Number of times the diaphragm move back and forth is 5.59×10^4
<u>Explanation:</u>
Given data,
ω=4.6 s
we have the formula
f=ω/2π
The number of times the diaphragm moves back and forth in 4.6 s is
Number of times= ft
Number of times= ft
=(ω/2π) t
=(7.54×10^4 rad/sec)(4.6 s)/2π
Number of time=5.59×10^4
Number of times the diaphragm move back and forth is 5.59×10^4
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Answer:
Answered
Explanation:
A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.
W= FS cosθ
θ= 90 ⇒cos90 = 0 ⇒W= 0
B) work done by tension
W= Tcosθ×S= 5cos30×2.30= 10J
C) Work done by friction force
W= f×s=1×2.30= 2.30 J
D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.
E) The net work done= Work done by tension in the rope - frictional work
=10-2.30= 7.7 J
V^2 = 700J/0.5*35kg
V = square root of 40
v = 6.324 m/s
Answer: B.move from the negatively charged body to the positively charged body.
Explanation: If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body. This happens because they have different charges and therefore electrons are more attracted to the positive or its opposite charge.
