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Lilit [14]
3 years ago
5

An electron moving parallel to a uniform electric field increases its speed from 2.0 ×× 1077 m/sm/s to 4.0 ×× 1077 m/sm/s over a

distance of 1.3 cmcm.
What is the electric field strength?
Physics
1 answer:
algol [13]3 years ago
4 0

Answer:

262 kN/C

Explanation:

If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:

v² = v0² + 2aΔS

Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:

(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013

16x10¹⁴ = 4x10¹⁴ + 0.026a

0.026a = 12x10¹⁴

a = 4.61x10¹⁶ m/s²

The electric force due to the electric field (E) is:

F = Eq

Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:

F = m*a

Where m is the mass, so:

E*q = m*a

The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:

E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶

E = 261,866.42 N/C

E = 262 kN/C

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Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (
Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}

9.28 + 8.3 x = 0

x = -1.12 m

5 0
3 years ago
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