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Shkiper50 [21]
3 years ago
10

Is 4/16 equal in measurement to 1/4

Engineering
2 answers:
Alja [10]3 years ago
5 0

Answer:yes

Explanation:

rodikova [14]3 years ago
4 0
Yes It is. The first fraction can be reduced to equal 1/4
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mH275 units

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Consider a 8-m-long, 8-m-wide, and 2-m-high aboveground swimming pool that is filled with water to the rim. (a) Determine the hy
Stolb23 [73]

Answer:

The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.

<u>Explanation:</u>

To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is \text {hydrostatic force}=\text {pressure} \times \text {area,}=\rho g h \times(l \times h)  

\equiv \rho g h^{2} l, we know ρ=density of fluid=1000 g / c m^{3},

g=acceleration due to gravity=9.81 m / s^{2}, h=height of the pool=2 m and l=length of the pool=8 m.  

hydrostatic force on each wall=1000 \times 9.81 \times 2^{2} \times 8 = 313920 N.

<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool. </em>

At 1 m from the ground this hydrostatic force is acted on each wall.  

The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation =\rho g h^{2} l = 1000 \times 9.81 \times 4^{2} \times 8 = 1255680 N. This is 4 times 313920 N.

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3 years ago
A steel ball of 20 mm diameter (k = 60 W/m·ºC, α = 177×10-6 m2/s, rho = 7854 kg/m3) is dropped in a large tank of water. The sph
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Answer:

attached below

Explanation:

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4 years ago
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Incomplete question. The options read;

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Technician B is correct because the technician highlighted valid reasons why draining the compressed air systems is important.

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Also, the reasons mentioned by Technician A are of course correct because it is generally believed that the application of lubricants such as oil helps to reduce wear and tear.

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