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3 years ago

How to find volume metric

1 answer:

8 0

You multiply the high length and width and if your using centimeters then divide by 500 and then there's your answer.hoped this helped.

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The diagram shows a person using a piece of gym equipment to lift weights.

ki77a [65]

Answer:

C. The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

Explanation:

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3 years ago

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How much is the velocity of a body when it travels 600m in 5 minutes?

maks197457 [2]

answer

so unit of velocity is m/s

displacement=600m

5minutes should be converted to seconds

5×60=300 seconds

so,

velocity= displacement÷time

= 600m ÷300s

=2m/s or 2ms^-1

4 0

1 year ago

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What type of configuration is used in the HEVs available in the US today?

Citrus2011 [14]

Answer:

Currently in the united states using parallel system

Explanation:

because you can walk with the twomodes with internal combustion engine or running on electric power.

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1 year ago

Scientific models can be used for a variety of different purposes. Which of the following is not a possible use for a scientific

alex41 [277]

Options a to c can be the reasons for scientific models.

But to primarily answer scientific questions,that would require an empirical and experimental approach and not use of models.

Though after getting the answers, models can be built to further explain the answers.

<span>d. answer scientific questions.</span>

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3 years ago

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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.,

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

6 0

3 years ago