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2 years ago

An object is moving from north to south what is the direction of the force of friction of the object

Physics

2 answers:

aev [14]2 years ago

6 0

Answer:

North

Explanation:

Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.

Murljashka [212]2 years ago

3 0

Answer:

South To North

Explanation:

Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north

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A capacitance C and an inductance L are operated at the same angular frequency.

PIT_PIT [208]

Answer:

(B) 7279.70 rad/sec (C) $X_L=37.1265\ ohm\ and \ X_C=37.1265\ ohm$

Explanation:

We have given $L=5.10mH$ and $C=3.70\mu F$

(B) The angular frequency is denoted as $\omega$ whose value is given by $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.10\times 10^{-3}\times 3.70\times 10^{-6}}}=7279.70\ rad/sec$

Capacitive reactance $X_C=\frac{1}{\omega C}=\frac{1}{7279.70\times 3.70\times 10^{-6}}=37.1265\ ohm$

6 0

3 years ago

Scientific laws explain_____.

saveliy_v [14]

A pattern in nature observed over and over

8 0

3 years ago

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1.(16 pts.) Find the volume of the solid obtained by revolving the region enclosed by y = xex , y = 0 and x = 1 about the x-axis

MrRa [10]

Answer:

<em>The Volume is 5.018 cubic units</em>

Explanation:

<u>Volume Of A Solid Of Revolution</u>

Let f(x) be a continuous function defined in an interval [a,b], if we take the area enclosed by f(x) between x=a, x=b and revolve it around the x-axis, we get a solid whose volume can be computed as

$\displaystyle V=\pi \int_a^bf^2(x)dx$

It's called the disk method. There are other available methods to compute the volume.

We have

$f(x)=xe^x$

And the boundaries defined as x=1, y=0 and revolved around the x-axis. The left endpoint of the integral is easily identified as x=0, because it defines the beginning of the region to revolve. So we need to compute

$\displaystyle V=\pi \int_0^1(xe^x)^2dx=\pi \int_0^1x^2e^{2x}dx$

We need to first determine the antiderivative

$\displaystyle I=\int x^2e^{2x}dx$

Let's integrate by parts using the formula

$\displaystyle \int u.dv=u.v-\int v.du$

We pick $u=x^2,\ dv=e^{2x}dx$

Then $du=2xdx,\ v=\frac{e^{2x}}{2}$

Applying by parts:

$\displaystyle I=x^2\frac{e^{2x}}{2}-\int 2x\frac{e^{2x}}{2}dx$

$\displaystyle I=\frac{x^2e^{2x}}{2}-\int xe^{2x}dx$

Now we solve

$\displaystyle I_1=\int xe^{2x}dx$

Making $u=x,\ dv=e^{2x}dx$

$\displaystyle du=dx,\ v=\frac{e^{2x}}{2}$

Applying by parts again:

$\displaystyle I_1=x\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}dx$

$\displaystyle I_1=\frac{xe^{2x}}{2}-\frac{1}{2}\int e^{2x}dx$

The last integral is directly computed

$\displaystyle \int e^{2x}dx=\frac{e^{2x}}{2}$

Replacing every integral computed above

$\displaystyle I=\frac{x^2e^{2x}}{2}-\left(\frac{xe^{2x}}{2}-\frac{1}{2}\frac{e^{2x}}{2}\right)$

Simplifying

$\displaystyle I=\dfrac{\left(2x^2-2x+1\right)\mathrm{e}^{2x}}{4}$

Now we compute the definite integral as the volume

$V=\pi \left[\dfrac{\left(2(1)^2-2(1)+1\right)\mathrm{e}^{2(1)}-\left(2(0)^2-2(0)+1\right)\mathrm{e}^{2(0)}}{4}\right]$

Finally

$V=\pi \dfrac{\mathrm{e}^2-1}{4}=5.018$

The Volume is 5.018 cubic units

8 0

3 years ago

Explain the following behaviour of molecules water rises up in a harrow tubes but mercury which is also a liquid falls in a narr

Mandarinka [93]

Answer: find the answer in the explanation

Explanation:

The capillarity of water molecules is different from the mercury molecules.

What is capillarity ?

This is the tendency of a liquid substance to rise in a capillary tube.

Molecules water rises up in a harrow tubes because of the force of adhesion between the water molecules and the tube molecules is greater than the force of cohesion between the water molecules. This helps water to wet the tube and rise. While mercury which is also a liquid falls in a narrow tubes to level below the outside surface because the force of cohesion between the mercury molecules is greater than the force of adhesion between the mercury molecules and the tube molecules. Mercury does not wet.

7 0

3 years ago

Undersea mountain ranges in the middle of the ocean floors are known as d͟e͟e͟p͟-͟o͟c͟e͟a͟n͟ ͟t͟r͟e͟n͟c͟h͟e͟s.

REY [17]

False

mid-ocean ridge

7 0

2 years ago

Read 2 more answers