Answer
a) charge of the sodium ion is,
q = n e
q = 2.68 x 10¹⁶ x 1.6 x 10⁻¹⁹
q = 4.288 x 10⁻³ C
charge of the chlorine ion is,
q' = n e
q' = 3.92 x 10¹⁶ x 1.6 x 10⁻¹⁹
q' = 6.272 x 10⁻³ C
the current



b) positive ion moves toward negative electrode hence direction of will be in the direction toward negative electrode.
Mainly because he was Johnny Carson's advisor and consultant
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In the question it is already given that the football player is 80 kg.
Then the mass of the football player = 80 kg
Velocity at which the football player is running = 8 m/s
<span>Kinetic Energy = 0.5 • mass • square of velocity
Now we have to put the known data in this equation to find the actual velocity of the footballer.
</span> <span></span>So
Kinetic Energy of the footballer = 0.5 * 80 * (8 * 8)
= 0.5 * 80 * 64
= 2560
So the Kinetic energy of the footballer is 2560 joules
The resultant force on the box is C) 10 Newtons to the left.
Explanation:
The figure of the problem is attached.
We see that there are two forces acting on the box along the horizontal direction:
- A force of 10 N acting to the right
- A force of 20 N acting to the left
By taking "to the right" as positive direction, we can rewrite the two forces as follows:

Where the second force is negative since it acts to the left.
Since the two forces are in line, we can find their resultant simply by calculating their algebraic sum. By doing so, we find:

And the negative sign tells us that the direction is to the left.
Therefore, the correct answer is
C) 10 Newtons to the left.
Learn more about forces here:
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Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:
(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>
where <em>v</em> is the velocity of the system. Solve for <em>v</em> :
<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)
<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s