Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
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Answer:
200 nm is the thinnest film that produces a strong reflection for green light with a wavelength of 500 nm
Explanation:
If two reflected waves interfere constructively ,strong reflection is produced. Two reflected waves will experience a phase change
For constructive interference
for thinnest film m=1
refractive index should be taken for film n=1.25
thickness of the thinnest film is
momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)
Answer:
- 256 lbs
Explanation:
The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero
; Then:
internal axial load at point D = Δ P
= -(P₂ - P₁)
= - ( 888 lbs - 632 lbs)
= - 256 lbs
