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3 years ago

HELP PLEASE!!! a= ? v= r=

Physics

1 answer:

ratelena [41]3 years ago

4 0

Answer:

$a=\frac{v^2}{r}$

$v=\sqrt{a*r}$

$r=\frac{v^2}{a}$

Explanation:

This is the formula for centripetal acceleration in terms of the tangential velocity (v) and the radius of the circular motion (r). The expression for the acceleration is already given, so simply type it as shown:

$a=\frac{v^2}{r}$

For the velocity (v) multiply by "r" both sides and then use the square root to solve for v:

$a*r=v^2\\v=\sqrt{a*r}$

For the radius multiply both sides by r and then divide both sides by the acceleration (a) in order to isolate r completely:

$a*r=v^2\\r=\frac{v^2}{a}$

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What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?

VikaD [51]

(a) $1.33\cdot 10^{-11} m$

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

$e \Delta V = \frac{hc}{\lambda}$

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

$e=1.6\cdot 10^{-19}C$ is the electron's charge

$\Delta V = 93.3 kV = 93300 V$ is the potential difference

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.00\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the emitted photon

Solving the formula for $\lambda$ , we find:

$\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m$

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

$E=\frac{hc}{\lambda}$

where

h is Planck constant

c is the speed of light

$\lambda=1.33\cdot 10^{-11} m$ is the wavelength of the photon, calculated previously

Substituting,

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J$

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

$E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV$

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

$e \Delta V = \frac{hc}{\lambda}$

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

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3 years ago

A 75kg Tibetan is trekking along flat, but icy ledge with his 450kg yak when he slips over the edge. Luckily, he is holding the

Tcecarenko [31]

Answer:

minimal coefficient of static friction: $\mu_s=0.1667$

Explanation:

Once the Tibetan is hanging from the strap, he is exerting a horizontal force on the yak equal to his weight which is the product of his mass times the acceleration of gravity (g) as written below:

$w = m\,*\,g= 75\,kg\,*\,g$

The other forces acting on the yak are (see attached diagram):

* the force of gravity on the yak (identified in blue color in the image as $F_g$ ,

* the normal force (indicated in green in the image and identified by the letter "n") of the ledge on the yak as reaction to the yak's weight

* the force of static friction between the yak's hooves and the ledge (pictured in red in the image and identified with $f_s$ )

Since the normal force and the force of gravity on the yak cancel each other (balance - the yak is not moving vertically), the only forces we need to analyse are the force of the Tibetan's weight via the strap, and the force of static friction which should at least be equal in magnitude so the Tibetan doesn't fall. We assume these two forces are acting horizontally (one to the right: the Tibetan's weight, and one to the left: the static friction).

As we said, we want them to be at least equal so thy are in balance.

We recall that the force of static friction is the product of the normal force (n) times the coefficient of static friction ( $\mu_s$ ), such that: $f_s=\mu_s\,*\,n$

In our case these are the forces at play:

$F_g= M\,*\,g=450\, kg \,*\,g\\n=F_g=450\,kg\,*\,g\\f_s=\mu_s\,*\,n=\mu_s\,*450\,kg\,*\,g\\w=m\,*\,g=75\,kg\,*\,g$

So we need to find what is the minimum coefficient of static friction that precludes the Tibetan from falling. We therefore proceed to make an equality between the force of static friction on the yak and the weight of the Tibetan:

$f_s=w\\\mu_s\,*450\,kg\,*g=75\,kg\,*\,g$

and proceed to solve for the coefficient of friction by dividing both sides by "g" (which by the way cancels out), and by the yak's mass:

$\mu_s\,*450\,kg\,*g=75\,kg\,*\,g\\\mu_s=\frac{75}{450} \\\mu_s=0.1667$

where we have rounded to four decimal places the periodic number that the quotient generates. Notice that as expected, the coefficient of friction has no units (they all cancelled out in the division).

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Answer:

Explanation:

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Nimfa-mama [501]

Answer:

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3 years ago

Show that on a hypothetical planet having half the diameter of the Earth but twice its density, the acceleration of free fall is

sergiy2304 [10]

Answer:

Steps include:

Find the ratio between the mass of this hypothetical planet and the mass of planet earth.
Calculate the ratio between the acceleration of free fall (gravitational field strength) near the surface of that hypothetical planet and near the surface of the earth.

Assumption: both planet earth and this hypothetical planet are spheres of uniform density.

Explanation:

<h3>Mass of that hypothetical planet</h3>

Let $M_0$ , $r_0$ , and $\rho_0$ denote the mass, radius, and density of planet earth.

The radius and density of this hypothetical planet would be $(1/2)\, r_0$ and $2\, \rho_0$ , respectively.

The volume of a sphere is $\displaystyle V = \frac{4}{3}\, \pi\, r^3$ .

Therefore:

The volume of planet earth would be $\displaystyle V_0 = \frac{4}{3}\, \pi\, {r_0}^{3}$ .

The volume of this hypothetical planet would be: $\displaystyle V = \frac{4}{3}\, \pi\, \left(\frac{r_0}{2}\right)^3 = \frac{1}{6}\, \pi\, {r_0}^3$ .

The mass of an object is the product of its volume and density. Hence, the mass of the earth could also be represented as:

$\displaystyle M_0 = \rho_0\, V_0 = \frac{4}{3}\, \pi\, {r_0}^{3}\, \rho_0$ .

In comparison, the mass of this hypothetical planet would be:

$\begin{aligned}M &= \rho\, V\\ &= (2\, \rho_0) \cdot \left(\frac{1}{6}\, \pi\, {r_0}^3\right) \\ &= \frac{1}{3}\, \pi\, {r_0}^{3}\, \rho_0 \end{aligned}$ .

Compare these two expressions. Notice that $\displaystyle M = \frac{1}{4}\, M_0$ . In other words, the mass of the hypothetical planet is one-fourth the mass of planet earth.

<h3>Gravitational field strength near the surface of that hypothetical planet</h3>

The acceleration of free fall near the surface of a planet is equal to the gravitational field strength at that very position.

Consider a sphere of uniform density. Let the mass and radius of that sphere be $M$ and $r$ , respectively. Let $G$ denote the constant of universal gravitation. Right next to the surface of that sphere, the strength of the gravitational field of that sphere would be:

$\displaystyle g = \frac{G \cdot M}{r^2}$ .

(That's the same as if all the mass of that sphere were concentrated at a point at the center of that sphere.)

Assume that both planet earth and this hypothetical planet are spheres of uniform density.

Using this equation, the gravitational field strength near the surface the earth would be:

$\displaystyle g_0 &= \frac{G \cdot M_0}{{r_0}^2}$ .

On the other hand, the gravitational field strength near the surface of that hypothetical planet would be:

$\begin{aligned}g &= \frac{G \cdot (M_0/4)}{{(r_0/2)}^2} = \frac{G \cdot M_0}{{r_0}^2}\end{aligned}$ .

Notice that these two expressions are equal. Therefore, the gravitational field strength (and hence the acceleration of free fall) would be the same near the surface of the earth and near the surface of that hypothetical planet.

As a side note, both $g$ and $g_0$ could be expressed in terms of $\rho_0$ and $r_0$ alongside the constants $\pi$ and $G$ :

$\begin{aligned}g &= \frac{G \cdot M}{r^2}\\ &= \frac{G \cdot (1/3)\, \pi\, {r_0}^3 \, \rho_0}{(r_0/2)^2} = \frac{4}{3}\,\pi\, G\, r_0\, \rho_0\end{aligned}$ .

$\begin{aligned}g_0 &= \frac{G \cdot M_0}{{r_0}^2}\\ &= \frac{G \cdot (4/3)\, \pi\, {r_0}^3 \, \rho_0}{{r_0}^2} = \frac{4}{3}\,\pi\, G\, r_0\, \rho_0\end{aligned}$ .

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3 years ago