Monica divorced Robert last year because she was not satisfied with the amount of money he earned and was distressed by their fi
2 answers:
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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
<h2>Answer:</h2>
<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>
<h2>Explanation:</h2>In the question,
Let us say the height from which the arrow was shot = h
Distance traveled by the arrow in horizontal = 61 m
Angle made by the arrow with the ground = 2°
So,
From the <u>equations of the motion</u>,
Now,
Also,
Finally, the angle made is 2 degrees with the horizontal.
So,
Final horizontal velocity = v.cos20°
Final vertical velocity = v.sin20°
Now,
u = v.cos20° (No acceleration in horizontal)
Also,
So,
We can say that,
<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>