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Ask QuestionPosted by Miss Akanksha 2 years, 10 months ago
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Shanika Sharma 2 years, 9 months ago
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Preeti Dabral 2 years, 10 months ago
- As AO = BO = CO = 1 m, hence we have
{tex}\left| \overrightarrow { F _ { A } } \right| = \left| \overrightarrow { F _ { B } } \right| = \left| \overrightarrow { F _ { C } } \right| = \frac { G M \cdot 2 M } { ( 1 ) ^ { 2 } } = 2 G M ^ { 2 }{/tex}
If we consider direction parallel to BC as x-axis and perpendicular direction as y-axis, then as shown in figure, we have
{tex}\vec { F } _ { A } = 2 G M ^ { 2 } \hat { j }{/tex}
{tex}\vec { F } _ { B } = \left( - 2 G M ^ { 2 } \cos 30 ^ { \circ } \hat { i } - 2 G M ^ { 2 } \sin 30 ^ { \circ } \hat { j } \right){/tex}
and {tex}\overrightarrow { F _ { C } } = \left( 2 G M ^ { 2 } \cos 30 ^ { \circ } \hat { i } - 2 G M ^ { 2 } \sin 30 ^ { \circ } \hat { j } \right){/tex}
Therefore, the net force on mass 2M placed at the centroid O is given by,
{tex}\vec { F } = \vec { F } _ { A } + \vec { F } _ { B } + \vec { F } _ { C } {/tex}
{tex}= 2 G M ^ { 2 } \left[ j + \left( - \frac { \sqrt { 3 } } { 2 } \hat { i } - \frac { 1 } { 2 } \hat { i } \right) + \left( \frac { \sqrt { 3 } } { 2 } \hat { i } - \frac { 1 } { 2 } \hat { j } \right) \right]{/tex}
{tex}=0{/tex}
Posted by Noyone Akter 2 years, 10 months ago
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Preeti Dabral 2 years, 10 months ago
The work obtained in bringing a body from infinity to a point in gravitational field is called gravitational potential energy. Force of attraction between the earth and the object when an object is at the distance a from the center of the earth. F=x2GmM.
Posted by Laxman Das 2 years, 8 months ago
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Preeti Dabral 2 years, 8 months ago
Distance to market s=2.5km=2.5×103 =2500m
Speed with which he goes to market =5km/h={tex}5 \frac{10^3}{3600}=\frac{25}{18} \mathrm{~m} / \mathrm{s}{/tex}
Speed with which he comes back = {tex}7.5 \mathrm{~km} / \mathrm{h}=7.5 \times \frac{10^3}{3600}=\frac{75}{36} \mathrm{~m} / \mathrm{s}{/tex}
(a)Average velocity is zero since his displacement is zero.
(b) (i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market: {tex}\frac{2.5}{5}=1 / 2 \mathrm{~h}=30 \text { minutes. }{/tex}
Average speed over this interval =5km/h
(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :{tex}7.5 \times \frac{1}{3}=2.5 \mathrm{~km}{/tex}
His average speed in 0 to 50 minutes: Vavg =distance traveled/time {tex}=\frac{2.5+2.5}{(50 / 60)}=6 \mathrm{~km} / \mathrm{h}{/tex}
(iii)In 40-30=10 minutes he travels a distance of : {tex}7.5 \times \frac{1}{6}=1.25 \mathrm{~km}{/tex}
{tex}\mathrm{V}_{\mathrm{avg}}=\frac{2.5+1.25}{(40 / 60)}=5.625 \mathrm{~km} / \mathrm{h}{/tex}
Posted by Danish Khan 2 years, 9 months ago
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Preeti Dabral 2 years, 9 months ago
As the depth increased the mass of the earth decreases. At the surface of earth this value will be maximum because radius will be maximum. When radius becomes less this value also decreases. Hence acceleration due to gravity decreases with increase in the depth.
Posted by Kanishka Parmar Kanishka 2 years, 9 months ago
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Preeti Dabral 2 years, 9 months ago
A solid cylinder rolling up an inclination is shown in the following figure.
Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, {tex}\theta = 30 ^ { \circ }{/tex}
Height reached by the cylinder = h
- Energy of the cylinder at point A will be purely kinetic due to the rotation and translational motion. Hence, total energy at A
= KErot + KEtrans
= {tex}\frac { 1 } { 2 } I \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 }{/tex}The energy of the cylinder at point B will be purely in the form of gravitational potential energy = mgh
Using the law of conservation of energy, we can write:
{tex}\frac { 1 } { 2 } I \omega ^ { 2 } +\frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
Moment of inertia of the solid cylinder, {tex}I = \frac { 1 } { 2 } m r ^ { 2 }{/tex}
{tex}\therefore \frac { 1 } { 2 } \left( \frac { 1 } { 2 } m r ^ { 2 } \right) \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
{tex}\frac { 1 } { 4 } I \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
But we have the relation, {tex}v = r \omega{/tex}
{tex}\therefore \frac { 1 } { 4 } v ^ { 2 } + \frac { 1 } { 2 } v ^ { 2 } = g h{/tex}
{tex}\frac { 3 } { 4 } v ^ { 2 } = g h{/tex}
{tex}\therefore h = \frac { 3 } { 4 } \frac { v ^ { 2 } } { g }{/tex}
{tex}= \frac { 3 } { 4 } \times \frac { 5 \times 5 } { 9.8 } = 1.91 \mathrm { m }{/tex}To find the distance covered along the inclined plane
In {tex}\Delta A B C{/tex}:
{tex}\sin \theta = \frac { B C } { A B }{/tex}
{tex}\sin 30 ^ { \circ } = \frac { h } { A B }{/tex}
{tex}A B = \frac { 1.91 } { 0.5 } = 3.82 \mathrm { m }{/tex}
Hence, the cylinder will travel 3.82 m up the inclined plane. -
{tex}v = \left( \frac { 2 g h } { 1 + \frac { K ^ { 2 } } { R ^ { 2 } } } \right) ^ { \frac { 1 } { 2 } }{/tex}
{tex}\therefore v = \left( \frac { 2 g A B \sin \theta } { 1 + \frac { K ^ { 2 } } { R ^ { 2 } } } \right) ^ { \frac { 1 } { 2 } }{/tex}
For the solid cylinder, {tex}K ^ { 2 } = \frac { R ^ { 2 } } { 2 }{/tex}
{tex}\therefore v = \left( \frac { 2 g A B \sin \theta } { 1 + \frac { 1 } { 2 } } \right) ^ { \frac { 1 } { 2 } }{/tex}
{tex}= \left( \frac { 4 } { 3 } g A B \sin \theta \right) ^ { \frac { 1 } { 2 } }{/tex}
The time taken to return to the bottom is:
{tex}t = \frac { A B } { v }{/tex}
{tex}= \frac { A B } { \left( \frac { 4 } { 3 } g A B \sin \theta \right) ^ { \frac { 1 } { 2 } } } = \left( \frac { 3 A B } { 4 g \sin \theta } \right) ^ { \frac { 1 } { 2 } }{/tex}
{tex}= \left( \frac { 11.46 } { 19.6 } \right) ^ { \frac { 1 } { 2 } } = 0.7645{/tex}
So the total time taken by the cylinder to return to the bottom is (2 {tex}\times{/tex} 0.764)= 1.53 s.as time of ascend is equal to time of descend for the following problem.
Posted by Tasbiya Pathan 2 years, 10 months ago
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Krishna Achu 2 years, 10 months ago
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Preeti Dabral 2 years, 10 months ago
The constitution and structure of matter at the scales of atoms and nuclei and their interaction with different elementary particles
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Preeti Dabral 2 years, 8 months ago
The mean value of the mass of the ring. Mmean = 2.52+2.53+2.51+2.49+2.545=2.52g.
Posted by Sakshi Singh 2 years, 11 months ago
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Preeti Dabral 2 years, 8 months ago
The diameter of a thread is so small that it cannot be measured using a metre scale. We wind a number of turns of the thread on the meter scale so that the turns are closely touching one another.
Measure the length ({tex}l{/tex}) of the winding on the scale which contains n number of turns. The diameter of thread = {tex}\frac{l}{n}{/tex}
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