Complete question :
A 12 m x 15 m house is built on a 12-cm-thick concrete slab.
What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C
Answer:
3kW
Explanation:
Given the following :
Dimension of house :
Length = 12m
Width = 15m
Thickness of concrete slab (t) = 12cm
t in metres :
100cm = 1m
12cm = (12/100)m
= 0.12m
Ground temperature (Tg) = 5°C
Interior temperature = (Th) = 25°C
Thermal conductivity of concrete (K) is approximately 1 Wm/k
Using the relation:
Q = KA * [ (Th - Tg) / d]
A = Length * width = (12 *15) = 180
Q = (1 * 180) * [(25°C - 5°C) / 0.12]
Q = 180 * (20/0.12)
Q = 180 * 16.6666
Q = 3,000W = 3kW
Answer:
4.04 s
Explanation:
h = vi + 1/2 a t ^2
HERE h = 80 m , vi = 0 , a =9.81 m/s^2
80 = 0 + 1/2 × 9.81 × t ^2
80 = 4.905 t^2
t^2 = 80/4.905
t ^2 = 16.30988
t = square root of 16.30988
t = 4.0385 s
t = 4.04 s
D= 9.7^3/(69)(4.2)^2
d=912.673/289.8^2
d=912.673/83984.04
d=0.01086721953362...
I hope this helped!
Answer:
Explanation:
The charge on 10μF capacitor = 10 x 12 x 10⁻⁶ = 120 μC
when it is connected with 20μF capacitor both acquires common potential whose value is
= 120 x 10⁻⁶ /( 10 +20) x 10⁻⁶ = 4 V.
Energy stored in 20μF capacitor =1/2 x 20 x 10⁻⁶ x 4 x 4 = 160 x 10⁻⁶ J.
