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Lelu [443]
3 years ago
9

Nellie pulls on a 10kg wagon with a constant horizontal force of 30N. If there are no other horizontal gorces what is the wagons

acceleration?
Physics
1 answer:
kicyunya [14]3 years ago
3 0

Answer:

a=3m/s^2

Explanation:

If we have a net force F acting on a body of mass m it will experiment an acceleration a. Newton's 2nd Law gives us the relation between these quantities: F=ma.

In our case, we want to calculate the acceleration, so we write:

a=\frac{F}{m}

With the values we have we get:

a=\frac{30N}{10Kg}=3m/s^2

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An electron in a TV is accelerated toward the screen. The potential between the back of the tube and the screen is 22,000 V (whi
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Answer:

The kinetic energy lost by the electron ΔK.E is 3.5244 × 10⁻¹⁵ J

Explanation:

Here we have;

Change in electric potential, ΔV, of the screen = 25000 V

The energy gained by the electron can be derived from the relation of change in electrical potential energy, ΔEPE, as follows;

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Therefore, since charge, q = -1.602 × 10⁻¹⁹ C we have;

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Also, since the electron is accelerated to the screen by the electrical potential energy, we have;

From the principle of conservation of energy in a given system;

The change in potential energy, ΔEPE of the electrons = Change in kinetic energy, ΔK.E, of the electrons3.5244 × 10⁻¹⁵ J

\Delta KE = \frac{1}{2} \cdot m \cdot v_f^2 - \frac{1}{2} \cdot m \cdot v_i^2

Where:

m = Mass of the electron = 9.11 × 10⁻³¹ kg

v_i = Initial velocity of the electrons = 0 m/s (electrons at rest)

v_f = Final velocity of the electrons

\therefore \Delta KE = \frac{1}{2} \cdot m \times (v_f^2 -  \cdot v_i^2)

\therefore \Delta KE = \frac{1}{2} \cdot m \times (v_f^2 -  0) =  \frac{1}{2} \cdot m \cdot v_f^2

Hence;

The kinetic energy lost by the electron ΔK.E = ΔEPE = 3.5244 × 10⁻¹⁵ J.

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