Answer:
3.14 × 10⁻⁴ m³ /s
Explanation:
The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.
Q = Area x velocity
Given:
Diameters of 3 sections of the pipe are given as
d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.
Speed in the first segment of the pipe is
v1 = 4 m/s.
From the equation of continuity the flow rate through different cross-sections remains the same.
Flow rate = Q = A1 v1 = A2 v2 = A3 v3.
Q = A1v1
=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s
Answer:
The car that is accelerating is B a car that rounds a curve at a constant speed
Explanation:
Although all of the cars are at a constant speed or not moving acceleration is the change in speed or the change of directions therefore making the only car changing directions your answer.
Answer:
4.98 m
Explanation:
Given that
Width of the mirror, d = 0.6 m
Organist distance to the mirror, s = 0.78 m
Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m
Width of north wall, D?
Using the simple relationship
D/S = d/s, on rearranging
D = dS /s
D = (0.6 * 6.48) / 0.78
D = 3.888 / 0.78
D = 4.98 m
Therefore, we can conclude that the Width of north wall is 4.98 m
