CBSE > Class 11 > Physics | CBSE Board | Homework Help

M.K.S. System = Base units in this system are Metre, Kilogram & Second System. C.G.S. System = Centimetre, Gram &Second System. F.P.S. System = Foot, pound & Second System. S.I. System = International System.

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Is adcc Academy good

Posted by Shatayu Ganvir 7 years, 9 months ago

CBSE > Class 11 > Physics

1 answers

Ranjan Kumar 7 years, 9 months ago

Maybe

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The time period of simple pendulum depends on length of pendulum ,and acceleration due to gravity derive it's formula

Posted by Sheshrao Dhawle 5 years, 11 months ago

CBSE > Class 11 > Physics

0 answers

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Why is a rainbow semi circle in shape and not in other shapes??

Posted by Koushik Viswanath 7 years, 9 months ago

CBSE > Class 11 > Physics

2 answers

Ʌɐpɐʎ Ɐuɐɥsɹɐp . 7 years, 9 months ago

for more just go through with S. chand 10th class physics

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Ʌɐpɐʎ Ɐuɐɥsɹɐp . 7 years, 9 months ago

Rainbow is not semicircular in shape . It appears semicircular bcs half of the rainbow is formed below the horizontal axis and the things below horizontal axis can't be seen .

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Why there is no neet section in this App

Posted by Devdutt Katata 7 years, 9 months ago

CBSE > Class 11 > Physics

5 answers

ब्राहम्ण Boy 7 years, 9 months ago

It is available here but included in our course so you have to concentrate on everything which is mentioned here...

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Ʌɐpɐʎ Ɐuɐɥsɹɐp . 7 years, 9 months ago

but there is only syllabus not study material

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Ʌɐpɐʎ Ɐuɐɥsɹɐp . 7 years, 9 months ago

it is available.. instead of cbse you have to choose entrance exam.. here you will get two option one is mains and other is Neet UG.

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Aman Kumar 7 years, 9 months ago

For neet , u have 2 download neet related apps like IIT neet app

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ब्राहम्ण Boy 7 years, 9 months ago

???

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A particle having initial velocity 10m/s moves with cinstant acceleration 5m/s2 for time 10 second. Find the displacement of the particle in the last 1 second.

Posted by Jasika Malik 7 years, 9 months ago

CBSE > Class 11 > Physics

1 answers

Tripti Sharma 7 years, 9 months ago

Is it 52.5m

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If f=x³, then relative error in f would be how many times the relative error in X?

Posted by Ʌɐpɐʎ Ɐuɐɥsɹɐp . 7 years, 9 months ago

CBSE > Class 11 > Physics

0 answers

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The time period of oscillation of a simple pendulum is recorded as 2.56s, 2.62s, 2.70s, 2.58s, 2.45s.find time period, absolute error in each observation and percentage error

Posted by Manpreet Kaur 7 years, 9 months ago

CBSE > Class 11 > Physics

2 answers

Manpreet Kaur 7 years, 9 months ago

Thanks for your kind help

1Thank You

Himanshu Jha 7 years, 9 months ago

A1=2.56 A2=2.62 A3=2.70 A4=2.58 A5=2.45 avg a=2.582 delta a=absolute error in the measurements are=(2.56-2.58)=-0.02 (2.62-2.62)=0 (2.70-2.62)=0.08 (2.58-2.62)=-0.04 (2.45-2.62)=-0.17 mean absolute errors=(-0.02+0+0.08-0.04-0.17)/5=-0.03 relative error=±0.03/2.58=±0.011 percentage error=1.16%

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????? ! ?: ᴀ ᴍᴀɴ ᴛʜƦᴏᴡS ʙᴀʟʟs ɪɴᴛᴏ ᴛʜᴇ ᴀɪʀ ᴏɴᴇ ᴀғᴛᴇʀ ᴛʜᴇ ᴏᴛʜᴇʀ . ᴛʜʀᴏᴡɪɴɢ ᴏɴᴇ ᴡʜᴇɴ ᴏᴛʜᴇʀ ɪs ᴀᴛ ᴛʜᴇ ʜɪɢʜᴇsᴛ ᴘᴏɪɴᴛ.ʜᴏᴡ ʜɪɢʜ ᴛʜᴇ ʙᴀʟʟs ʀɪsᴇ ɪғ ʜᴇ ᴛʜʀᴏᴡs ᴛᴡɪᴄᴇ ᴀ sᴇᴄᴏɴᴅ? ?????? ?????? ?? ???? !

Posted by Archita Bhattacharjee 7 years, 9 months ago

CBSE > Class 11 > Physics

5 answers

Archita Bhattacharjee 7 years, 9 months ago

My q.was : a man throws balls into the air one after the other throwing one when other is at the highest point how far the ball rise if he throws twice a second

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Himanshu Jha 7 years, 9 months ago

No ,, what iu have wrote

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Archita Bhattacharjee 7 years, 9 months ago

ᴰᴵᴰ ᵁ ᵁᴺᴰᴱᴿˢᵀᴼᴼᴰ ᵀᴴᴱ Qᵁᴱˢᵀᴵᴼᴺ ?

