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3 years ago

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A particle is at the position (x,y,z) = (1.0, 2.0, 3.0) m. it is traveling with a vector velocity (-5.3, -4.8, -3.9) m/s. its ma

ss is 7.6 kg. what is its vector angular momentum about the origin?

1 answer:

Alisiya [41]3 years ago

5 0

<1, 2, 3> x (7.6 <-5.3, -4.8, -3.9>) x stands for cross product

What does the above notation mean?

Multiply 7.6 to each component of the velocity vector to obtain the linear momentum vector. Find the cross product of the position vector and the linear momentum vector. That gives you the angular momentum vector.

7.6 * -5.3 = -40.28
7.6 * -4.8 = -36.48
7.6 * -3.9 = -29.64

Cross Multiplication
-40.28 1 = -80.56 - (-36.48)
-36.48 2

-40.28 1 = -120.84 - (-29.64)
-29.64 3

-36.48 2 = -109.44 - (-59.28)
-29.64 3

answer is (-91.2) - (50.16) + 117.04 = -24.32

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a 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

Lesechka [4]

The final velocity is 3.47 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the two players before and after the collision must be conserved:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 91.5 kg$ is the mass of the first player

$u_1 = 3.73 m/s$ is the initial velocity of the first player (we take east as positive direction)

$m_2 = 63.5 kg$ is the mass of the second player

$u_2 = 3.09 m/s$ is the initial velocity of the second player

$v$ is their final combined velocity after the collision

Re-arranging the equation and substituting the values, we find:

$v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(3.73)+(63.5)(3.09)}{91.5+63.5}=3.47 m/s$

So, their velocity afterwards is 3.47 m/s east.

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3 years ago

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).

3 0

3 years ago

A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separa

Alenkasestr [34]

Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

p = q s

s = p / q

let's calculate

s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

s = 0.625 10⁻²¹ m

s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

6 0

3 years ago