1 kilometre is equal to 1000m
and 4.1 minutes is equal to 246 seconds
thus 1000/246 = 4.065 m/s
and the direction is towards the west
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>
Transmission electron microscope
Answer:
At light intensity I = 3, is P a maximum
Explanation:
Given:
now differentiating the above equation with respect to Intensity 'I' we get
or
or
or
Now for the maxima 
thus,
or
or
or
or
I = 3
thus, <u>for the value of intensity I = 3, the P is maximum</u>
at I = 3
or
or
All these resistors are in series so we can take the sum of them by:
Rtotal = R1 + R2 + R3......
So...
Rtotal = 2 + 3 + 4 + 6
Rtotal = 15
So now the total resistance in the circuit is 15 ohms and the potential difference applied to the circuit is 45 volts
Now we can use:
V = IR
Isolate for I
V/R = I
45/15 = I
I = 3 amps (A)
