Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the inc
line. The coefficient of kinetic friction between both blocks and the inclined planes is μk=0.10.
1 answer:
vₐ = 0.771 ft/s
vb = -1.54 ft/s
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<u>Explanation:</u>
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Block A:
F = ma
Nₐ = - 60 cos60° = 0
Nₐ = 30 lb
Fₐ = 0.1 X 30 = 3lb
Block B:
F = ma
Nb = - 40 cos30° = 0
Nb = 34.64lb
Fb = 0.1 X 34.64 = 3.464lb
T1 + ∑U = T2
(0 + 0) + 60 sin 60° |Δsₐ| - 40 sin 30° |Δsb| - 3|Δsₐ| - 3.464 |Δsb|
= 1/2 (60/32.2) vₐ² + 1/2 (40/32.2) vb²
and
2vₐ = - vb
On solving, we get
vₐ = 0.771 ft/s
vb = -1.54 ft/s
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Answer: maximum length of the nanowire is 510 nm
Explanation:
From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K
Thermal conductivity of silicon carbide k = 30 W/m.K
Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m
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m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )
m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04
now lets find the value of h/mk
h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353
lets consider the value θ/θb by using the equation
θ/θb = (T - T∞) / (T - T∞)
θ/θb = (3000 - 8000) / (2400 - 8000)
= 0.893
the temperature distribution at steady-state is expressed as;
θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]
so we substitute
0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]
L = 510 × 10⁻⁹m
L = 510 nm
therefore maximum length of the nanowire is 510 nm
