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New answer posted

2 months ago

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Vishal Baghel

Contributor-Level 10

Maximum range is obtained at 45?
u²/g = 1.6 or u = 4m/s
T = (2u sin 45? )/g = (2*4* (1/√2)/10 = 0.4√2s
Number of jumps in a given time, n = t/T = 2√2 / 0.4√2 = 5

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Gold band is equal to 5% tolerance

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

x = 8R/2, y = 5R/2
x/y = 1.60

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

a? = v? ²/4r
a_A? = (v? ²/r²) * r = v? ²/r
a_A = 3v? ²/4r

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

-mv cos 60? + 2mu = 0 => v = 4u
½m [v² + u² + 2uv cos 120? ] + ½mu² = mgx sin 60?
=> v² = (8/7)√3gx => ar = (4√3/7)g
∴ t = √ (2L * 7)/ (4√3g)

New answer posted

2 months ago

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Vishal Baghel

Contributor-Level 10

z² * (13.6) (1 - ¼) = 3 * (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) * h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 * 10.2 - Φ) = 1/5.25
=> Φ = 3eV

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Ui = ½C (V/3)² + ½C (V/3)² + ½C (2V/3)²
Uf = ½CV²
Wb = CV²/3
ΔH = [Wb - (Uf - Ui)] = CV²/6 = 0.30 mJ

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Is? = 4I?
Is? = I?
∴ I_max/I_min = (√Is? + √Is? )² / (√Is? - √Is? )² = (2+1)²/ (2-1)² = 9/1

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

N = [n + n/10 + n/100 + .]
= n/ (1 - 1/10) = 10/9

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