Answer:
E=1760.93 cal
Explanation:
a) First we find the total time
t=x/v
t=189000/12.5
t=15120 seconds
Now we multiply the time by his total wattage
E=t*W
E=(15120)(6.5)(75)=7371000 joules
As,
1 joule = 2.389 x 10-4
So, in term of nutritional Calories
E=7371000 * 2.389*
E=1760.93 Cal
Answer:
The average gauge pressure inside the vein is 110270.58 Pa
Explanation:
This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.
Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:
As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:
Answer:
Q = 165.95 cm³ / s, 1) v = , 2) v = 2.05 m / s,
3) d₂ = 1.014 cm
Explanation:
This is a fluid mechanics exercise
1) the continuity equation is
Q = v A
where Q is the flow rate, A is area and v is the velocity
the area of a circle is
A = π r²
radius and diameter are related
r = d / 2
substituting
A = π d²/4
Q = π/4 v d²
let's reduce the magnitudes
v = 0.55 m / s = 55 cm / s
let's calculate
Q = π/4 55 1.96²
Q = 165.95 cm³ / s
If we focus on a water particle and apply the zimematics equations
v² = v₀² + 2 g y
where the initial velocity is v₀ = 0.55 m / s
v =
v =
2) ask to calculate the velocity for y = 0.2 m
v =
v = 2.05 m / s
3) We write the continuous equation for this point 2
Q = v₂ A₂
A₂ = Q / v₂
let us reduce to the same units of the SI system
Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s
A₂ = 165.95 10⁻⁶ / 2.05
A₂ = 80,759 10⁻⁶ m²
area is
A₂ = π/4 d₂²
d₂ =
d₂ =
d₂ = 10.14 10⁻³ m
d₂ = 1.014 cm