Answer:
so heat loss = 4312 W
cost of heat loss daily is $8.28 per day
Explanation:
given data
slab length L = 11 m
slab wide W = 8 m
thickness t = 0.20 m
temperature top T1 = 17°C
temperature bottom T2 = 10°C
thermal conductivity k = 1.4 W/m-K
efficiency ηf = 0.90
priced Cg = $0.02 / MJ
to find out
rate of heat loss and daily cost of the heat loss
solution
we calculate here heat loss by heat transfer
so apply here formula that is
q = (thermal conductivity × area × temperature difference) / thickness
put here all these value we get heat loss
q =
q =
q = 4312 W
so heat loss = 4312 W
and
cost of the heat loss is express as
cost of heat loss =
put here all these value
cost of heat loss is = × 24 hr/day × 3600 s/hr
cost of heat loss = 8.279
so cost of heat loss daily is $8.28 per day
Given Information:
Angular velocity = ω = 4 rad/s
Angular acceleration = α = 5 rad/s²
Center deceleration = a₀ = 2 m/s
Required Information:
Acceleration of point A at this instant = ?
Answer:
Acceleration of point A at this instant = 5.94 m/s²
Explanation:
Refer to the attached diagram of the question,
The acceleration of point A is given by
a = a₀ + rα - rω²
Where r is the radial distance between the center and point A, a₀ is the deceleration of center, α is the angular acceleration and ω is the angular velocity.
a = -2i + 0.3j*5k - 0.3j*4²
a = -2i + 1.5(j*k) - 0.3j*16
a = -2i + 1.5(-i) - 4.8j
a = -2i - 1.5i - 4.8j
a = -3.5i - 4.8j
The magnitude of acceleration vector is
a = √(-3.5)² + (-4.8)²
a = √35.29
a = 5.94 m/s²
Therefore, the acceleration of point A is 5.94 m/s²
The angle is given by
θ = tan⁻¹(y/x)
θ = tan⁻¹(-4.8/-3.5)
θ = 53.9°