What two properties show that the drink is a fluid ( girl drinking fruit punch out of a clear glass cup with a straw.)

Physics

2 answers:

jeka944 years ago

5 0

Answer:

The answer to your question is given below

Explanation:

The properties that shows that the drink is a fluid includes:

1. She is able to drink the fluid with a straw. Fluid has capacity to flow. Since the drink can flow through the straw to get to the girl's mouth, then, the drink can be said to be a fluid.

2. The drink occupy a specific volume of the clear glass cup. We can easily see the volume occupied by the fluid through the clear glass cup. Since the drink occupies a particular volume in the clear glass cup, the drink is obviously as fluid

Ainat [17]4 years ago

3 0

It takes the shape of the cup and it can be sucked through a straw

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What is the approximate mass of air in a living room 4.5m×3.4m×2.9m? the density of air is 1.29 kg/m3?

topjm [15]

First we have to calculate the volume of the living room:
V = L x W x H = 4.5 m * 3.4 m * 2.9 m
V = 44.37 m³
We know that Density = 1.29 kg/m²
D = m / V
m = D · V
m = 1.29 kg/m³ · 44.37 m³
m = 57.2373 kg ≈ 57.2 kg
Answer: The approximate mass of air in living room is 57.2 kg.

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3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .