a-rat-has-a-birth-mass-of-8-mathrmg-and-when-fully-grown-its-mass-is-280-g-using-equation-11-26-with

Name: a rat has a birth mass of 8 mathrmg and when fully grown its mass is 280 g using equation 11 26 with
Uploaded: 2020-06-16
Duration: 6 min 28 s
Channel: Numerade
Description: A rat has a birth mass of $8 \mathrm{g},$ and when fully grown its mass is 280 g. Using equation $(1.1 .26)$ with $a=0.25$ determine how many days it will take for the rat to reach $75 \%$ of its fully grown size.

Question

A rat has a birth mass of $8 \mathrm{g},$ and when fully grown its mass is 280 g. Using equation $(1.1 .26)$ with $a=0.25$ determine how many days it will take for the rat to reach $75 \%$ of its fully grown size.

Michael Jacobsen · Accepted Answer

for the set up of this problem. We have a rodent with the birth mass off eight grams. We know that this adult will have a mass of 280 grams. We&#x27;re going to use this model to find the time needed to reach 75% of the adult mass. When a is set to be equal to 0.25 to start out, let&#x27;s start assigning what the variables it should be uppercase m that appears here is the adult mass. So that would be 280 grams in this problem. The next symbol is m not I m not is going to be the birth mass. So that&#x27;s eight grams. In this situation, we have the variable a given to be 0.25 here and if we say t it is time to reach 75% of adult mass, that would then mean that M A T should be 0.75 times 280 so m of tea would be equal to 210 grams. This tells us our goal is to solve M of tea equals 210 14. Once we found this expression for tea will have solved this problem. So let&#x27;s start out by making a substitution ample sides of this model, so we&#x27;ll have 210 on the left hand side is equal to uppercase M, which is given to be 280 grams times one minus one minus. I am not, which is eight grams. Divide by uppercase M or 280 Alta power of 1/4 times the Basie exponential to the power of negative A or negative 0.25 times T divide by the quantity four times uppercase m or four times 280 to the power of 1/4 and the entire quantity, once again to the power of four. To solve this equation for tea will be using the order of operations to simplify each of these quantities. Step by step pushed out. Let&#x27;s start out with this quantity. Here we can say on the next step that 210 is going to be equal to 280 times the quantity. One minus zero point 5889 times the Basie exponential And let&#x27;s do on the next step, the simplification of negative 0.25 Divide by this quantity in our calculator that would turn into the quantity negative 0.1 52 nine times T Well, once again raised to the power of four. What we can do in the next step is divide both sides by this quantity to 80. So that will take to 10 divide by 2 80 We also know that from this expression here that should result in 0.75 So now we have 0.75 equals one minus 0.5889 times E to the power of negative is your a 0.15 to 90 all raised to power for on the very next step, we can take the fourth root of both sides Know the words and my calculator, or raise your 0.75 to power 1/4 raising the right hand side to power. 1/4 gives a cancellation of this outer exponents with the 1/4 next in my calculator, 0.75 to the 1/4 power or the four through to that quantity is 0.93 06 which is now set equal to one minus 0.5 89 Times e to the power negative. 0.152 nine Times t. Next, we can subtract one from both sides of this equation. If we do that. Left hand side is negative. 0.6 939 which is equal to negative zero point 5889 either the power of negative 0.15 to 9 t for next step. Let&#x27;s divide both sides of this equation. By that coefficient upon that division will now obtain zero point 1178 equals e to the power negative zero point 015 to 9 t for this exponential equation. If we take the natural laudable sides, then the Basie exponential would be cancelled. So first, the natural log of zero point 1178 turns into negative 2.1384 Taking the natural log of the right hand side cancels the Basie. So we have just the exponents negative 0.15 to 9 t. Finally, we can solve this equation by dividing by the coefficient of tea. So we will obtain that T is approximately equal to 139.9 days. In other words, using the provided my bottle with that provide information, we know that this rodent will reach 75% of its adult mass in just about 240 days.

Michael Jacobsen · Answer

here we&#x27;re dealing with the bird that we know at the weight at which they hatch is three grams. We also know that this bird is meant to away to 2700 grams when they&#x27;re an adult. Our jobs to determine what the massive this bird should be 30 days after hatching, using the following model with the parameter A set equal to 1.5. To solve this problem. Using the provide model, we will write M for Mass at a time of 30 days so Emma, 30 will be equal to the following values nor to carefully evaluate this formula. Let&#x27;s take a moments, too. I figure out what each of the quantities ought to be. First, uppercase M is the adult way to the bird, which is 2700 grams. You have a lower case M, but that&#x27;s going to be the hatch weight or birth weight and other problems. So am not. And this problem is going to be three grams and we know the value of the parameter A is 1.5 and the Time T is going to be set equal to 30 days. With that information were ready to begin evaluating. So it&#x27;s right first uppercase M or 2700 times the Group one minus the Group one minus I m not, which is three grams or just three. Divide by uppercase M 2700 to the quantity of or the power of 1/4 is a quantity that&#x27;s multiplied by the Basie exponential to the power of a or negative 1.5 times T, which is set to 30 days. Divide by the quantity four times uppercase M, which is 2700 to the power of 1/4 and this entire quantity is raised to the power off. Four. When this expression is carefully put into a calculator, we find that the mass after 30 days is approximately equal to 1271 0.18 grams, and this solves the current problem.

