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andre [41]
3 years ago
9

A proton accelerates from rest in a uniform electric field of 664 N/C. At some later time, its speed is 1.46 106 m/s. (a) Find t

he magnitude of the acceleration of the proton. m/s2 (b) How long does it take the proton to reach this speed? µs (c) How far has it moved in that interval? m (d) What is its kinetic energy at the later time? J
Physics
1 answer:
Katarina [22]3 years ago
6 0

Answer:

Explanation:

In an electric field E force on charge q

F = Eq , acceleration a = Eq / m

a = 664 x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷

= 636.16 x 10⁸ m /s²

b )

initial velocity u = 0

final velocity v = 1.46 x 10⁶ m/s

v = u + at

1.46 x 10⁶ = 0 + 636.16 x 10⁸ x t

t = 2.29 x 10⁻⁵ s

c )

s = ut + 1/2 a t²

= 0 + .5 x 636.16 x 10⁸ x (  2.29 x 10⁻⁵ )²

= 1668 x 10⁻²

= 16.68 m

d )

Kinetic energy = 1/2 m v²

= .5 x 1.67 x 10⁻²⁷ x ( 1.46 x 10⁶ )²

= 1.78 x 10⁻¹⁵ J .

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Calculate A, E, μ, cv and S for 1 mole of Kr at 298 K and 1 atm (assuming ideal behavior)
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Answer:

internal energy E 3716.35 j

cv = 12.47 J/K

S = 12.47 J/K

A =  0.29 J

\mu =3716.35 J/mole

Explanation:

given data:

Kr  atomic number = 36

degree of freedom = 3

1) internal energy E = \frac{f}{2} n R T

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2) cv = \frac{E}{T}

         = \frac{3716.35}{298} = 12.47 J/K

3) S = cv =\frac{E}{T} = 12.47 J/K

4) A, Halmholtz free energy = E -TS = 37146.35 - 12.47*298 = 0.29 J

5)chemcial potential \mu = \frac{energy}{mole} = 3716.35 J/mole

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What does newton's first law describes​
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Earlier, we stated Newton's first law as “A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless acted on by a net external force.” It can also be stated as “Every body remains in its state of uniform motion in a straight line unless it is compelled to change that state by forces
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The equation itself comes from splitting up the forces acting on the block into components pointing parallel or perpendicular to the incline. The only forces acting on the block in the perpendicular direction are the normal force and the perpendicular component of the block's weight.

Solve for <em>n</em> :

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