Answer:
Explanation:
Given that
Q= 5 L/min
1 L = 10⁻³ m³/s
1 min = 60 s
Q=0.083 x 10⁻³ m³/s
d= 6 μm
v= 1 mm/s
So the discharge flow through one tube
q = A v
A=2.8 x 10⁻¹¹ m²
v= 1 x 10⁻³ m/s
q= 2.8 x 10⁻¹⁴ m³/s
Lets take total number of tube is n
Q= n q
n=Q/q
Surface area A
A= π d L
Metalloids are in Group 13 to 16
2.c
3.b
1.a
......................................................................................................................................................
<span>radiation, hydrogen, and helium </span>
Answer:
7.72 Liters
Explanation:
normal body temperature = T_body =37° C
temperature of ice water = T_ice =0°c
specfic heat of water = c_{water} =4186J/kg.°C
if the person drink 1 liter of cold water mass of water is = m = 1.0kg
heat lost by body is Qwater =mc_{water} ΔT
= mc{water} ( T_ice - T_body)
= 1.0×4186× (0 -37)
= -154.882 ×10^3 J
here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories
= 286×4186J
so number of liters of ice water have to drink is
n×Q_{water} =Q_{walk} n= Q_{walk}/ Q_{water}
= 286×4186J/154.882×10^3 J
= 7.72 Liters
