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3 years ago

Can someone explain this to me please will give brainliest

1 answer:

3 0

The answer is A. for this, you have to set up a system of equations. the first one will be the area equation. since you know area=length x width, your equation will be LxW=50. the next equation is L=2W, since the length is two times the width. then, plug in 2W for the L in the other equation and you get 2W^2=50. divide by 2 and get W^2=25. square root both sides and you get W=5. plug back into the other equation to find L=10. Then, add the sides of the rectangle for the perimeter. 10+5+10+5=30.

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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

3 years ago

Which is steeper: a road with a 12% grade or a road with a pitch of 1 in 8?

11Alexandr11 [23.1K]

Answer:

A road with a pitch of 1 in 8 is steeper.

Step-by-step explanation:

Let us convert these to the same units so that we can better compare them.

$\frac{1}{8} = 0.125$

0.125=12.5 %

As 12.5% is greater than 12%, the road with a pitch of 1 in 8 will be steeper.

7 0

4 years ago

The time needed for checking in at a hotel is to be investigated. Historically, the process has had a standard deviation equal t

WITCHER [35]

Answer:

See Attachment

Step-by-step explanation:

3 sigma= 3×0.146

= 0.438

for upper limit= sample mean + 3 sigma

for lower limit= sample mean - 3 sigma

8 0

3 years ago

The length of triangle base is 26. A line, which is parallel to the base divides the triangle into two equal area parts. Find th

Yanka [14]

Answer:

Step-by-step explanation:

It is given that the length of triangle base is 26, then let ABC be the triangle and BC be the base of the triangle=26.Let DE be the parallel line to the base that divides triangle ABC into two equal area parts.

Now, Let AD=a, DB=b, DE=c, AE=d and EC=e, then

Since, triangle ABC is similar to triangle ADE, thus using basic proportions, we get

$\frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}$