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3 years ago

Plz do it all i will give brainlest and thanks to best answer.

Physics

2 answers:

Natasha_Volkova [10]3 years ago

7 0

It’s a because metal is a good heat conductor and I don’t see any other questions on there

maxonik [38]3 years ago

4 0

Its a, metal is a good conductor of heat so yea
Hope this helps :)

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By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what

Vinil7 [7]

Explanation:

1. Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}$

$V=\dfrac{h^2}{2q_pm_p\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}$

V = 45.83 volts

2. Mass of the electron, $m_p=9.1\times 10^{-31}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}$

$V=\dfrac{h^2}{2q_em_e\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}$

$V=6.92\times 10^{34}\ V$

V = 84109.27 volt

Hence, this is the required solution.

7 0

3 years ago

a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of wh

pychu [463]

Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!

7 0

3 years ago

If we have less power, we most likely have

katrin [286]

<span>the same amount of work being done over a longer period of time.</span>

3 0

3 years ago

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A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$ , 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°

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3 years ago

Which law of motion accounts for the following statement?

strojnjashka [21]

I know that its not the second law. I'm almost positive its the first one. Please let me know if I'm wrong. This sentence makes no sense when you put it with the third law. So, the first law is my guess...

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3 years ago

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