Emf = d (phi-B) / dt
<span>B dA/dt, where dA/dt is the area swept out by the wire per unit time. </span>
<span>0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so </span>
<span>L = (0.88 J/C) / [ (0.075 N s/C m)(4.2 m/s) ] = about 3 meters</span>
Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).
I=1/(d*d)
Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.
L1/L2=(D2*D2)/(D1*D1)
L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>
Answer:
50 N/m
Explanation:
Elastic energy = kinetic energy
EE = KE
½ kx² = ½ mv²
½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²
k = 50 N/m
