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New Question

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

x 1 = 5 s i n ( 2 π t + π 4 )

x 2 = 5 2 ( s i n 2 π t + c o s 2 π t )

= 1 0 s i n ( 2 π t + π 4 )

x 2 , m a x = 2 . x 1 , m a x

New Question

8 months ago

0 Follower 1 View

R
Rohini Rawat

Contributor-Level 8

In the introduction part of LOR for UK, the recommender must state how long he has known the applicant and in what capacity. Basically, the association/ relationship between the recommender and the applicant should be clearly established. Then, one line can be written appreciating the candidate and stating the recommender's overall impression of the applicant.

New Question

8 months ago

0 Follower 3 Views

R
Ranjeeta Shukla

Contributor-Level 8

Generally, the number of required LORs for UK universities vary from 1 to 2. This choice is specifically defined by each university for their esteemed courses. Hence, Indian students must check the strict requirements of the course they ae going to apply for. For instance, the University of Cambridge requires 1 LOR and the University of Oxford asks for two.

New Question

8 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

[ λ × 3 × g ] × 3 2 [ λ × 1 × g ] × 1 2 = K         

=> λ m a s s l e n g t h

=> λ g ( 9 2 1 2 ) = K

3 3 × 1 0 × 4 = K

K = 40 J

New Question

8 months ago

0 Follower 1 View

A
Abhishek Mehra

Contributor-Level 8

The LOR format for UK is no different from those for US or any other country. The LOR should be one page or approximately 500 words in length. The font and size should be consistent and readable, for instance, Times New Roman 11 or 12 point size. The LOR can be academic or professional, showcasing the eligibility of the candidate.

The recommender must establish his association with the candidate right in the beginning of the LOR and later discuss at least three qualities of the applicant and provide suitable examples to back his claims.

New Question

8 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

As image coincides with the object, image formed by convex lens should be at the centre of curvature of the convex mirror. If the convex mirror is removed, then, also, image formed the convex lens is at the same location.

Distance of image from object = 12 + 8 + 2 × 15 = 50 cm

New Question

8 months ago

0 Follower 7 Views

New Question

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  p 1 p 2 = 4 T r 1 . . . . . . . ( 1 )

p 2 p 0 = 4 T r 2 . . . . . . . ( 2 )

(1) + (2), p1 – p0 = 4T ( 1 r 1 + 1 r 2 )

4 T r e q = 4 T ( 1 r 1 + 1 r 2 )

r e g = r 1 r 2 r 1 + r 2 = 3 × 6 3 + 6 = 2 c m

             


New Question

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

L e t , E m a g b T c

So, [ E ] = [ m ] a [ g ] b [ T ] c  

M L 2 T 2 = M a ( L T 2 ) b T c  

Thus, E = k mg2 T2

Δ E E = 2 . Δ g g + 2 . Δ T T   

Percentage error in E = 2 × 4 + 2 × 3 = 14%

New Question

8 months ago

0 Follower 6 Views

S
Sejal Baveja

Contributor-Level 10

Yes, the application form for Arena Creative Campus, Jayanagar admission can be rejected if the details filled by the candidate are incorrect.

New Question

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

For first fringe

d y 1 D = 1 × λ 1 [ n = 1 ] λ 1 Red wavelength

d y 2 D = 1 × λ 2 [ n = 1 ] λ 2 violet wavelength

d D ( y 1 y 2 ) = λ 1 λ 2           

y 1 y 2 = D d ( λ 1 λ 2 )        

= 0 . 3 × 1 0 3 1 . 5 ( 3 . 5 2 ) × 1 0 3

= 300 × 10-9

= 300 nm

New Question

8 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

2nqV = 1 2 m v 2

n = m v 2 4 q V

= 1 . 6 7 × 1 0 2 7 × ( 0 . 5 × 1 0 8 ) 2 4 × 1 . 6 × 1 0 1 9 × 1 2 × 1 0 3 = 5 4 3  

             

New Question

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

l 2 c o s 2 ? = 3 l 8

? c o s ? = 3 2

? ? = 3 0 °

New Question

8 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

  τ 2 = B 1 M 2 S m 9 0 °

= B1 M2

= ( μ 0 4 π × m 1 r 3 ) × m 2        

= 1 0 7 1 1 3 × 1

= 10-7 N – m

New Question

8 months ago

0 Follower 12 Views

N
Nitesh Gulati

Contributor-Level 10

Around 20% of students enrolled for litigation and assisted Senior Counsel in various High Courts. Further,  28% of students chose to pursue higher studies. In addition to this,  47% of the batch was placed in Law firms and 33% were selected in corporate houses. Kochhar & Co., Pepsico, ICICI Lombard, and many other leading companies participated in Symbiosis Law School Hyderabad placements 2023.

New Question

8 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

  F m a x 1 + 2 = 0 . 5 × 1 0 F m a x = 1 5 N

New Question

8 months ago

0 Follower 20 Views

V
Vishal Baghel

Contributor-Level 10

= l x x ' = [ l y y ' + M ( 5 r 2 ) 2 ] + [ l Z Z ' + M ( 1 3 r 2 ) 2 ]

M ( 3 6 r 2 1 2 ) + M × 2 5 r 2 4 = M r 2 2 + M × 1 6 9 M r 2 4

= 52 Mr2

New Question

8 months ago

0 Follower 4 Views

Shiksha Ask & Answer
Indrani Choudhury

Contributor-Level 10

Admission to Symbiosis Law School, Symbiosis International, Hyderabad, is based on an entrance exam. The institute offers three courses at the UG and PG level. Aspirants must complete Class 12/graduation as per the course requirement. Candidates need to meet the basic eligibility criteria set by the college. 

New Question

8 months ago

0 Follower 6 Views

S
Sejal Baveja

Contributor-Level 10

Joining Arena Creative Campus, Jayanagar for BSc course is generally considered a decent option. They can apply online or offline for admission at Arena Creative Campus, Jayanagar.

New Question

8 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

f ' = 3 4 0 2 0 3 4 0 + 2 0 × f 0

1 8 0 0 = 3 2 0 3 6 0 × f 0

f0 = 2025 Hz

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