Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
Answer:
<em>Gravity</em><em>.</em><em> </em><em>The</em><em> </em><em>weight-force</em><em> </em><em>or</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>an</em><em> </em><em>object</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>because</em><em> </em><em>of</em><em> </em><em>Gravity</em><em>,</em><em> </em><em>which</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>object</em><em> </em><em>attracting</em><em> </em><em>it</em><em> </em><em>towards</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>earth</em><em>.</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x</em>
Known :
l = 7 cm
w = 4 cm
Asked :
h = ...?
Answer :
V = B triangle × h (long)
35 = ½ × 4 × h × 7
35 = ½ × 28 × h
35 = 14 h
h = 35 ÷ 14
h = 2,5 cm
Sorry if I am wrong, I only study
No, aluminum has a density near 2.7 g/cm^3
<span>7.8 g/cm^3 is near the density of iron (or in the case of a fork, steel).
this is it
</span>
Answer:
2442.5 Nm
Explanation:
Tension, T = 8.57 x 10^2 N
length of rope, l = 8.17 m
y = 0.524 m
h = 2.99 m
According to diagram
Sin θ = (2.99 - 0.524) / 8.17
Sin θ = 0.3018
θ = 17.6°
So, torque about the base of the tree is
Torque = T x Cos θ x 2.99
Torque = 8.57 x 100 x Cos 17.6° x 2.99
Torque = 2442.5 Nm
thus, the torque is 2442.5 Nm.