We have that the value for y is
From the Question we are told that
Relaxed length of 35 cm (0.35 m)
Spring stiffness is 12 N/m.
Glue a 66 gram block (0.066 kg)
Total length is l_t=16 cm
Block during a 0.24-second interval
Generally the equation for the Force of spring of block is mathematically given as
Generally
Generally,Net Force
Generally,Velocity of block
Where
Therefore
Generally the equation for the Velocity of block is mathematically given as
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Answer:
Explanation:
The amplitude of the oscillation under SHM will be .5 m and the equation of
SHM can be written as follows
x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.
x = .5 cosωt
given , when t = .2 s , x = .35 m
.35 = .5 cos ωt
ωt = .79
ω = .79 / .20
= 3.95 rad /s
period of oscillation
T = 2π / ω
= 2 x 3.14 / 3.95
= 1.6 s
b )
ω = 
ω² = k / m
k = ω² x m
= 3.95² x .6
= 9.36 N/s
c )
v = ω
At t = .2 , x = .35
v = 3.95 
= 3.95 x .357
= 1.41 m/ s
d )
Acceleration at x
a = ω² x
= 3.95 x .35
= 1.3825 m s⁻²
Given,
The momentum of the object A before the collision, p₁ =80 Ns
The momentum of the object B before the collision, p₂=-30 Ns
Given that the objects stick together after the collision.
From the law of conservation of momentum, the total momentum of a system should always remain the same. Thus the total momentum of the objects before the collision must be equal to the total momentum of the objects after the collision.
Thus,
Where p is the total momentum of the system at any instant of time.
On substituting the known values,
Therefore the total momentum of the system is 50 Ns
Thus the momentum of the object AB after the collision is 50 Ns
<span>F = m*a = 7000kg * 9.8N/kg = 68,600 N.
68,600 N is your answer
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Answer:
That is an example of Radiational transfer
Explanation:
Radiation is a direct transfer of exposer
