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4 years ago

15

Help .................

1 answer:

Tema [17]4 years ago

5 0

Answer:

the answer is a

Step-by-step explanation:

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Question is shown in picture

zalisa [80]

It is showing growth.

8 0

3 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

4 years ago

I continue to be bad at math please help me with it

Aleksandr-060686 [28]

Answer:

24

Step-by-step explanation:

There are 4 cups in one US quart, 4*6 is 24.

7 0

3 years ago

11. Max is designing a hot tub for a local spa.

Bezzdna [24]

Answer:

Area required for circular hot tube = 6,218 inch² (Approx.)

Step-by-step explanation:

Given:

Diameter of circular hot tube = 89 inches

Value of π = 3.14

Find:

Area required for circular hot tube

Computation:

Radius of hot tube = Diameter / 2

Radius of hot tube = 89 / 2 inches

Area required for circular hot tube = Area of circle

Area of circle = πr²

Area required for circular hot tube = πr²

Area required for circular hot tube = (3.14)(89/2)²

Area required for circular hot tube = (3.14)(7921 / 4)

Area required for circular hot tube = (3.14)(1,980.25)

Area required for circular hot tube = 6,217.985

Area required for circular hot tube = 6,218 inch² (Approx.)

5 0

3 years ago

Find the perimeter of the triangle with vertices (−2,2), (0,4), and (1,−2).

balu736 [363]

By distance formula
the perimeter would be
the sum of 5+root5+root8

please thank me and mark as the brainiest answer

3 0

4 years ago