Why should we convert units within the metric system?

Physics

1 answer:

postnew [5]4 years ago

3 0

Because we should convert units within the metric system

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In the process of loading a ship, a shipping container gets dropped into the water and sinks to the bottom of the harbor. Salvag

jolli1 [7]

Answer:

57.885.8 kg weight of the container

Explanation:

The volume of the balloon * density of water = buoyant force of balloon

volume of a sphere = 4/3 pi r^3

= 4/3 pi * (1.5)^3 = 14.14 m^3 <===balloon volume

Now, find the buoyant force on the container ALONE ....

5.8 * 2.6 * 2.8 * 1027 = 43 364 kg <===== buoyant force

Now add the buoyant force of the balloon to find the weight

43 364 + 14.14 * 1027 = 57885.8 kg

3 0

2 years ago

The gage pressure in a liquid at a depth of 3 m is read to be 50 kPa. Determine the gage pressure in the same liquid at a depth

Andrei [34K]

Answer:

The Gauge pressure at 9 meters depth is $150 \, kPa$

Explanation:

Gauge pressure is the difference between absolute pressure and some reference pressure, most commonly atmospheric pressure. The increment in pressure caused by a static fluid is given by:

$\Delta P = \rho g d$ where $\rho$ is the density of the liquid, g is the accleration due to gravity and d is the depth.

Now, we see that $\Delta P$ is linearly proportional to d, and we can assume that $\rho$ remains constant, because liquids are usually not compressible.

Given that the greater depth is simply 3 times the smaller depth:

$d_2=3\cdot d_1\\9\,m= 3 \cdot 3\,m$

$\Delta P$ at $9\, m$ of depth will also be three times the gauge pressure at $3 \,m$ of depth.

We could also have calculated $\rho$ ny using:

$\Delta P = \rho \,g \,d\\\\\rho = \frac{\Delta P}{g \, d}\\\\\rho = \frac{\Delta P}{g \, d}= \frac{30 \,kPa}{9.8 \frac{m}{s^2} \, 3\,m}=1020.41 \frac{kg}{m^3}$