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4 years ago

11

You have gathered information about which websites your classmates visit to research science questions. What type of graph would

best represent this information?
A. pie chart
B. bar graph
C. line graph
D. scatter plot

1 answer:

swat324 years ago

8 0

A bar graph would be best so B. Bar graph.

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3. What will happen to the black bass and blue gill as the floor of the ponds fills with organic

yKpoI14uk [10]

Answer: die

Explanation: oyxagan all goon bc of all dat suffs

8 0

3 years ago

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete

Sonbull [250]

Answer:

<em>work done in pumping the entire fuel is 466587 J</em>

<em></em>

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical,<em> we assume that the fuel within also takes the cylindrical shape.</em>

<em>Also, we assume that the fuel completely fills the tank.</em>

volume of a cylinder = $\pi r^{2}l$

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x $1.5^{2}$ x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = <em>466587 J</em>

8 0

3 years ago

If two variables have a non-related relationship and one of the variables is changed, how will the other variable change?

PolarNik [594]

If one of the variables is changed, that tells nothing about what happens to the other one, or IF anything happens, or when, or how long it lasts. Because they are UN-RELATED. You just said so yourself.

None of the choices says this.

6 0

3 years ago

A particle moves according to a law of motions=f(t),t≥0 Wheretismeasured in seconds andsin feet.a) Find the velocity at timet.b)

Ray Of Light [21]

Answer:

a) $v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) 3s

c) t < 3s

d) 1.8ft

e) particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

Explanation:

a)Suppose the equation for motion is:

$f(t) = \frac{9t}{t^2+9}$

Then the velocity is the derivative of the motion function

$v(t) = \frac{df(t)}{dt} = (9t(t^2+9)^{-1})^'$

From here we can apply product rule

$v(t) = (9t)^{'}(t^2+9)^{-1} + (9t)((t^2+9)^{-1})^'$

$v(t) = \frac{9}{t^2+9} - \frac{9t(2t)}{(t^2+9)^2}$

$v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) The particle is at rest when v(t) = 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} = 0$

$\frac{9}{t^2+9} = \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 = 2t^2$

$t^2 = 9$

$t = 3s$

(c) The particle is moving in positive direction when v(t) > 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} > 0$

$\frac{9}{t^2+9} > \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 > 2t^2$

$t^2 < 9$

$t < 3s$

(d) As particle is moving in positive direction when t < 3s and negative direction when t > 3s, we can calculate the distance it's moving up to 3s and then after 3s

$f(3) = \frac{9*3}{3^2+9} = \frac{27}{18} = 1.5 ft$

At 3s, particle is changing direction to negative, so its position at 6s is

$f(6) = \frac{9*6}{6^2+9} = \frac{54}{45} = 1.2 ft$

Therefore from 3s to 6s it would have moved a distance of 1.5 - 1.2 = 0.3 ft

Then the total distance it has moved in the first 6 s is 1.5 + 0.3 = 1.8 ft

e) Acceleration is the derivative of velocity function:

$a(t) = \frac{dv(t)}{dt} = (\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2})^'$

$a(t) = -\frac{9(2(t^2+9))(2t)}{(t^2+9)^2} - \frac{36t}{(t^2+9)^2} + \frac{18t^2(2(t^2+9))(2t)}{(t^2+9)^3}$

$a(t) = -\frac{36t}{t^2+9} - \frac{36t}{(t^2+9)^2} + \frac{72t^3}{(t^2+9)^2}$

$a(t) = -\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2}$

Particle is speeding up when a(t) > 0:

$-\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2} > 0$

$\frac{72t^3 - 36t}{(t^2+9)^2} > \frac{54t}{t^2+9}$

as $t \& (t^2 + 9)^2 \geq 0$ we can multiply/divide both sides by it:

$8t^2 - 4 > 6(t^2+9)$

$8t^2 > 6t^2 + 58$

$t^2 > 29$

$t > \sqrt(29) \approx 5.385 s$

so particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

8 0

3 years ago

Which of the following factors can negatively affect development?

PtichkaEL [24]

The answer would be D. It's D because if there was corruption and instability, how will we fund the project? If there was lack of resources, how will the house be completed? If there was a lack of education, how will buildings be able to be engineered?

8 0

4 years ago