How big is the international space station?
1 answer:
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Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
Answer:
(B) 1.41 x 10^{10} N/m^{2}
Explanation:
diameter of wire (d) = 0.6 mm = 0.0006 m
radius of wire (r) = 0.0006/2 = 0.0003 m
tension (force F) = 20 N
percentage strain = 0.5 %
what is the Young's modulus?
Young's modulus =
where
- stress = force/area
stress =
stress =
stress =70735530.3 N/m^{2}
- strain = percentage strain ÷ 100
strain = 0.5 ÷ 100 = 0.005
therefore Young's modulus = =
= 1.41 x 10^{10} N/m^{2}