Answer:
L= 1.468 m
Explanation:
The moment of inertia of the rod about its center is (1/12)m_RL^2
The moment of inertia of each of the two bodies about the described axes is m_B(L/2)2
Hence, the moment of inertia of three body system is (1/12)m_RL^2+ 2×m_B(L/2)^2 which is given to be equal to I_T
=> L^2[(1/12)m_R+m_B/2] = I_T
=> L2 = IT/(mR/12+ mB/2)
=> L = sqrt( 12I_T/(m_R+6m_B))
now putting the value of m_R= 3.47 kg
m_B= 0.263 kg
I_T = 0.97 kg/m^2
L= 1.468 m is the length of the rod be so that the moment of inertia of the three-body system
Answer:
A = 0.22 m
Explanation:
The spring with the block and the pebble forms an oscillatory system, this system is described by the expression
x = A cos (wt + φ)
w = √ (k / m).
The data they give us is the amplitude of the movement (A = 10 cm), the oscillation mass is equal to the block mass plus the mass of the pebble
m = m + M
m = 0.031 + 0.108
m = 0. 139 kg
To find the spring constant let's use Hooke's law
F = k X
The force is the weight of the pebble and the additional elongation is x = 4.9 cm, let's calculate
k = F / x
k = mg / x
k = 0.031 9.8 / 0.049
k = 6.2 N / m
Let's look for angular velocity
w = √ (6.2 / 0.139)
w = 6,670 rad / s
Let's write the oscillation equation with this data
x = 0.10 cos (6,670 t)
For the pebble to remain in contact with the block, the acceleration of the spring system plus block with pebble must be less than the acceleration of gravity in the descending oscillation
Let's look for system acceleration
a = d²x / dt²
dx / dt = - A w sin (wt + Ф)
d²x / dt² = - A w² cos (wt+Ф)
To find the maximum value cos (wt) = ±1
g = A w²
A = g / w²
A = 9.8 / 6.67²
A = 0.22 m
When the amplitude of the oscillation exceeds this value the pebble is delayed with respect to the block
All 15 patients will most likely have the same test results reguarding the ages would make the results vary
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Answer:
Wt= 0.73kJ, Wg=0.21kJ Wn= 0J
(B) Wf= 0.096kJ
Explanation:
See attachment below.
Answer:
Momentum, p = 5 kg-m/s
Explanation:
The magnitude of the momentum of an object is the product of its mass m and speed v i.e.
p = m v
Mass, m = 3 kg
Velocity, v = 1.5 m/s
So, momentum of this object is given by :
p = 4.5 kg-m/s
or
p = 5 kg-m/s
So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.