 years ago
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Aadya Singh 4 years ago
Posted by Angel Akku 4 years ago
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Posted by Rashida Rasshi 4 years ago
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Ankit Sharma 4 years ago
Yogita Ingle 4 years ago
All the three are units of physical quantities, where
nm stands for nanometer, and1 nm =10-9 m,
mN stands for milli - newton, and 1 mN = 10-3 N,
Nm stands for Newton metre.
Posted by Michael Chawngthu 4 years ago
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Posted by Vansh Jain 4 years ago
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Rishabh Yadav 4 years ago
Posted by Vedha ..... 4 years ago
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Posted by Vedha ..... 4 years ago
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Posted by Prmod Ray 4 years ago
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Yogita Ingle 4 years ago
The centre of mass of a body or a system of particles is defined as “A single point at which the whole mass of the body or system is imagined to be concentrated and all the applied forces acts at that point.”
Posted by Vedha ..... 4 years ago
- 3 answers
Yogita Ingle 4 years ago
Let uniform acceleration be a
Acceleration is defined as rate of change of velocity
......................(1)
where v is velocity of an object at time t
By integrating above ewqn.(1) , we get
where u is velocity at time t = 0 and v is velocity after time t seconds
from above integration we get , v - u = a t or v = u + ( a t ) ...................... (2)
velocity is defined as rate of change of displacement
...................(3)
Where S is distance travelled.
By integrating above equation (2) , we get
Where xo is initial position and x is position after time t
x - xo = u t + (1/2) a t2 .................. (4)
If the object travelled a distance S staring from initial velocity u and attained final velocity v after travelling distance S in time t , then we have
S = (1/2) ( u+v ) t ...............(5)
Let us substitute t from eqn.(2) in eqn.(5) as t = [ ( v - u ) / a ] , then we have
S = (1/2) ( u + v ) [ ( v - u ) /a ] or v2 = u2 + ( 2 a S ) .................. (6)
Eqn.(2) , eqn.(4) and eqn.(6) are equations of motion
Posted by Vedha ..... 4 years ago
- 1 answers
Yogita Ingle 4 years ago
When the ball is thrown vertically upward its velocity decreases as it moves upward due to the gravitational force acting on it downwards and thus the graph slopes downwards with a negative gradient.
When it reaches a certain height, that is at a stationary point, the velocity becomes zero and at this point the graph cuts the x axis.
The ball then comes back with an increasing velocity from zero to its initial position and at this point the graph slopes downwards from the x axis.
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Aseem Mahajan 4 years ago
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