The normal force of the force given is calculated through the equation,
Fn = F(sin θ)
where Fn is the normal force, F is the force, and θ is the angle.
Fn = (25 N)(sin 60°) = 21.65 N
The x-component of the force applied is,
Fx = (25 N)(cos 60°) = 12.5 N
The value of the coefficient of static friction is calculated through the equation,
F = μFn
μ = Fx / Fn = 12.5 N / 21.65 N = 0.577
The answer is 50 kg plz mark as brainliest
Because the acceleration of gravity is the acceleration of gravity.
It doesn't matter what the mass of a falling object is, and it doesn't
matter whether a falling object is solid or liquid. ALL falling objects
fall with the same acceleration, reach the same speed, and hit the
ground at the same time.
If there was no air in the way, then a feather, a school bus, and a
battleship would accelerate at the same rate, fall together and hit
the ground at the same time.
When you drop a cup full of water that has holes in it, the cup and
the water fall with the same acceleration, reach the same speed,
and hit the floor at the same time. Then, THAT's the time to go
and get the mop.
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
