This would be known as a sedimentary rock. SEDIMENTary rock
Answer:
Let:
r= radius= 3cm= 0.03m
∅1= initial temperature= 30°C
∅2= final temperature= 100°C
∆∅= change in temperature= ∅2 - ∅1= 70°C
L.E.= linear expansivity= 0.000019k^-1
¥= 3L.E.= volume expansivity= 3(0.000019)= 0.000057k^-1
The initial volume of the metal sphere= 4/3(πr³)
= 4/3(π×0.03³)
= 4(π×0.03²×0.01)
= 4(3.142×0.0003×0.01)
= 4(0.000009426)
= 0.000037704
= 3.7704 × 10^-5m³
¥= V2 - V1/V1∆∅
V2= ¥V1∆∅ + V1
= (1.5044 × 10^-7) + (3.7704×10^-5)
= 0.00003785
= 3.785 × 10^-5m³
Therefore, the initial and final volume of the metre sphere are 3.7704 × 10^-5 and 3.785 × 10^-5 respectively.
Answer:
Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42
The image is a virtual and upright image.
Explanation:
The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.
The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.
Given f = -33.0cm, v = -19.0cm
using thr mirror formula to get the object distance u, we have;
To calculate the image height, we will use the magnification formula
M =
M =
Given Hi = 7.0cm
v = 19.0cm
u = 44.8cm
HI = 7*44.8/19
HI = 16.5cm
The object height is 16.5cm
Magnification = v/u = 19.0/44.8 = 0.42
SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image
Answer:
aquarium vs the value of both obstacles