Answer:
The answer is A. on edgen.
Explanation:
A. adding in the boxes an arrow that points from Qh to Qc
Answer 1) The electric field at distance r from the thread is radial and has magnitude
E = λ / (2 π ε° r)
The electric field from the point charge usually is observed to follow coulomb's law:
E = Q / (4 π ε° 
)
Now, adding the two field vectors:
= {2.5 / (22 π ε° X 0.07 ) ; 0}
Answer 2) 
= {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))
Adding these two vectors will give the length which is magnitude of the combined field.
The y-component / x-component gives the tangent of the angle with the positive x-axes.
Please refer the graph and the attachment for better understanding.
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em>
</em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em>
</em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em>
</em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em>
</em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em>
</em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em>
</em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em>
</em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em>
</em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
<span>The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.</span>
