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Natali5045456 [20]
3 years ago
6

Panuto: nibabahagi ang sariling ideya tungkol sa kahalagan ng akdang tinalakay.Umiisip ng mga pangyayari sa kasalukuyan (sa naba

sa, narinig at napanood) na maaari mong iugnay sa nakalahad na pangyayari sa akda.
Mga Pangyayari sa Ibong Adarna

1. Sina Don Pedro at Don Diego ay nahumaling sa kinang at ganda ng puno ng Piedras Platas kaya’t sila’y nabiktima ng taglay nitong panganib na nakakubli sa kagandahan.

2. Nagdasal at humingi ng gabay si Don Juan sa Panginoon bago siya sumuong sa kanyang mapanganib na misyon.

Kaugnay na Pangyayari sa Kasalukuyan

1.



2.
Physics
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

ubmj nycni itvjug big nh.

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Which of these will be the correct relationship between work input and work output?
dsp73

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8 0
2 years ago
Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
6 0
3 years ago
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