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New answer posted
5 months agoContributor-Level 10
v = 0.5t² i + 3t j + 9 k m/s ⇒ a = dv/dt = (ti + 3j) m/s²
At t=2sec, v = 2i + 6j + 9k m/s and a = (2i + 3j) m/s²
The direction of acceleration of mosquito after 2s is given by the angle θ with the y-axis, where tanθ = a? /a? = 2/3.
So, the direction is tan? ¹ (2/3) from the y-axis.
New answer posted
5 months agoContributor-Level 10
E_K - E_L = hc/λ_Kα
E_K - E_L = (4.14*10? ¹? * 3*10? )/0.071*10? = 17500 eV = 17.5 keV.
E_L = E_K - 17.5 = 27.5 - 17.5 = 10 keV.
New answer posted
5 months agoContributor-Level 10
Range R = v*t = √ (2gh) * √ (2 (H-h)/g) = 2√ (h (H-h).
For R to be max, dR/dh = 0.
h (H-h) must be max. d/dh (Hh-h²)=H-2h=0.
h=H/2 = 12/2=6m.
New answer posted
5 months agoContributor-Level 10
E = -dφ/dt = - (20t + 20) mV.
At t=5s, E = - (100+20) = -120mV.
I = |E|/R = 120mV/2Ω = 60mA.
New answer posted
5 months agoContributor-Level 10
v = ∫a dt = (1/M)∫F dt
v = (F? /M) ∫ [1 - (t-T)²/T²] dt from 0 to 2T
= (F? /M) [t - (t-T)³/3T²]? ²?
= (F? /M) [ (2T - T³/3T²) - (0 - (-T)³/3T²) ]
= (F? /M) [ 2T - T/3 - T/3 ] = (F? /M) [ 4T/3 ] = 4F? T/3M.
New answer posted
5 months agoContributor-Level 10
Δy/y = 2Δm/m + 4Δr/r + |x|Δg/g + (3/2)Δl/l
18 = 2 (1) + 4 (0.5) + |x|p + (3/2)4
18 = 2 + 2 + |x|p + 6 = 10 + |x|p
8 = |x|p
From the options, if x=8, p=±1. If x=16/3, p=±3/2. If x=5, p=±8/5. If x=4, p=±2.
Option B gives x=16/3, and p is not among the options.
Option A gives x=5, p not among options.
Option C gives x=8, p not among options.
Option D gives x=4, p not among options.
There must be a typo in the question or options. The solution gives x=16/3 and p=3/2. Let's check.
8 = (16/3) * (3/2) = 8. So B is correct.
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