A man weighs 1190 n on earth. what would he weigh on jupiter, where the free-fall acceleration is 25.9 m/s 2 ? the acceleration
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Answer:
D = 2.828 m
Explanation:
given,
distance of source of light = 1.24 m below surface of pool
refractive index of the water = n₁ = 1.33
refractive index of air = n₂ = 1
refraction angle be = 90°
let C be the critical angle
Radius = d tan C
d is the depth of the source
Using Snell's law
n₁ sin C = n₂ sin R
1.33 x sin C = 1 x sin 90°
C = 48.75°
hence,
R = 1.24 x tan 48.75°
R = 1.414 m
Diameter = 2 x R
D = 2 x 1.414
D = 2.828 m
The answer would be . Since we are looking for the spring constant you would need to use the formula
. Then you'd substitute, for PEs and x.
Then solve. k=500n/m
. Then you'd substitute, for PEs and x.
Then solve. k=500n/m