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jarptica [38.1K]
4 years ago
8

Which statement about van der Waals forces is true?

Physics
2 answers:
Helen [10]4 years ago
7 0

The answer is,

<u>A. When the forces are weaker, a substance will have higher volatility.</u>

Just took the test (:

kobusy [5.1K]4 years ago
3 0
The answer is A. When the forces are weaker, they will not be able to hold the particles of the substances together; therefore, the substance will be observed as being volatile.
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Two forces,
serg [7]

First compute the resultant force F:

\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N

\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N

\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N

Then use Newton's second law to determine the acceleration vector \mathbf a for the particle:

\mathbf F=m\mathbf a

(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a

\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}

Let \mathbf x(t) and \mathbf v(t) denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so \mathbf v(0)=0. Then the particle's velocity vector at <em>t</em> = 10.4 s is

\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du

\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}

\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}

If you don't know calculus, then just use the formula,

v_f=v_i+at

So, for instance, the velocity vector at <em>t</em> = 10.4 s has <em>x</em>-component

v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}

(b) Compute the angle \theta for \mathbf v(10.4\,\mathrm s):

\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ

so that the particle is moving at an angle of about 313º counterclockwise from the positive <em>x</em> axis.

(c) We can find the velocity at any time <em>t</em> by generalizing the integral in part (a):

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

Then using the fundamental theorem of calculus again, we have

\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du

where \mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m is the particle's initial position. So we get

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}

\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m

So over the first 10.4 s, the particle is displaced by the vector

\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

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1 Cl2 + 3 F2---> 2 ClxFy

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