Physics, 29.02.2020 01:39 jamarengle2

A cutting tool under microprocessor control has several forces acting on it. One force is F⃗ =−αxy2j^, a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is α = 3.00 N/m3 . Consider the displacement of the tool from the origin to the point x= 2.85 m , y= 2.85 m .a. Calculate the work done on the tool by F⃗ if this displacement is along the straight line y=x that connects these two points.

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Answer from: kimsouther2

The work done is 49.48 J.

The tool is acted upon by a force F = $-axy^{2} j$

a = 3.00 $\frac{N}{m^{3} }$

Now the tool is displaced along the straight line y = x

Then the force F follows by the line:

F = $-ay^{3} j$ , putting x = y so that it satisfies the given straight line equation y = x

So we need only y-coordinate to calculate the work done W.

$W=\int\limits {F} \, ds$ , here ds is the displacement element.

$W=\int\limits^{2.85}_0 {-ay^{3} } \, dy =-3*[y^{4}/4 ]^{2.85}_0$

W = -3 × $[(2.85)^{4}/4 ]$

W = -49.48 J

Learn more about work done:

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Answer from: 2021CanadyRaniya

Explanation:

Given that

F=-αxy²j

α=3N/m³

Displacement from x=2.85 to y=2.85

Since, y=x

Then dy=dx

ds=dx i + dy j

a. Work done

W=∫ F.ds from (0,0) to (2.85,2.85)

W=∫(-αxy²j).(dx i + dy j) (0,0)—(2.85,2.85)

Note, i.i=j.j=1 i.j=j.i=0

W=∫-αxy²dy from (0,0)—(2.85,2.85)

W=-α∫xy²dy from (0,0)—(2.85,2.85)

x=y

W=-α∫y³dy from (0,0)—(2.85,2.85)

W=-αy⁴/4. from (0,0)—(2.85,2.85)

W=-3[(2.85)⁴/4-0]

W=-49.48J

The work done on the tool by the force is -49.48J

Answer from: Quest

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

step-by-step explanation:

we are given

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}$

we have to simplify it

firstly, we will factor it

$2y^2-5y-7=\left(y+1\right)\left(2y-7\right)$

$6y^2+10y+4=2\left(y+1\right)\left(3y+2\right)$

$9y^2+6y=3y(3y+2)$

$=\frac{\left(y+1\right)\left(2y-7\right)}{2\left(y+1\right)\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can cancel like terms

$=\frac{\left(2y-7\right)}{2\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can combine it

$=\frac{3y(2y-7\right)}{6y\left(3y+2\right)}-\frac{7\times 2}{6y(3y+2)}$

$=\frac{3y(2y-7\right)-14}{6y\left(3y+2\right)}$

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

so, we get

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

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