Forces are never isolated because they come in... *
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Answer:48.2 Joules
Explanation:
Given
two masses of 0.2 kg and 0.4 kg collide with each other
after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east
Velocity of 0.2 kg mass is
Velocity of 0.4 kg mass
Thus total Kinetic energy
Kinetic energy=48.2 J
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
Answer:
90.78 rev/min
Explanation:
In first place, we have to do the force balance to determine the minimum angular speed required to avoid slipping. The forces acting here are friction and the force due to circular movement, that is centripetal force. Then, we have:
μmg=mRω^2
ω=
Then, replacing the given values in the expression we have the following result:
ω=1.51 rev/s*60s=90.78 rev/min