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Mnenie [13.5K]
3 years ago
10

An electron moving along the x axis has an initial speed of 1 × 106 m/s at the origin. Its speed is reduced to 5 × 105 m/s at th

e point xP , 3 cm away from the origin. Calculate the magnitude of the potential difference between this point and the origin. The mass of the electron is 9.109 × 10−31 kg and the elemental charge is 1.602 × 10−19 C. Answer in units of V.
Physics
1 answer:
mina [271]3 years ago
5 0

Answer:

-2.13 V

Explanation:

Given parameters are:

v_i = 1 * 10^6 m/s

v_f = 5 * 10^5 m/s

q = -e = 1.602 * 10^{-19} C

m = 9.109 * 10^{-31} kg

By conservation of energy principle, we know that the potential difference between two points is equal to the change in the kinetic energy between these points.

-\Delta U = \Delta KE\\\\-e\Delta V = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2

Hence,

1.602*10^{-19}\Delta V = \frac{1}{2}*9.11*10^{-31}((5*10^5)^2 - (1*10^6)^2)

⇒ \Delta V = -2.13 V

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The time elapsed is 1.5 seconds, so the acceleration is

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2. We know, from the previous point, that the lift travelled 20m from the first floor. Since it returns to the first floor after the ascent, it must travel again those same 20m, just in reverse (descending instead of ascending). So, the total distance travelled is 20+20=40 meters.

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3 years ago
What is the strength of the electric field of a point charge of magnitude +4.8 × 10^-19 C at a distance of 4.0 × 10^-3 m? A. 3.6
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3 years ago
A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s aft
OLEGan [10]

Answer:

0.90291 m/s

0.45055 m/s

Explanation:

m_1 = Mass of canon = 2090 kg

m_2 = Mass of ball = 16.7 kg

v_1 = Velocity of canon

v_2 = Velocity of ball = 113 m/s

In this system the momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{16.7\times 113}{2090}\\\Rightarrow v_1=0.90291\ m/s

The velocity of the cannon is 0.90291 m/s

Applying energy conservation

\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2=\dfrac{1}{2}m_2v^2\\\Rightarrow m_1v_1^2+m_2v_2^2=m_2v^2\\\Rightarrow v_2=\sqrt{\dfrac{m_1v_1^2+m_2v_2^2}{m_2}}\\\Rightarrow v_2=\sqrt{\dfrac{2090\times 0.90291^2+16.7\times 113^2}{16.7}}\\\Rightarrow v_2=113.45055\ m/s

The ball would travel 113.45055-113 = 0.45055 m/s faster

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Are the two expressions equvalent 7(8x+5) and 48x + 35
Evgesh-ka [11]
Yes they are equivalent because 7x5=35 and 8x x 5=48x
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