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3 years ago

How many protons are there in any te atom

Physics

1 answer:

Andreyy893 years ago

4 0

Answer:

Explanation:

Tellurium is the element of the periodic table with atomic number 52.

The atomic number of a chemical element represents the number of atoms contained in the nucleus of the atom: therefore, this means that an atom of tellurium contains exactly 52 protons in its nucleus.

Tellurium is classified as a metalloid, having properties in between metals and non-metals, and it appears with a silver color.

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What is an independent variable?

Natali5045456 [20]

Answer:

The answer is A

Explanation:

Independent variables don't have to depend on other factors of the experiment because they're independent

4 0

3 years ago

A baseball with a mass of 0.15 kilograms collides with a bat at a speed of 40 meters/second. The duration of the collision is 8.

Vedmedyk [2.9K]

Answer: 1687.5 N

Explanation:

From the second law of motion given by Newton, Force is the rate change of momentum.

$F = \frac{dp}{dt}=\frac {m dv}{dt} = \frac{m (v_f-v_i)}{dt}$

Mass of the baseball, m = 0.15 kg

Initial velocity, $v_i=-40 m/s$ (negative because direction of initial velocity is opposite to the final velocity)

Final velocity, $v_f=50 m/s$

The duration of collision, $dt= 8.0 \times 10^{-3} s$

Force, $F = \frac{0.15 kg (50-(-40) m/s)}{8.0 \times 10^{-3} s}=1687.5 N$

Hence, the value of force is 1687.5 N.

8 0

3 years ago

Find the magnitude of this vector: 174 m N 188.4 m HELP FAST

Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

$a^2+b^2=c^2$

Plug values in

$88.4^2+174^2=c^2$

Simplify

$7814.56+30276=c^2$

Add the two values

$38090.56=c^2$

Take the square root of both sides

$195.168\approx195.168$

8 0

2 years ago

A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table

MA_775_DIABLO [31]

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3

consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m

C 2.2 m

3 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

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