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EastWind [94]
3 years ago
8

When waves of equal amplitude from two sources are in phase when they interact, it is called ________.?

Physics
1 answer:
8090 [49]3 years ago
3 0
The answer is constructive interference. At the point when two waves meet such that their peaks line up together, then it's called productive obstruction. The subsequent wave has a higher adequacy. In dangerous obstruction, the peak of one wave meets the trough of another, and the outcome is a lower add up to adequacy.
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Need help with this question!
statuscvo [17]

Answer:(note that values will be on top in small)

cobalt :- 60Co

potassium :- 40K

neon :- 24Ne

lead :- 208Pb

7 0
2 years ago
Change the following as indicated in the brackets.<br> 8m (km,cm)​
Jet001 [13]

metres to kilometres = 1/1000

8 m ⇒ 0.008 km

metres to centimetres = × 100

8 m ⇒ 800 cm

5 0
2 years ago
Read 2 more answers
Which is the correct order of the academic pathway of a pulmonologist?
saul85 [17]

Answer:

B

Explanation:

on edge 2021

6 0
3 years ago
Read 2 more answers
Queremos un cilindro de simple efecto que utilice en su funcionamiento un volumen de aire a presión atmosférica de 13,122 litros
zhuklara [117]

Answer:

1) El diámetro es de aproximadamente 913,987 cm.

2) La fuerza del cilindro es 5576850 kgf

Explanation:

1) Los parámetros dados son;

El volumen del aire = 13,122 litros = 13122000 cm³

La presión de trabajo = 8.5 kgf / cm²

La longitud del cilindro = 20 cm.

Por lo tanto, tenemos;

El área de la base del cilindro = π · r² = 13122000 cm³ / (20 cm) = 656100 cm²

r = √ (656100 / π) ≈ 456,994 cm

El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm

El diámetro ≈ 913,987 cm

2) La fuerza del cilindro = El área de la base del cilindro × La presión de trabajo

∴ La fuerza del cilindro = 656100 cm² × 8.5 kgf / cm² = 5576850 kgf

La fuerza del cilindro = 5576850 kgf

3 0
2 years ago
Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball
Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

              y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e)    y = 0 + ½ g t²

     t = √ (2y / g)

     t = √(2 125 / 9.8)

     t = 5.05 s

6 0
2 years ago
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