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andrew11 [14]
3 years ago
6

What factor account for the change of viscosity in fluid

Physics
1 answer:
netineya [11]3 years ago
5 0

Answer:

A fluid's viscosity strongly depends on its temperature. Along with the shear rate, temperature really is the dominating influence

Explanation:

The higher the temperature is, the lower a substance's viscosity is. Consequently, decreasing temperature causes an increase in viscosity.  

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Raina is running for student body president and doesn't want to forget her campaign speech. She practices her speech over and ov
mezya [45]

The answer is intentional.

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3 years ago
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1.20 Newton force is working on a 250 gram object. What is the acceleration?
Leokris [45]

Answer:

The answer is B

Explanation:

250g = 0.25kg

F = m × a

a = F/m

= 1.2/0.25

= 4.8m/s²

7 0
3 years ago
How is temperature related to the physical change of a<br>substance​
blagie [28]

Answer:

~~Now, you have left your question very open ended and didn't ask for any particular kind of answer so I'll do my best to get what you're looking for.~~

A physical change in a substance doesn't change what the substance is. It can possibly melt or freeze an object. I mean heat makes things expand while cooling makes them retract.... In chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

4 0
3 years ago
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Una caja de 5.0kg de masa se acelera desde el reposo a través del piso mediante una fuerza a una tasa de 2.0 /s2 durante 7.0s en
Nady [450]

Responder:

<h2>490 julios </h2>

Explicación:

Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;

Workdone = Fuerza * Distancia

Como Fuerza = masa * aceleración,

Workdone = masa * aceleración * distancia

Masa dada = 5.0kg, aceleración = 2.0m / s² d =?

Para obtener d, usaremos una de las leyes del movimiento,

d = ut + 1 / 2at²

u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s

d = 0 + 1/2 (2) (7) ²

d = 49m

Workdone = 5 * 2 * 49

Workdone = 490 Julios

4 0
3 years ago
What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as
Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h  to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

tan^-1(\frac{v^2}{rg}) = \frac{29.2^2}{(1200)(9.8)} = 4.14°

8 0
2 years ago
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