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New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

Time period of Oscillation, T = 2 π I M B

1 4 = 2 π 9 . 8 * 1 0 6 M * 0 . 0 4 9

1 1 6 = 4 π 2 * 9 . 8 * 1 0 6 M * 4 9 * 1 0 3

M = 4 π 2 * 9 . 8 * 1 0 6 4 9 * 1 0 3 * 1 6

= 4 π 2 * 9 . 8 * 1 6 * 1 0 3 4 9

= 1 2 . 8 π 2 * 1 0 3 * 1 0 2 * 1 0 2

=1280π2*105Am2

New answer posted

11 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

Sol. Before collision

It undergoes completely inelastic collision

Using conservation of linear momentum

Initial momentum = Final momentum

m v 1 = m v 2 + m v 2

m v 1 = 2 m v 2

v 1 v 2 = 2 1

New answer posted

11 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

The potential V at any point, at distance r from centre of dipole = KPcosθ r 2

At axial point where  θ=0? , V=KPr2=9*109*4*10622=9*103 V

At axial point where  θ=180? , V=KPr2=9*103 V

New answer posted

11 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

x = 2 t 1

v=dxdt=2 m s1

P = F · v

=2*5=10 W

New answer posted

11 months ago

0 Follower 41 Views

A
alok kumar singh

Contributor-Level 10

Given circuit is balanced Wheatstone bridge

C A B = 1 + 1

= 2 ? F

 

New answer posted

11 months ago

0 Follower 27 Views

A
alok kumar singh

Contributor-Level 10

In the case of pure rolling,

The topmost point will have velocity 2 v while point Q i.e. the lowest point will have zero velocity. Hence point P moves faster than point Q .

 

New answer posted

11 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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According to given truth table, output is independent on value of A

Output Y = B ¯

New answer posted

11 months ago

0 Follower 3 Views

J
Jaya Sharma

Contributor-Level 10

Air resistance resists the motion of an object. In this case, the net acceleration is lesser than 'g' and it shrinks as the speed increases. This makes the object to speed up more slowly. Ultimately, it reaches a constant terminal velocity which is lower for large-area ones and higher for heavy and streamlined ones.

New answer posted

11 months ago

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J
Jaya Sharma

Contributor-Level 10

Suppose the position-time graph is a straight line, in this case, the velocity is constant. This means that there is no acceleration.
If the graph is curved, velocity is changing, which means that there is acceleration. If the graph is concave, the slopes will get more positive with time. This means that there is positive acceleration. If the graph is cap-shaped, the slope will become more negative with time. This is known as negative acceleration.

New answer posted

11 months ago

0 Follower 1 View

J
Jaya Sharma

Contributor-Level 10

To graph motion in a straight line, you need to visualise the relationship between different kinematic quantities like position, velocity and time. Suppose an object moves with a constant velocity, the position-time graph will be a straight line with constant slope. If the object accelerates, the slope of position-time graph will change with time and result in a curved line.

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