Class 11th

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New answer posted

3 months ago

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S
Syed Aquib Ur Rahman

Contributor-Level 10

Elastic deformation leads to the temporary change in a material's shape that is also reversible. It occurs for energy storage in devices, such as springs. It's also essential for understanding the flexibility of structural components like beams in bridges. This principle is used in engineering design. You can think vehicle suspension systems and even the resilience of buildings against wind forces.

New answer posted

3 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Time to reach at max height t = ugno. of balls thrown in 1 sec = n.

So, time taken by each ball to reach maxm height, = 1nsec

i.e. ug=1nu=gn So, hmax = u22g

g22gn2

=g2n2

New answer posted

3 months ago

0 Follower 2 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

When we talk about elastic materials, we're referring to those that really show off their elasticity. Think of natural rubber, synthetic polymers, including spandex (also known as Lycra) and nylon, and even metals such as spring steel when they're within their elastic limits. These materials are quite resilient. They can store potential energy and bounce back to their original shape after being stretched or compressed. This behaviour is determined by the Hooke's Law.

New answer posted

3 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

tangent at (2t2, 4t) is ty = x + 2t2,

It passes through (5, 7)

2 t 2 7 t + 5 = 0 t = 1 , 5 2                

P ( 2 t 2 , 4 t ) will be (2, 4), ( 2 5 2 , 0 )  

New answer posted

3 months ago

0 Follower 1 View

S
Syed Aquib Ur Rahman

Contributor-Level 10

Elastic energy is stored in objects that can deform. They have several applications, including springs in vehicle suspension systems and wind-up clocks. This potential energy, present in items like stretched rubber bands, trampolines, and an archer's bow, is converted to kinetic energy upon release.

New answer posted

3 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

S 5 S 9 = 5 1 7 d = 4 a  

110 < a15 < 120

110 < a + 14d < 120

110 < 57a < 120

->a = 2, d = 8

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Use binomial theorem

2023 = 7 * 289

2 0 2 1 2 0 2 2 + 2 0 2 2 2 0 2 1 = ( 2 0 2 2 2 ) 2 0 2 2 + ( 2 0 2 3 1 ) 2 0 2 1

= 7 P 1 + 2 2 0 2 2 + 7 P 2 1

= 7 ( P 1 + P 2 ) + 1 + 7 P 3 1                          

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

For R1, take a = 1, b = 0, c = -1

a b 0 , b c 0 but AC < 0

So R1 is not an equivalence relation.

For R2, it will not be symmetric.

So R2 is also not an equivalence relation.

 

New answer posted

3 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

A. Torque   (iii) Nm

B. Stress  (iv) Nm-2

C. Latent Heat  (ii) J kg-1

D. Power  (i) Nm S-1

New answer posted

3 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Let the equation of circle be

x ( x 1 2 ) + y 2 + λ y = 0               

x 2 + y 2 1 2 x + λ y = 0

Radius = 1 1 6 + λ 2 4 = 2  

λ 2 = 6 3 4 ( x 1 4 ) 2 + ( y + λ 2 ) 2 = 4   

? This circle and parabola

y α = ( x 1 4 ) 2 touch each other, so

α = λ 2 + 2 α 2 = λ 2 ( α 2 ) 2 = λ 2 4 = 6 3 1 6  

( 4 α 8 ) 2 = 6 3  

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