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Size: 68 Kb Pages: N/A Date: 2012-07-16
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Ans8 1 pdf
Ans8 1 pdf

(you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun.Why ? 1. Solution: (a)No,we cannot shield a body by forming any kind of gravity screen as in case of electrical forces the force varies as per the medium but in case of gravitational force,the force is always attractive.

Size: 68 Kb Pages: N/A Date: 2012-07-16

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Dipole 1 PDF
Dipole 1 pdf

DIPOLE The dipole source is a basic theoretical acoustic source consisting of two adjacent monopoles 180° out of phase. Mathematically this approximates a single source in translational vibration, which is what we really want to model. − + − + = Tom Penick [email protected] www.teicontrols.com/notes/readme.html Dipole-1.PDF 11/

Size: 13 Kb Pages: N/A Date: 2011-08-18

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mata pelajaran kimia kelas x tkj semester 1 pdf pdf
mata pelajaran kimia kelas x tkj semester 1 pdf pdf

Mata Pelajaran Kimia Kelas X Tkj Semester 1 Pdf 1/1

Size: 9 Kb Pages: N/A Date: 2012-08-17

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Ans8 7 pdf
Ans8 7 pdf

NCERT/CBSE MATHEMATICS CLASS 11 textbook http://www.TutorBreeze.com MISCELLANEOUS EXERCISES Answers to NCERT/CBSE MATH (Class XI)textbook 8.BINOMIAL THEOREM 7.(0.99)5 = (1 − .01)5 = 5C0 (.01)0 − 5C1(.01)1 + 5C2 (.01)2 − 5C3 (.01)3 + 5C4 (.01)4 − 5C5 (.01)5 ≈ 1 − 5(.0001) + 10(−00001) = 1 − .05 + .001 ≈ .951 Please do not copy the answer given here Write to us for help

Size: 64 Kb Pages: N/A Date: 2012-04-22

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Ans8 16 pdf
Ans8 16 pdf

Formatted:Font:Bold,Lowered by 74 pt 2Radius of earth=R(let)Acceleration due to gravity at given height h,g h = g (1 − d )R gh = g (1 − 1) 2 ⇒ gh = 1g 2Weight of body at height h,Wh = mgh = mg ( 1 ) = 125N 2 ©TutorBreeze.comPlease do not copy the answer given hereWrite to us for help in understanding the solution

Size: 82 Kb Pages: N/A Date: 2011-10-29

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Ans8 15 pdf
Ans8 15 pdf

Formatted:Font:Bold,Lowered by 74 pt 2Radius of earth=R(let)Acceleration due to gravity at given height h,g h = g (1 + h )−2R gh = g (1 + 1 )− 2 2 ⇒ gh = 4g 9Weight of body at height h,Wh = mgh = mg ( 4 ) = 28N 9 ©TutorBreeze.comPlease do not copy the answer given hereWrite to us for help in understanding the solution

Size: 82 Kb Pages: N/A Date: 2013-02-08

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Ans8 12 pdf
Ans8 12 pdf

Mass of sun,Ms = 2 ×1030 kgFormatted:Font:Bold,Lowered by 18 ptMass of earth,Me = 6 ×1024 kgOrbital radius,r=1.5 ×1011 mLet gravitational force on rocket becomes 0 at a distance of x from earth's centre.At this point,Electric field due to earth=Electric field due to sun ⇒Me =Ms 2 x2 (r − x) ⇒ r − x = (M s )0.5 xMe 30 ⇒ r −1 = ( 2 ×10 24 s )0.5 x 6 ×10 ⇒x= 3 r 1735Putting value of r and solving, x=2.6 ×108 m ©Tutor

Size: 84 Kb Pages: N/A Date: 2013-07-17

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Ans8 4 pdf
Ans8 4 pdf

Formatted:Font:Bold,Lowered by 13 ptPeriod of orbit,T=1.769 days=1.769 × 86400 s=15.2841×104 sRadius of orbit,r=4.22 ×108 mG=6.67 ×10-11Nm2kg −2Using the relation:M j = ω2 r3 2 2 3 ⇒M j = ω r 3 = 4π2 rGTGSubsituting the values,M j = 2 ×1027 kgM j 2 ×1027 =M s 2 ×1030HenceM j =M s 1000 ©TutorBreeze.comPlease do not copy the answer given hereWrite to us for help in understanding the solution

Size: 85 Kb Pages: N/A Date: 2011-12-06

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Ans8 21 pdf
Ans8 21 pdf

Mass of spheres,M=200 kgRadius of each sphere,R=0.1 mDistance between the two spheres,r=1mA=B=0.5 mLet a body of mass m is placed atODue toA,Gravitational force,FA = 100GmA 0.25Gravitational potential,A = −100G 0.5Due toB,Gravitational force,FB = 100GmB 0.25Gravitational potential,B = −100G 0.5TutorBreeze.comPlease do not copy the answer given here

Size: 73 Kb Pages: N/A Date: 2013-07-21

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Ans8 10 pdf
Ans8 10 pdf

C   3x 2  +   1  ( 2ax ) − 2  3x2  ( 2ax )   3a 2  + 3

Size: 70 Kb Pages: N/A Date: 2012-11-04

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