Answer:
charges of the beads is 1.173 ×
C
Explanation:
given data
mass = 3.8589 g = 0.003859 kg
spring length = 5 cm = 0.05 m
extend spring x = 1.5747 cm = 0.15747 m
spring's extension = 0.0116 m
to find out
charges of the beads
solution
we know that force is
force = mass × g
force = 0.003859 × 9.8
force = 0.03782 N
so we know force for mass
force = -kx
so k = force / x
put here force and x value
k = -0.03782 / 0.1575
k = -0.24 N/m
and
force for spring's extension
force = -kx
force = -0.24 ( 0.0116) = 0.002784 N
so here
total length L = 0.05 + 0.0116 = 0.0616
so charges of the beads = force × L² / ke
charges of the beads = 0.002784 × (0.0616)² / (9 ×
)
so charges of the beads = 1.173 × C
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Answer:
100 ÷ 9.58 = 10.44 (approximate answer)
Answer:
796.18 Hz
Explanation:
Applying,
Maximum velocity = Amplitude×Angular velocity
Therefore,
V' = A(2πf)............... Equation 1
Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie
make f the subject of the equation
f = V'/2πA................ Equation 2
From the question,
Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,
Constant: 3.14.
Substitute these values into equation 2
f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)
f = 796.18 Hz
