Answer:
low freezing point. high vapour pressure.
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>
Answer:
The current in the primary is 0.026 A
Explanation:
Using the formula
I1 = (V1/V2)*I2
we have
I1 = (6.4/120)*0.500
I1 = 0.026 A
<u>Answer</u>:
Effort is the unaltered force. Load is the altered force.
Answer:
Option A
Solution:
As per the question:
The distance covered by the woman in the North direction, d = 3000 m
Time taken to travel in North direction, t = 25.0 min = 1500 s
Velocity of woman in the south direction, v = 2.00 m/s
Time taken in the south direction, t' = 60.0 min = 3600 s
Now,
The distance covered in the south direction, d' = vt' = 
Now, the total displacement is given by:
D = d' - d = 7200 - 3000 = 4200 m in South
(a) Average velocity of the woman in the whole journey is given by:

≈ 0.824 m/s South
Answer:

Explanation:
Hi!
The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.
The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

Then the electric field at the point of interest is estimated as:
