Answer:
m1=914.9kg
m2=604.9kg
m3=864.75kg
Explanation
I think we are suppose to find the mass of the crate.
The effective force that moves the body in positive x direction is 3615N
ΣFx = Σma
Then Fx=3615N
Then the masses be m1, m2 and m3
Then,
ΣF = Σ(ma)
3615=(m1+m2+m3)a
Given that a=1.516
The masses are
m1+m2+m3=, 2384.56. Equation 1
Between mass 1 and mass 2 is, F12=1387.
The effective force that pull mass 1 is 1387.
F12=m1 ×a
Therefore,
m1=F12/a
m1=1387/1.516
m1=914.9kg.
The effective force that pulls crate 1 and crate 2 is F23
F23=(m1+m2)a
Therefore
2304=(m1+m2)a
Therefore, since a=1.516
m1+m2=2304/1.516
m1+m2=1519.8kg
Since m1=914.9kg
So, m2=1519.8-m1
m2=1519.8-914.9
m2=604.9kg
Also from equation 1
m1+m2+m3=2384.56
Since m1=914.9kg and m2=604.9kg
Then, m3=2384.56-604.9-914.9
m3=864.75kg
D I think is the correct answer
If the cylinder is slightly
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
-- impurities in the water
-- air pressure is higher than standard