Answer:
6222.22 sec
Explanation:
Given data the power input to the refrigerator is 450 W
The COP of refrigerator is 1.5
Temperature 
mass of watermelon =10 kg
specific heat =4.2 KJ/kg°C
The amount of heat removed from 5 watermelon
We know that 
W=0.675 KW
so time required to cool the watermelon is
Answer:
The number of germanium atoms per cubic centimeter for this germanium-silicon alloy is 3.16 x 10²¹ atoms/cm³.
Explanation:
Concentration of Ge (
) = 15%
Concentration of Si (C
) = 85%
Density of Germanium (ρ
) = 5.32 g/cm³
Density of Silicon (ρ) = 2.33 g/cm³
Atomic mass of Ge (A)= 72.64 g/mol
To calculate the number of Ge atoms per cubic centimeter for the alloy, we will use the formula:
No of Ge atoms/cm³=[Avogadro's Number*]/([*A/ρ)+(C*A/ρ)]
= (6.02x10²³ * 15%) / [(15% * 72.64/5.32)+(72.64*85%/2.33)]
= (9.03x10²²)/(2.048+26.499)
= (9.03x10²²)/(28.547)
No of Ge atoms/cm³ = 3.16 x 10²¹ atoms/cm³
Answer:
71.17°C
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
15.4 g/cm³, 17.4 g/cm³
Explanation:
The densities can be calculated using the formula below
ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)
fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8
a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³
b) density after infiltration with silver
fraction occupied by silver = 20 / 100 = 0.2
density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³
Answer:
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0 it is a incompressible flow.
c) ap = (16/3) i + (32/3) j + (16/3) k
Explanation:
Given
V = xy² i − (1/3) y³ j + xy k
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0
then
∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z
⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.
c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z
⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)
⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k
⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k
At point (1, 2, 3)
⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k
⇒ ap = (16/3) i + (32/3) j + (16/3) k
