Class 11th
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New answer posted
3 months agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoNew question posted
3 months agoNew answer posted
3 months agoContributor-Level 10
If
Y = 3K (1 - 2
and Y = 2
With the help of equations (1) and (2), we can write
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