Answer:
d) -4.0
Explanation:
The magnification of a lens is given by

where
M is the magnification
q is the distance of the image from the lens
p is the distance of the object from the lens
In this problem, we have
p = 50 cm is the distance of the object from the lens
q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct
Also, q is positive since the image is real
So, the magnification is

Answer:
according to this question best answer is C
The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 12.0V (assuming circuit resistance is negligable) and it has a capacitance of 18.0μf or 18.0x10^-6f, therefore charge equals (18.0x10^-6)x12=2.16x10^-4C (Coulombs).
Answer:
a) 0.022%
b) 10014.32 lb
Explanation:
a) Percentage uncertainty would be

Percent uncertainty is 0.022%
b) For 1 kg uncertainty mass in kg would be

Mass in pounds would be

Mass in pound-mass is 10014.32 lb
Answer:1). Distance of far point x=0.9m
Therefore, since the image is virtual
-f=-x = -0.9m
Power of the concave lenses = 1/f = 1/-0.9
= -1.11D
2 ) near point is 21cm = 0.21m
Power = 4-1/near point
= 4/0.21
= 14.2D.