Juan Pablo Olloqui · Answer

for this question. We want to make determine aretz mess and we&#x27;re where you need in in space. So we&#x27;re told that are FG is of 0.46 mutants and ah, the cages Mess embassy is of 320 Gramps, which is equal to 0.32 kilograms. Okay. And we are also told that the acceleration is of 0.4 meters over second squared on. This is how much the cage incinerator accelerated after he was put, uh after it was hanged from the spring scale. So how we want to set up this equation is that FG is equal to the total mess times the acceleration. So the total mass is gonna be the massive gauge plus the mass of the red. Remember that they would want to solve for the message read. So we divide close site of decoration by a and we get this. So now we can just move EMC to the other side and we get that the mass of the red It&#x27;s equal to have Jean divided by a minus the massive cage. Now I have all these values and I console for the mess of the red. Uh, if she We said it was zero point 46 Newtons. Acceleration was zero point for meters sober, second squared and the mass of the cage was syrup 0.32 kilograms. Now you can plug this into your calculator and you get that. The mess of the red is equal to zero point 83 telegraphs. So this is our answer for the most of the Rhett.

Ryan Eaton · Answer

in this problem we&#x27;re giving the radio activity of. So for 35 I on, we&#x27;re being asked to find the amount of so for ion present in the sample in terms of milligrams and units and milligrams. They give us the given information that the half life of sulphur 35 is 87.37 days and we can look up in appendix B, the atomic mass of so for 35 which is 34.969032 atomic mass units. This is all the information we need to solve this problem because the activity of a radioactive substance scales with its decay constant times the amount of ah, substance there is in terms of number of nuclei. So we can rearrange these factors and write this in terms of number of nuclei from the activity and decay constant Um, the decay constant is the reciprocal of the dick decay Constant. Rather, is the half life over the natural log of two and we can plug in our, um activity, which is given a decays per second. It&#x27;s a given in the problem and, um, were given 1/2 life in terms of in units of days. We need to convert that two seconds to match the activity given in this problem. So we&#x27;re using a, uh, conversion factor of 86,400 seconds per day. In order to do that, plugging those numbers in, we get, um, 4.66 times 10 to the 11th nuclei in our sample. But recall that were asked for the amount of sample in terms of mass. So to do that, we need to multiply this, um, number of nuclei by the atomic mass of the substance which is in atomic mass units. Then we need to convert that, um, we need to multiply it again by another conversion factor. And that is the amount of mass in grams per atomic unit. And that will give us an end result in grams, which turns out to be 2.71 times 10 to the negative 11th grams of so for 35

Amy Jiang · Answer

given peachy is equal. Joan One for you, 50 over one plus 29 e t o *** of 0.62 Yeah, with 60 T rest souvenirs. Tent. How many days will it take the culture to reach 75% of its carrying capacity? Well caring capacity of 142 50 It&#x27;s time that time&#x27;s up about 75% you get. That&#x27;s approximately 10 687.5. So for this equation, in case of a 10 687.5 plus 29 times 10 687.5 give me three girls I 937.5 each. In the Gozo 0.62 tea is equal to 14 to 50 subtracting 14 to 50 I mean subjecting 10687.5 on what sides we get. One. There are 687.5, divided by 309937 45. Take the eligible sides and I divide my natives a 450.62 This gives me t and plug. It wasn&#x27;t your calculator. A lot of 10.10 06 or 10687.5. What about trees are all night? 937 45. But by negative 450.62 it gives me five, approximately five points for three days or 5.4 days.

Katie Mcalpine · Answer

Okay, so this one&#x27;s pretty similar to ones before it. You&#x27;re given the half life and, um, the activity, and you wanna know the mass of the sample. So let&#x27;s s I mean, the crux of this problem is just we&#x27;re gonna be you relating the number of particles to the read and then the decay rate and then and then selling for the number of particles and converting it to mass. Um, it takes a long time because these numbers are so many numbers and unit conversions involved. So that&#x27;s what I&#x27;m gonna spend the majority of the time doing, so Eighties. So the half life is 87.32 days. I&#x27;m gonna double check that number. Yeah, and let&#x27;s just go ahead and convert that two seconds. So Google it six 400 seconds per day goes 7.54 times 10 to the sixth, um, seconds and pilot reactivity is roof 3.65 times 10 to the fore decays per second. And finally, let&#x27;s write down the molar mass of this sulfur isotope, which is 36 0.985 grams per man. All right? And so basically so there&#x27;s are sort of ingredient prep area. Um, so now for the actual math. So we activity rate is three initial number or initially, an actual activity rate jumps zooming this would&#x27;ve which would they give us? Is this initial number times lambda And then Lambda is equal to the Ellen two divided by the half life. And you can get that from the definition of our which is D N d t evaluating it ti ful zero and then an equals the initial number e to the minus lambda tm, which is in the textbook. So okay, And then now our goal is to sulfur, and not so just algebraic Lee. Rearranging this I get its are not at times t to the 1/2 divided by the log of two. And yeah, so if I plug in this are here in this tea here in the right by Elena too. Um, I get the original number is 3.97 times 10 to the 11 particles. And now I need to convert that to amass. So the mass is gonna be and not divided by 6.22 times 10 to the 23 particles per no times the more mass. Um, And then when I did that, I obtain to point for four times 10 to the minus 11 grams.

Problem 33 Hard Difficulty

A rat has a birth mass of $8 \mathrm{g},$ and when fully grown its mass is 280 g. Using equation $(1.1 .26)$ with $a=0.25$ determine how many days it will take for the rat to reach $75 \%$ of its fully grown size.

Answer

$T\cong 139.9$ days

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$T\cong 139.9$ days

Differential Equations and Linear Algebra

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Michael J.

Integration Review - Intro

Definite vs Indefinite

Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

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Catherine R.

Kristen K.

Michael J.

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra