Joint replacement are often made of the element titanium. Which type of matter is titanium

A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its

dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

   <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

   (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

      (150 million) x (1 / 3,262,000) x (70 km/sec) =

   <em>3,219 km/sec </em>in the direction away from us (rounded)

4 0

3 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

6 0

3 years ago

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

$F_s = \mu_s*N$

Where $F_s$ is the static friction force, $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, $N = 110\text{ N}$ , to make the puppy moving we need to use a force of 80 N, therefore, $F_s = 80 \text{ N}$ , so we can solve for the coefficient as shown below:

$80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\$

The coefficient of static friction between the puppy and the floor is 0.7273.

5 0

3 years ago

Fg =G m1m2/r2 solver for G

Blababa [14]

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Explanation:

Solving for G simply implies that we make G the subject of the formula:

Given equation:

F $_{g}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$

To make G the subject of this expression follow these steps:

Multiply both sides of the equation by $r^{2}$

F $_{g}$ x $r^{2}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$ x $r^{2}$

This gives:

F $_{g}$ $r^{2}$ = G $m_{1} m_{2}$

Multiply both sides by $\frac{1}{m_{1} m_{2} }$

F $_{g}$ $r^{2}$ x $\frac{1}{m_{1} m_{2} }$ = $\frac{1}{m_{1} m_{2} }$ x G $m_{1} m_{2}$

Therefore:

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Learn more:

Solving formula brainly.com/question/2998489

#learnwithBrainly

6 0

3 years ago

While studying physics at the library late one night, you noticethe image of the desk lamp reflected from the varnished tabletop

Natalija [7]

<h2>Answer: 1.73</h2>

Explanation:

The situation described here is known as polarization by reflection. This was discovered by Scottish physicist David Brewster and then formulated the law that bears his name:

<em>"When a beam of light hits the surface that separates two non-conducting media characterized by different electromagnetic characteristics (electrical permittivity and magnetic permeability), part of it is reflected back to the source medium, and part is transmitted to the second medium."</em>

This polarization happens when the light incides at a specific angle, called the Brewster angle ( $\theta_{B}$ ), which is given by the following formula (<u>taking into account that generally the magnetic permeabilities of the two media involved do not vary):</u>

$tan\theta_{B}=\frac{n_{2}}{n_{1}}$ (1)

Where $n_{2}$ is the index of refraction of the second medium (the varnish in this case) and $n_{1}=1$ is the index of refraction of the first medium (the air).

Now, if we are told the angle between the incident and reflected rays is $120\°$ , this means the incident angle is the half ( $60\°$ ), which is the Brewster angle in this case.

So, $\theta_{B}=60\°$ (2)

Rewriting (1) with this incident ray angle:

$tan(60\°)=\frac{n_{2}}{1}$ (3)

$n_{2}=tan(60\°)$

Finally we obtain the index ofrefraction of the varnish:

$n_{2}=1.732$

5 0

3 years ago