Can some one please help me

An oil tank is leaking into a lake at a rate of 0.1 m3/day. The oil slick forms a semicircular disk whose center is at the leak

gregori [183]

Answer:

17.9 m/s

Step-by-step explanation:

Volume of the slick = 0.5 x π r² h--------------------------------- (1)

Where r = radius of slick

h = thickness of slick, 10⁻⁶m

If 0.5m³ of oil leaked, then the radius of the semicircular slick can be calculated from equation (1)

V = 0.5 x π r² h

0.5= 0.5 x π x r² x 10⁻⁶

r² = 10⁶/ π

r = 10³/√π

dV/dt = πrh dr/dt + 0.5π r² dh/dt----------------------------------- (2)

Asumming the film thickness is constant , equation (2) becomes

dV/dt = πrh dr/dt-------------------------------- (3)

dV/dt = 0.1m³/day

r= 10³/√π

dr/dt= rate of expansion of the slick

Substituting into (3);

0.1 = π x 10³/√π x 10⁻⁶ x dr/dt

dr/dt = 0.1 x 10⁶/ ( π x 10³/√π)

= 17.9479 m/s

≅ 17.9 m/s

8 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago

Mary estimated that it would take 3 hours to write a book report, but it actually took her 5 hours. What is the percent error fo

Eduardwww [97]

Answer:

The percent error is 40%.

Step-by-step explanation:

5 (actual time) - 3 (estimated time) = 2

2 divided by 5 (actual time) = 0.4

0.4 x 100 = 40%

6 0

3 years ago

Which expression has the same GCF as 15x2 - 21x?

Mice21 [21]

Answer:

$15x^2 +21 x$

Step-by-step explanation:

We have the following expression:

$15x^2 -21 x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are -21,-7,-3,-1,1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

If we consider the new expression:

$15x^2 +21x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are 1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

And as we can see both GCF are equal. So then we satisfy the condition required.

5 0

3 years ago

Read 2 more answers

Miranda made 6.88 pounds of trail mix. She used 4.96 pounds of granola for the mix. Miranda wrote the equation w + 4.96 = 6.88 t

joja [24]

Step 1: to solve this equation first subtract 4.9 from both sides.

w + 4.9 - 4.9 = 6.88 - 4.9

When you reduce that your answer should be w = 1.92

Therefore the weight of ingredients other than granola in Miranda's trail mix is 1.92

5 0

3 years ago