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IrinaVladis [17]
3 years ago
11

A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J

at point A. The particle subsequently passes through point B which has an electric potential of -2.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.
Physics
1 answer:
Vesna [10]3 years ago
3 0

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

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aalyn [17]
<h2>Answer: It becomes an Ion </h2>

When an atom has gained or lost electrons (negative charge), it becomes an ion.

In this sense:

<h2>Ions are atoms that have <u>gained or lost</u> electrons in their electronic cortex. </h2><h2> </h2>

If a neutral atom <u>loses electrons</u>, it remains with an excess of positive charge and transforms into a positive ion or <u>cation</u>, whereas if a neutral atom <u>gains electrons</u>, it acquires an excess of negative charge and transforms into a negative ion or <u>anion</u>.

It is then how ions form bonds with other atoms differently depending on the number of electrons they have.

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A sailboat picks up a gust of wind and accelerates to a speed of 6m/s in 16 seconds.If the initial velocity 1 m/s,what is the ac
noname [10]

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3 years ago
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
dybincka [34]

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau
Tamiku [17]

Answer:

7.0\cdot 10^{-13}C

Explanation:

The magnitude of the electrostatic force between two charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

r=0.070 cm =7\cdot 10^{-4} m is the distance between the charges

q_1=q_2=q since the charges are identical

F=9.0\cdot 10^{-9}N is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C

8 0
2 years ago
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