Answer:
Heat energy required (Q) = 3,000 J
Explanation:
Find:
Mass of water (M) = 200 g
Change in temperature (ΔT) = 15°C
Specific heat of water (C) = 1 cal/g°C
Find:
Heat energy required (Q) = ?
Computation:
Q = M × ΔT × C
Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)
Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C
Heat energy required (Q) = 3,000 J
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
<em><u>Support Cy:</u></em>
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
<em><u>Support Ay:</u></em>
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
P = IV
I = P/V = 30 / 120 = 0.25 A.
Current = 0.25A
Answer:
D...............................
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