An electric kitchen range has a total wall area of 1.40 m2 and is insulated with a layer of fiberglass 4.0 cm thick. The inside
surface of the fiberglass has a temperature of 175 ∘C and its outside surface is at 35 ∘C. The fiberglass has a thermal conductivity of 0.040 W/(m⋅K).What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 m2?What electric-power input to the heating element is required to maintain this temperature?
1 answer:
Answer: Heat current through the insulator=196W
Electric power= 196W
Explanation: Given: Kglass = 0.040W/m
Temperature of inside glassTi=175°C
Temperature of outside glass To= 35°C
Area=1.4m^2 , L= 4×10^-2
Heat current(H)= K ×A× (Ti - To)/L
Substituting the values into the equation
H = 0.04 × K × 1.4 ×(175-35)/4×10^-2
H= 196W.
The electric power = Heat current =196W
The electric power is the magnitude of heat current
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Answer:
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
Explanation:
Given data
initial circumference = 165 cm
rate = 12.0 cm/s
magnitude = 0.500 T
tome = 9 sec
to find out
emf induced and direction
solution
we know emf in loop is - d∅/dt ........1
here ∅ = ( BAcosθ)
so we say angle is zero degree and magnetic filed is uniform here so that
emf = - d ( BAcos0) /dt
emf = - B dA /dt ..............2
so area will be
dA/dt = d(πr²) / dt
dA/dt = 2πr dr/dt
we know 2πr = c,
r = c/2π = 165 / 2π
r = 26.27 cm
c is circumference so from equation 2
emf = - B 2πr dr/dt ................3
and
here we find rate of change of radius that is
dr/dt = 12/2π = 1.91 cm/s
so when 9.0s have passed that radius of coil = 26.27 - 191 (9)
radius = 9.08 cm
so now from equation 3 we find emf
emf = - (0.500 ) 2π(9.08 ) 1.91
emf = - 0.005445
and magnitude of emf = 0.005445 V
so
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
Answer:
a.6.5025 J
b.6.5025 J
Explanation:
We are given that
Mass of pellet,m=0.27 g=
1 kg=1000 g
Spring constant,k=1800 N/m
x=8.5 cm=
1m=100 cm
a.Potential energy stored in the compressed spring is given by
P.E=
b.By using law of conservation of energy
P.E of spring=K.E of the pellet
K.E of the pellet=6.5025 J