Answer:
The rate at which temperature changes is proportional to the rate at which heat is transferred.
Explanation:
The temperature of a sample changes more rapidly if heat is transferred at a high rate and less rapidly if heat is transferred at a low rate
Example:
<span>Suppose that a ball has an initial velocity v0 that has magnitude 20.0 m/s and is directed 60.0 degrees above the negative x axis.
Then, the x-component is,
20.0 * cos(-60) = 20.0 * (1/2) = 10.0</span> m/s
Answer:
<h2>S.A. = 402.62 m²</h2>
Step-by-step explanation:
We have:
two right triangles with legs 4m and 6m
The area:
three rectangles
22m × 6m
The area:
22m × 7.21m
The area:
22m × 4m
The area:
The Surface Area:
Substitute:
Answer:
(a) Q = (1.27)*10⁻⁴ (m³/s)
(b) v₂ = 18 m/s
Explanation:
Flow Equation
Q = v*A
where:
Q = Flow in (m³/s) Formula (1)
A is the surface of the cross sections (m²)
v: is the speed of the fluid ( m/s)
Continuity equation
The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.
Q₁ = Q₂
where :
Q₁ : Flow in the point 1 of the hose
Q₂: Flow in the point 2 of the hose
Data
D₁= 2.25 cm = 0.0225 m
v₁ = 0.320 m/s
D₂ = 0.30 cm = 0.003 m
Area calculation
A = (π*D²)/4
A₁ = (π*D₁²)/4 = (π*(0.0225)²)/4 = 3.976*10⁻⁴ m²
(a) Volume flow rate (Q)
Q = v₁*A₁
Q = (0.320 m/s ) (3.976*10⁻⁴ m²)
Q = (1.27)*10⁻⁴ (m³/s)
(b) Speed of the water (v₂) emerging from a 0.30 cm diameter nozzle.
A₂ = (π*D₂²)/4 = (π*(0.003)²)/4 = 7.07*10⁻⁶ m²
Continuity equation
Q=Q₁ = Q₂
(1.27)*10⁻⁴ = v₂A₂
(1.27*10⁻⁴) /A₂ = v₂
v₂ = (1.27)*10⁻⁴ / (7.07)*10⁻⁶
v₂ = 18 m/s
