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m_a_m_a [10]
3 years ago
8

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

sion tube. It is constantly accelerated while travelling 0.01 m, and leaves the gun at 5.4 x 106 m/s. What was the acceleration of the electron
Physics
1 answer:
Crank3 years ago
7 0

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

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2 years ago
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Grace [21]

Answer:

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Explanation:

From the question we are told that

    The mass of the girl is  m_g  = 35.4 \ kg

     The mass of the cart is  m_c  = 15.23 \ kg

      The speed of the cart and  kid(girl) is  v = 4.25 \ m/s

     The final velocity of  the girl is v_g  = 3.06 \  m/s

Let assume that velocity eastward is  positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

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substituting values    

      p__{T2}} = (35.4 * 3.06 ) +(  15.23 * v_c )

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       p__{T1}} =p__{T2}}

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=>      v_c = 7.02 \  m/s

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