Answer:
1.25 m
0.5 s
Explanation:
Given:
v₀ = 5 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (5 m/s)² + 2 (-10 m/s²) Δy
Δy = 1.25 m
Find: t
v = at + v₀
(0 m/s) = (-10 m/s²) t + (5 m/s)
t = 0.5 s
Answer:
magnification is - 159
objective distance is 3.85 cm
Explanation:
Given data
focal length f1 = 1.40 cm
focal length f2 = 2.20 cm
separated d = 19.6 cm
to find out
angular magnification and How far from the objective
solution
we know magnification formula that is
magnification = ( - L / f1 ) (D/f2)
here D = 25 cm put all value
magnification = ( - 19.6 / 1.40 ) (25/2.20)
magnification = - 159
and
now we apply lens formula
i/f = 1/q + 1/p
p = f2 = 2.20
so
q = f2 p / p -f2
q = 1.4(2.20) / ( 2.2 - 1.4 )
q = 3.85 cm
so objective distance is 3.85 cm
All of the above. they would all decrease due to a lack of energy from using it in the beginning.
For a floating object, the buoyancy force is equal to the gravity force on the object. Hence, the buoyancy force doesn't change with a denser fluid. Instead the displaced volumedecreases to cancel out the effect of the increased fluid density.
