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KATRIN_1 [288]
3 years ago
13

How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0

Answer:

-2.00876\times 10^{-18}\ J

Explanation:

v = Speed of electron = 2.1\times 10^6\ m/s (generally the order of magnitude is 6)

m = Mass of electron = 9.11\times 10^{-31}\ kg

Work done would be done by

W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J

The work required to stop the electron is -2.00876\times 10^{-18}\ J

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ryzh [129]

Answer:

a. The horizontal component of acceleration a₁ = 0.68 m/s²

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Explanation:

The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ =  17.0i - 1.00j m/s

a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²

The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. The direction of the acceleration relative to the unit vector i,

tanθ = a₂/a₁ = -0.11/0.68 = -0.1618

θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis

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