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kirill [66]
3 years ago
15

An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N

/C. What is the kinetic energy of the electron at the end of the motion
Physics
1 answer:
algol133 years ago
4 0

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

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Explanation:

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We know that torque is given by \tau =F\times d=45.5\times 0.0245=1.11475N-m

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At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of
photoshop1234 [79]

The recoil velocity of cannon is (4) 5.0 m/s

Explanation:

We can find the recoil velocity from the law of conservation of momentum.

The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.

Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec

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So recoil velocity of cannon v2 is given by,

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Find the wavelength of the balmer series spectral line corresponding to n = 15.
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Answer:

371.2 mm

Explanation:

The Balmer series of spectral lines is obtained from the formula

1/λ = R(1/2² -1/n²) where λ = wavelength, R = Rydberg's constant = 1.097 × 10⁷ m⁻¹

when n = 15

1/λ = 1.097 × 10⁷ m⁻¹(1/2² -1/15²)

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    = 1.097 × 10⁷ m⁻¹ 0.245556

    = 2.693 10⁶ m⁻¹

So,

λ  = 1/2.693 10⁶ m⁻¹

    = 0.3712 10⁻⁶ m

    = 371.2 mm

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