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In-s [12.5K]
3 years ago
5

A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. How mu

ch kinetic energy does it gain? The mass of a proton is 1.67262 × 10−27 kg, the mass of a neutron 1.67493 × 10−27 kg and the charge on an electron −1.60218 × 10−19 C. Answer in units of J\
Physics
1 answer:
Nonamiya [84]3 years ago
6 0

Complete Question:

A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. b) How much kinetic energy does it gain? The mass of a proton is 1.67262 × 10−27 kg, the mass of a neutron 1.67493 × 10−27 kg and the charge on an electron −1.60218 × 10−19 C. Answer in units of J\

b) what is the speed?

Answer:

a) the kinetic energy gained =  6.42 * 10⁻¹⁶ J

b) the speed of the particle, v = 619328.3 m/s

Explanation:

q = 1.602 *10⁻¹⁹C

V = 4.01 kV  = 4.01 * 10³ V

Work done by the deuteron = qV

Work done by the deuteron = 1.602 * 10⁻¹⁹ *  4.01 *10³

Work done = 6.42 * 10⁻¹⁶ J

Kinetic Energy gained = work done

Kinetic Energy gained by the deuteron =  6.42 * 10⁻¹⁶ J

B) The formula for Kinetic Energy is given by:

KE = 1/2 Mv²

Let the mass of the proton be m₁ = 1.67262 × 10⁻²⁷kg

Let the mass of the neutron be m₂ =   1.67493 × 10−27 kg

M = m₁ + m₂

KE = 1/2 ( m₁ + m₂)v²

Let v = speed of the deuteron

From part (a)

KE = 6.42 * 10⁻¹⁶ J

 1/2 ( m₁ + m₂)v²=  6.42 * 10⁻¹⁶

0.5 * (1.67262 + 1.6749) *10⁻²⁷ * v² =  6.42 * 10⁻¹⁶

v = 619328.3 m/s

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Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

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now, form the third equation of motion;

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we substitute

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v_{x}² = 100 + 30

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for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

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d_{y} = 56.5 m

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we substitute

|d| = √( (20)² + (56.5)² )

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2 years ago
Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For
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Answer:

1.27\times 10^{12}\Omega/m

Explanation:

We are given that

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Using 1\mu m=10^{-6} m

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Resistance =R=2\times 10^5\Omega cm^2

Internal specific resistance=r=100 ohm cm=100\times \frac{1}{100}\Omega-m=1\Omega m

Using 1 m=100 cm

Internal resistance per unit length=\frac{r}{A}=\frac{1}{\pi r^2}=\frac{1}{3.14\times (0.5\times 10^{-6})^2}=1.27\times 10^{12}\Omega/m

Using \pi=3.14

Internal resistance per unit length=1.27\times 10^{12}\Omega/m

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a = -1 m/s^2

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