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Himanshu Jha 7 years, 9 months ago

What's that ?????

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Archita Bhattacharjee 7 years, 9 months ago

???? ᵍ ⁼ 10 ᵐ/ˢ^2

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If unit of force, velocity and energy are 100dyne, 10 cm/s and 400ergs. What is the maximum error in kinetic energy obtained by the measurement

Posted by Manpreet Kaur 7 years, 9 months ago

CBSE > Class 11 > Physics

1 answers

Himanshu Jha 7 years, 9 months ago

The Dimensional formulae of force = [ MLT^-2 ] energy = [ML²T^-2] velocity = [LT^-1] Thus We have [MLT^-2] = 10N.......(i) [ML²T^-2] = 100J .....(ii) [LT-1] = 5ms^-1 .......(iii) and Dividing equation (ii) by eq.(i) ,we have [L] = 100J/10N = 100Nm/10N = 10m. Substituting the unit of [L] in eq. (iii) , we have [T^-1] = 5ms^-1/[L] = 5ms^-1/ 10m [ T ] = 10m/5ms^-1 = 2s. Substituting the units of [L] and [ T] in eq.( i) we have [M] = 10N / [L] [T^-2] = 10N × 4s² / 10m = 4kg. Hope it helps !!!☺☺

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Notes on system of units

Posted by Manpreet Kaur 7 years, 9 months ago

CBSE > Class 11 > Physics

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ŴĤÁŤ ĨŚ MĔÁŃŤ ßŶ ĨMPĔĎÁŃČĔ ?

Posted by Archita Bhattacharjee 7 years, 9 months ago

CBSE > Class 11 > Physics

3 answers

Himanshu Jha 7 years, 9 months ago

Mention not ☺☺

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Archita Bhattacharjee 7 years, 9 months ago

ᵀʰaⁿᵏˢ a ˡoᵗ ᴴᴵᴹᴬᴺˢᴴᵁ

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Himanshu Jha 7 years, 9 months ago

resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance... ☺☺

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Hey friends

Posted by Siddharth Gahalot 7 years, 9 months ago

CBSE > Class 11 > Physics

1 answers

Siddharth Gahalot 7 years, 9 months ago

Does any one know aqsha

1Thank You

ANSWER

a particle moves a distance x in time t according to equation x=(t+5)^-1 . the acceleration of particle in proportional to

Posted by Nancy Rajput 7 years, 9 months ago

CBSE > Class 11 > Physics

1 answers

Aman Kumar 7 years, 9 months ago

X= (t+5)^-1.... V= dx/dt = -1(t+5)^-1-1..... =>V = -(t+5)^-2 ... a= d^2x/dt^2= 2(t+5)^-3... => a= -2(v)^3/2.. => a directly proportional to v^3/2

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open an expression for the centripetal force f acting on a particle of mass M moving with velocity V in a circle of radius r a dimensionless constant K is equal to 1

Posted by Abhyuday Patel 7 years, 9 months ago

CBSE > Class 11 > Physics

1 answers

Nancy Rajput 7 years, 9 months ago

F=mV^2/r

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A body of mass 50kg is kept on a horizontal surface of coefficients of static friction 0.5. Find the least horizontal force required to start motion. If now the body is slightly disturbed and the same force continues to act, find the acceleration of the body.

Posted by Parshant Sharma 7 years, 9 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

the distance travelled by a particle starting from rest and moving with an acceleration 4/3 m/s2 , in the third second is??

Posted by Nancy Rajput 7 years, 9 months ago

CBSE > Class 11 > Physics

2 answers

Nancy Rajput 7 years, 9 months ago

thanks

0Thank You

Aman Kumar 7 years, 9 months ago

Distance travelled by the nth second is given by t = u +a/2(2n-1)... Put u= 0 , a= 4/3 m/s^2 , n=3.. D= 0+4/3×2(2×3-1)= 4/6×5= 10/3 m..

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What is log

Posted by Mohit Yadav 7 years, 9 months ago

CBSE > Class 11 > Physics

0 answers

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How we can say that motion of an insect is in two dimension

Posted by Bishal Murasing 7 years, 9 months ago

CBSE > Class 11 > Physics

0 answers

ANSWER

Why do the electric field lines not form closed loops?

Posted by Ʌɐpɐʎ Ɐuɐɥsɹɐp . 7 years, 9 months ago

CBSE > Class 11 > Physics

1 answers

Aman Kumar 7 years, 9 months ago

The field lines start from positive field lines and terminates on negative field lines.If electric field lines form a closed loop , these lines must originate and terminate on the same charge which is not possible. So , electric field does not form closed loop.

1Thank You

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