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2 years ago

An 4.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, starting

Physics

1 answer:

mina [271]2 years ago

5 0

Answer:

2.64N

Explanation:

Force = mass * acceleration

Given

mass = 4kg

distance = 1.9m

Time t = 2.4s

Get the acceleration using the equation of motion

S = ut + 1/2at²

1.9 = 0 + 1/2a(2.4)²

1.9 = 5.76a/2

1.9 = 2.88a

a = 1.9/2.88

a = 0.66m/s²

Get the magnitude of the force

Force = 4 * 0.66

Force = 2.64N

Hence the net force acting on the fish is 2.64N

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Part Two: Criteria, Constraints, and Prioritizations

AlexFokin [52]

Answer:

Usually, a solution can have several criteria and constraints. Even though all are important, some criteria are more important than others. The same holds true for constraints. But what do you do if it's impossible for a solution to cover every criterion while avoiding every constraint? In cases like this, you can use prioritization. Listing criteria and constraints based on priority shows the relative importance of each. You will need to prioritize the criteria and constraints for each sub-problem so that you can design a solution for each one individually. Prioritization can help you compare two different possible solutions. For example, the criterion that cars travel at 15 mph through the neighborhood might be a higher priority than the constraint that homeowners are only willing to spend $10,000 on this issue. If this is the case, you would want to generate solutions that also follow the priority in mind. All criteria are important, but engineers must sometimes make a trade-off, which is a compromise or change in one or more criteria or constraints so that they can be met at the same time. This is where prioritization comes in handy as it helps determine the trade-offs. A solution that is doing a better job of meeting one criterion may result in not completely meeting another criterion. Prioritization will help you choose which solution to go with.

Explanation:

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2 years ago

Based on the replacement reaction, what would the products of the reaction be?

cricket20 [7]

Answer:

$\rm Be(OH)_2$ and $\rm (NH_4)_2 SO_4$ . The missing ion would be $\rm OH^{-}$ .

Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

$\rm BeSO_4$ , and
$\rm NH_4 OH$ .

In particular,

$\rm BeSO_4$ is made up of $\rm Be^{2+}$ ions and $\rm {SO_4}^{2-}$ ions, while
$\rm NH_4 OH$ is made up of $\rm {NH_4}^{+}$ ions and $\rm OH^{-}$ ions.

In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

$\rm Be^{2+}$ is a cation. In $\rm BeSO_4$ , $\rm Be^{2+}$ was bounded $\rm {SO_4}^{2-}$ anions. During the reaction, it bonds with $\rm OH^{-}$ anions to produce $\rm Be(OH)_2$ .

$\rm {NH_4}^{+}$ is also a cation. In $\rm NH_4 OH$ , $\rm {NH_4}^{+}$ was bounded to $\rm OH^{-}$ ions. During the reaction, it bonds with $\rm {SO_4}^{2-}$ anions to produce $\rm (NH_4)_2 SO_4$ .

Hence, the two products will be $\rm Be(OH)_2$ and $\rm (NH_4)_2 SO_4$ .

Note that charges on the ions must balance. For example, a $\rm Be^{2+}$ ion carries twice as much charge as an $\rm {NH_4}^{+}$ ion. As a result, each $\rm Be^{2+}$ ion would bond with twice as many $\rm OH^{-}$ ions as $\rm {NH_4}^{+}$ would in $\rm NH_4 OH$ .

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3 years ago

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What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?

PolarNik [594]

Answer:

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$n_i=n\ and\ n_f=n-1$

Thus solving it, we get:

$\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

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c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m$

So,

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

Also, $\Delta E=h\times \nu$

So,

$h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}$

$\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

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3 years ago

Sodium has an atomic mass of 23 and an atomic number of 11, so the

horrorfan [7]

Answer:

12 Neutrons

Explanation:

So the mass of sodium is 22.990. You round it up to get 23(as stated in the problem). So, <em>what exactly is atomic mass?</em>

Atomic Mass is the total amount of neutrons and protons added up to form a total mass. So when you subtract 23-11 you get 12 Neutrons.

<u>Tip: </u>Don't know if you need this but-

The neutrons and protons are typically close in number (unless it's an isotope). So say that you subtract and the numbers of protons and neutrons aren't close at all. Well if that's the case, it's probably wrong.

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2 years ago