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Fiesta28 [93]
3 years ago
7

____is the distance traveled during a specific unit of time

Physics
1 answer:
qaws [65]3 years ago
6 0
I think speed is the answer
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Draw a neat labelled diagram for a particle moving in a circular path with a constant speed. In your diagram show the direction
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Explanation:

Particle moving in a circular path with a constant speed.

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A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o
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Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

F = qv \times B = qvBsin(\theta)     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

K = \frac{1}{2}mv^{2}

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

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I'd answer that but I can't text graphs and tables...
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Explanation:

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Answer:

D.

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A solar system is a collection of planets, their moons, and other objects in orbit around a central star.

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