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myrzilka [38]
3 years ago
6

At t=0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a con

stant acceleration of (2.0i - 4.0j) m/s2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle?
a. 10 m/s
b. 16 m/s
c. 12 m/s
d. 14 m/s
e. 26 m/s
Physics
1 answer:
fiasKO [112]3 years ago
5 0

Answer:

e.26m/s

Explanation:

Vf=Vi+at      (1)

Vf=9j+(2i-4j)t

X= X₀+at

now, in the i direction

15=O+2t or t=7.5 when x position is 15

Lets put that into the (1) equation, solve for Vf.

 Vf=9j+(2i-4j)7.5

Vf= 15i - 21j

Speed= \sqrt{xcomponent^2 + ycomponent^2}

Vf= 25.8 m/s

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Calculate the de Broglie wavelength for (a) an electron with a kinetic energy of 100eV, (b) a proton with a kinetic energy of 10
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The equation given by Broglie relates the momentum of a particle with its wavelength.

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In addition, kinetic energy is related to the amount of movement

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      p = mv

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      h = 6,626 10-34 Js

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       λ1 = 6.6 10⁻³⁴ / √(2 9.1 10⁻³¹ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 5.39 10⁻²⁴

       λ1 = 1.22 10⁻¹⁰ m

b) Proton

       λ2 = 6.6 10-34 / √(2 1.67 10⁻²⁷ 1.6 10⁻¹⁷) = 6.6 10⁻³⁴ / 2.3 10⁻²²

       λ2 = 2.87 10⁻¹² m

c) Bohr's first orbit

       En = 13.606 / n2 [eV]

       n = 1

       E1 = 13.606 eV

       E1 = 13,606 ev (1.6 10⁻¹⁹ / 1eV) = 21.77 10⁻¹⁹ J

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       λ3 = 0.774 10⁻¹¹ m = 7.74 10⁻¹² m

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