Answer:
C
Explanation:
a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer
In this problem we have the electric field intensity E:
E = 6.5 ×
newtons/coulomb
We have the magnitude of the load:
q = 6.4 ×
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 ×
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 ×
)(6.5 ×
)(1.2 ×
)
PE = 5.0 x
joules
None of the options shown is correct.
Because even though the moon is smaller, therefore a weaker gravitational pull, the moon is much closer to the earth than the sun, thus having a greater gravitational pull
Answer:
7.50 cm
Explanation:
The formula
1/v + 1/u = 1/f
Is used.
where.
u is the object distance.
v is the image distance.
f is the focal length of the lens.
1/v + 1/15 = 1/5
1/v = 1/5 - 1/15
1/v = (3-1)/15
1/v = 2/15
2v = 15
V = 15/2
V = 7.5 cm
For focal length, f in lens is always taken as negative for concave and positive for convex. ... And for image distance, V in lens it is taken as positive in Convex lens since image is formed on +X side. It is taken as negative in Concave lens since image is formed in -X side of the Cartesian.
Answer:
<em>Because </em><em>of </em><em>the </em><em>given </em><em>stranded</em><em> </em><em>wires </em><em>is </em><em>that </em><em>it's </em><em>thinner </em><em>there </em><em>are </em><em>even </em><em>more </em><em>air </em><em>gaps </em><em>and </em><em>a </em><em>greater </em><em>surface</em><em> </em><em>area </em><em>in </em><em>the </em><em>individual</em><em> </em><em>stranded</em><em> wires</em><em> </em><em>then </em><em>therefore </em><em>it </em><em>carries </em><em>less </em><em>current </em><em>than </em><em>similar </em><em>solid </em><em>wires </em><em>can </em><em>with</em><em> </em><em>each</em><em> </em><em>type </em><em>of </em><em>wire </em><em>,</em><em> insulations</em><em> </em><em>technologies </em><em>can </em><em>greatly</em><em> </em><em>assist </em><em> </em><em>in </em><em>reducing</em><em> </em><em>power </em><em>dissipation</em><em>.</em>