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wariber [46]
3 years ago
13

Cations are formed by gaining protons losing protons gaining electrons losing electrons

Physics
1 answer:
krek1111 [17]3 years ago
7 0
Cations are formed by losing electrons

Cations are basically positive ions


It's not lose or gain protons because atoms can't lose or gain protons. The protons don't move.
It's not gain electrons because electrons are negative. More electrons=more negative

If you can't remember cations are positive
CAT-ions are PAWS-itively charged
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While traveling along a highway a driver slows from 31 m/s to 15 m/s in 8 seconds. What is the automobile’s acceleration? (Remem
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Answer:

The automobile's acceleration in that time interval is -2 m/s^2

Explanation:

The acceleration is defined as the rate of change of the velocity.

The average acceleration in a given lapse of time is calculated as:

A = (final velocity - initial velocity)/time.

In this case, we have:

initial velocity = 31 m/s

final velocity = 15 m/s

time = 8 seconds.

Then the average acceleration is:

A = (15m/s - 31m/s)/8s = -2 m/s^2

8 0
2 years ago
Review. From a large distance away, a particle of mass 2.00 g and charge 15.0σC is fired at 21.0 i^ m/s straight toward a second
MissTica

(a)

Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.

Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.

m1v1 + m1v2 = (m1+m2)v

The second particle's starting velocity is zero, so:

m1v1  = (m1+m2)v

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v = 6i m/s

(b)

Since the two particle system is also energy-isolated, we may use the energy-conservation principle.

dK + dU = 0

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Substituting the values,

1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf

The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:

1/2m1v1^2i  = 1/2m1v1^2 + 1/2m2v2^2 +ke q1q2/rf

After solving we get,

rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2

Substituting the values we get,

rf = 3.64m

(c)

v1f = (m1-m2 / m1 + m2) v1i

v1f  = -9i m/s

(d)

v2f =  (2m1/ m1 +m2) v1i

After substituting the values,

v2f = 12i m/ s

Question :

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}

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b

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C. A cold front that came through does not affect the climate of an area. 
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