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hodyreva [135]
3 years ago
5

A company ships sand to customers for their aquariums. To ship the sand, they use tubes with a 4 cm radius and cut the tubes to

whatever height is needed. The sand they sell has a density of 1.41 grams per cubic centimeter. If Jerry buys 2,100 grams of sand, then approximately how high will the tube be that the sand is shipped in?
Mathematics
1 answer:
Akimi4 [234]3 years ago
4 0

we know that

volume of a cylinder (tube)=pi*r²*h

in this problem

r=4 cm

Volume of a cylinder (tube)=pi*4²*h------> V=16*pi*h cm³-----> equation 1


density=mass/volume------> solve for volume

volume=mass/density

mass=2,100 gr

density=1.41 gr/cm³

so

V=2,100/1.41--------> V=1,489.36 cm³-------> equation 2


equate equation 1 and equation 2

16*pi*h=1,489.36------- h=1,489.36/(16*pi)-------> h=29.63 cm


therefore


the answer is

the high of the tube will be 29.63 cm


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Not sure if any of this is correct, but it’s what I got so far
Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

==================================================

Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote \frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}

I would write it as \frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}

==================================================

For problem 3, we first need to convert the runner's speed from mph to feet per second.

17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}

Since the runner needs to travel 90-12 = 78 ft, this means\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

-------------

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0
3 years ago
A 5% destroys solution is given intravenously. Suppose a total of 2.64 L of the solution is given over a 19-hour period. What is
kow [346]

Answer:

The flow rate is 138.95 mL per hour.          

Step-by-step explanation:

We are given the following in the question:

Volume of solution = 2.64 L

1 L = 1000 mL

Volume of solution in mL =

2.64\times 1000 = 2640\text{ ml}

Time period = 19 hour

Unit rate of flow =

=\dfrac{\text{Volume}}{\text{Time period}}\\\\=\dfrac{2640}{19}\\\\=138.95\text{ mL per hour}

Thus, the flow rate is 138.95 mL per hour.

5 0
3 years ago
Triangles ABC and EDC are outlined on a bridge. The triangles share vertex C and angles D and B are right angles. A new bridge s
dexar [7]

CHECK THE ATTACHED FIQURE FOR THE BRIDGE

Answer:

60 ft

Step-by-step explanation:

From the question we know that triangles ABC and EDC are in a 1:1 with this given ratio it implies that

triangles ABC and EDC are congruent then we can say

side EC = side AC

3x + 9 = 5x - 5

Then we can simplify to know value of x

3x + 14= 5x

2x = 14

x = 7

But we know that AC= 5x - 5 , then substitute value of x into it

AC = 5x + 5 = 5(7) - 5

= 35 - 5

AC= 30 ft

Also EC= 3x + 9 then substitute value of x into it

EC = 3x + 9 = 3(7) + 9

= 21 + 9

EC= 30 ft

Then the the distance between the top and bottom of the bridge, in feet, = EC+AC

= 30 + 30 = 60 ft

4 0
3 years ago
Why is it important to include a scale when displaying a box plot?
KIM [24]
Because then it would be easier to get an answer for a problem
6 0
3 years ago
Simplify: <br><br> 3c-5d+5d<br><br> Pls
Arturiano [62]

Answer:

3 c

Step-by-step explanation:

Simplify the following:

5 d - 5 d + 3 c

Hint: | Combine like terms in 5 d - 5 d + 3 c.

5 d - 5 d = 0:

Answer: 3 c

5 0
3 years ago
Read 2 more answers
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