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kherson [118]
3 years ago
6

Rounded to the nearest whole number what is the atomic mass of platinum?

Physics
1 answer:
NISA [10]3 years ago
6 0

195 is the rounded atomic atomic mass of platinum

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The world record for the 100.0 meter dash is 9.580 seconds. Find the speed of the runner in miles per hour.
grigory [225]

Answer:

233.1 miles per hours

Explanation:

Speed: This is defined as the ratio of distance to time. The S.I unit of speed is m/s. speed is a vector quantity because it can only be represented by magnitude only. Mathematically, speed can be expressed as,

S = d/t ....................... Equation 1

Where S = speed of the runner, d = distance covered, t = time.

Given: d = 100 meter , t = 9.580 seconds

Conversion:

If, 1 meter =  0.00062 miles

Then, 100 meters = (0.00062×100) miles = 0.62 miles.

Also

If, 3600 s = 1 h

Then, 9.580 s = (1×9.580)/3600 = 0.00266 hours.

Substitute into equation 1

S = 0.62/0.00266

S = 233.1 miles per hours.

Hence the runner speed is 233.1 miles per hours

7 0
3 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
When you changed from low to high power, how did the change affect the working distance of the lens?
Basile [38]

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about  working distance

brainly.com/question/13551539

#SPJ4

4 0
1 year ago
A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces up
Serjik [45]

Answer:

0.906

Explanation:

Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second

\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW

So the efficiency of the water turbine is the ratio of output power over input power:

\frac{234}{258.307} = 0.906

3 0
4 years ago
A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced s
Hatshy [7]

Answer:

Explanation:

fringe separation or fringe width is directly proportional to wavelength .

Δ₁ / Δ₂ = wave length of first laser / wavelength of second laser

Δ₁ is fringe separation in first case and Δ₂ is fringe separation in second case.

putting the given values

632.8 / wavelength of second laser = 4.5 / 4.64

wavelength of second laser  = 632.8 x (4.64 / 4.5)

= 652.48 nm

652 nm

3 0
3 years ago
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