Answer:
Position =
behind the mirror
Nature = Virtual and Erect
Size =
: Diminished
Explanation:
Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.
Object distance = u = -20 cm
Focal length = f = Radius of curvature/2 = 30/2 = 15 cm
We have to use mirror formula to find image distance.

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.
Magnification = 

Height of the object = 5 cm
Height of the image = 
Since the height of the image is positive and less than the size of object,it is erect and diminished.
Answer:
Explanation:
If a number of less than 1, then the number has a decimal point like
0.085, 0.008 e.t.c.
The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.
But if there a zero between the none zero e.g. 0.0087056
Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6
But if the zero is not in between the none zero digit, then the zero is insignificant
E.g 0.05800
The last two zero is insignificant, the significant number is 5 and 8
So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.
Answer:
Explanation:
We shall apply law of refraction which is as follows
sin i / sinr = μ , where i is angle of incidence , r is angle of refraction and μ is refractive index
here i = θa = 22.5°
r = θb
μ = 1.77
sin22.5 / sinθb = 1.77
.3826 / sinθb = 1.77
sinθb = .216
θb = 12.5 °.
Energy consumed in doing the work = 300 Joules
Force applied on the object = 75 N
Let the distance moved by the object be d.
Work done by the force is determined by the force applied and the displacement happened in the direction of the force applied.
Work done = Force x displacement
300 = 75 x d

d = 4 m
Hence, the maximum distance moved by the object = 4 meters