Physics
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New answer posted
a year agoContributor-Level 10
V = I? (G + R? )
1 = I? (G + R? )
2 = I? (G + R? + R? )
2 = 1 + I? R?
I? R? = 1
R? = G + R?
New answer posted
a year agoContributor-Level 10
dW? = Eqdx
∫ dU? = ∫ kQ/x² dxq
W? = kQq (-1/x)|? ^ (R+y)
W? = kQq (y)/ (R) (R + y)
W? + W? = 1/2 mv²
V² = 2/m (kQqy/ (R) (R+y) + mgy)
V² = 2y (kQq/ (m (R) (R+y) + g) ; k = 1/ (4πε? )
New answer posted
a year agoContributor-Level 10
1/4 m (210)² = m (0.03) x (4.2) x 1000 x ΔT ; Q = mSΔt
ΔT = (210) (210)/ (4) (4.2) (0.03) (1000) = 87.5°C
New answer posted
a year agoContributor-Level 10
Photodiode operate in reverse bias. The photocurrent increases initially and saturates fi
New answer posted
a year agoContributor-Level 10
f = V (n)/ (4 (1 - 17/100) (as closed from end)
f = (n - 2) (v)/ (4 (1 - 24.5/100)
nV (100)/4 (83) = (n - 2) (v) (100)/ (4) (75.5)
75.5n = 83 (n - 2)
75.5n = 83 (M - 2)
7.5n = 166
n = 22 (approx)
f = (330) (22) (100)/ (4 (83) = 2200 Hz
New answer posted
a year agoContributor-Level 10
As Torque net = 0
Hence, L = constant
Iω = (3I + I)ω'
ω' = ω/4
Loss in K.E. = 1/2 Iω² - 1/2 (I + 3I)ω²/16 = 1/2 Iω² (1 - 1/4)
Fractional loss = (3/4 * 1/2 Iω²)/ (1/2 Iω²) = 3/4
New answer posted
a year agoContributor-Level 10
As it just floats B = mg
(4πR³/3) (ρ? ) (g) = (4πR³/3 - 4πr³/3) (ρ? ) (g)
R³ = (R³ - r³) (27/8)
On solving we get,
r = 8/9 R (approx)
New answer posted
a year agoContributor-Level 10
Refractive index (n) is a dimensionless quantity that has no units. Since both numerator and denominator are in meters per seconds (m/s), both units cancel each other. This makes refractive index simply a number.
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