Class 11th

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New answer posted

3 months ago

0 Follower 12 Views

R
Raj Pandey

Contributor-Level 9

For ball (a): I? = Δp? = 2mu
For ball (b): I? = Δp? = 2mu cos 45°
I? /I? = 2mu / (2mu cos 45°) = √2

New answer posted

3 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

By t = mx² + nx
dt/dx = 2mx + n
V = dx/dt = 1/ (2mx + n)
a = v (dv/dx) = V (d/dx (1/ (2mx+n) = V [-2m/ (2mx+n)²] = -2mV³
So, Retardation will be (2mv³)

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

According to question
|x| = |y| and
|x - y| = n|x + y|
x² + y² - 2x·y = n² (x² + y² + 2x·y)
(1 - n²) (x² + y²) = (1 + n²)2x·y
(1 - n²) (x² + y²) = (1 + n²)2xy cosθ
(1 - n²) (2x²) = (1 + n²) (2x²)cosθ
cos θ = (1 - n²)/ (1 + n²)
θ = cos? ¹ (1 - n²)/ (1 + n²) = cos? ¹ (- (n²-1)/ (n²+1)

New answer posted

3 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Displacement in 4? second
= S? - S?
= (1/2)g [2n - 1]
= (1/2)g [2*4 - 1]
= (1/2) * 9.8 * 7
= 34.3m
As this distance matches with data given in question for position of next drop. So drops are falling at the rate of 1 drop/second.

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Body is dropped from height 75m with initial velocity (upward) 10m/s
So, s = ut + ½ at²
-75 = 10t - ½ gt²
5t² - 10t - 75 = 0
By solving t = 5 sec.
In 5sec. balloon covered
h = 10 * 5 = 50m
Now height of balloon = 75 + 50 = 125m

New question posted

3 months ago

0 Follower 3 Views

New answer posted

3 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

From triangle law of addition, we can clearly write answer.

New question posted

3 months ago

0 Follower 8 Views

New answer posted

3 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

f ( x ) = x 4 4 x + 1 = 0

f ' ( x ) = 4 x 3 4

= 4 ( x 1 ) ( x 2 + 1 + x )

Two solution

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

If   μ is Poisson's ratio,

Y = 3K (1 - 2 μ )      ……… (1)

and Y = 2 n ( 1 + μ )   ……… (2)

With the help of equations (1) and (2), we can write

3 Y = 1 η + 1 3 k K = η Y 9 η 3 Y      

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