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anzhelika [568]

3 years ago

14

If the bottom of a ship is 15.8 below the surface of the sea, calculate the pressure at the bottom of the ship. The density of t

he seawater is 1030kg/m3

1 answer:

Zolol [24]3 years ago

7 0

Given:
Depth, h= 15.8
Density of seawater= 1030kg/m3

We need to find:
Pressure

Answer:
P= ρhg
P=(1030)(15.8)(9.81)= 159647.94 Pa

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4 divided by 2/9? If you could, please also explain.

shtirl [24]

You do 4/1 divided by 2/9

4 2
_ divided by _

1 9

Equals 18

(You have to multiply by the reciprocal)

4/1 times 1/4
And
2/9 times 1/4

Hope I helped!!!

7 0

3 years ago

In the diagram, the length of segment VS is 39 units.

Darina [25.2K]

Answer:

$TV=38\ units$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

The figure shows a kite

The kite has two pairs of consecutive and congruent sides and the diagonals are perpendicular

That means

TS=VS

TQ=VQ

TR=RV

we have

$VS=39\ units$

$TS=6x-3$

so

$6x-3=39$

solve for x

$6x=39+3\\6x=42\\x=7$

<em>Find the value of RV</em>

$RV=2x+5$

substitute the value of x

$RV=2(7)+5=19\ units$

Remember that

$TV=TR+RV$ ---> by segment addition postulate

we have

$TR=RV=19\ units$

so

$TV=19+19=38\ units$

5 0

2 years ago

The figure shows a carpeted room. How many square feet of the room is carpeted?

alukav5142 [94]

Answer:

40

Step-by-step explanation:

So split it into two shapes. Two rectangles.

Rectangle 1 = 4 * 5 = 20

Rectangle 2 = 2 * 10 = 20

20 + 20 = 40

6 0

3 years ago

Read 2 more answers

PLS TRY TO ANSWER ASAP.

Harrizon [31]

Answer:

<u>0.075 m/min</u>

Explanation:

You need to use derivatives which is an advanced concept used in calculus.

<u>1. Write the equation for the volume of the cone:</u>

$V=\dfrac{1}{3}\pi r^2h$

<u />

<u>2. Find the relation between the radius and the height:</u>

r = diameter/2 = 5m/2 = 2.5m

h = 5.2m

h/r =5.2 / 2.5 = 2.08

<u>3. Filling the tank:</u>

Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

The ratio x/y is the same r/h

x/y=r/h

y = x . h / r

The volume of water inside the cone is:

$V=\dfrac{1}{3}\pi x^2y$

$V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3$

<u>4. Find the derivative of the volume of water with respect to time:</u>

$\dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}$

<u>5. Find x² when the volume of water is 8π m³:</u>

$V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\ 11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062$ m²

<u>6. Solve for dx/dt:</u>

$1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=0.03596m/min$

<u />

<u>7. Find dh/dt:</u>

From y/x = h/r = 2.08:

$y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min$

That is the rate at which the water level is rising when there is 8π m³ of water.

4 0

2 years ago

What is the length of the line segment whose endpoints are A(-1,9) and B(7,4) in the simplest radical form?

monitta

length : $\sf \sqrt{89}$

Explanation:

use the distance formula : $\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

using the formula:

$\sf \rightarrow \sf \sf \sqrt{(7--1)^2+(4-9)^2}$

$\sf \rightarrow \sf \sf \sqrt{(8)^2+(-5)^2}$

$\sf \rightarrow \sf \sf \sqrt{64+25}$

$\sf \rightarrow \sf \sf \sqrt{89}$

8 0

1 year ago

Read 2 more answers