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Vsevolod [243]
3 years ago
10

What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength

in angstroms (x 10-10 m) to 2 sig figs.
Physics
1 answer:
ad-work [718]3 years ago
6 0

Answer:

1.2826 x 10^-13 m

Explanation:

\lambda  = \frac{h}{\sqrt{2 m K}}

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

\lambda  = \frac{6.63 \times  10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}

λ = 1.2826 x 10^-13 m

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Under state law, what is the blood-alcohol limit for legally operating a motor vehicle?
Ymorist [56]

Answer:

HOPE IT HELPS....

Explanation:

CORRECT ANSWER IS 0.08

THANK YOU,

PLZ MARK ME AS BRAINLIST

5 0
3 years ago
Analyzing a blood sample usually involves which type of separation method?
schepotkina [342]

Answer:(D.)

Explanation:

4 0
3 years ago
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef
navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0
3 years ago
We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in
leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

B=\mu_0\, \frac{N}{L} I

Then, if we assign the subindex "1" to the quantities that define the magnetic field (B_1) inside solenoid 1, we have:

B_1=\mu_0\, \frac{N_1}{L_1} I_1

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then L_2=L_1/2

2) twice as many turns as solenoid 1. Then N_2=2\,N_1

3) three times the current of solenoid 1. Then I_2=3\,I_1

we obtain:

B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1

5 0
4 years ago
Which is NOT one of the four principals of Dalton’s atomic theory?
Gre4nikov [31]

Answer:

Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>

Explanation:

Hope it helps..

Just correct me if I'm wrong, okay?

But ur welcome!!

(;ŏ﹏ŏ)(◕ᴗ◕✿)

8 0
3 years ago
Read 2 more answers
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