Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
The problem is basically asking us to find a way to find the sound intensity I, in terms dependent on the sound level and the reference intensity 
.For this purpose we can start from the unit used in the scale logarithmic decibel, that is
Where
I = Acoustic intensity on the linear scale
Hearing threshold
Using the logarithmic properties of the exponents the above expression can be described as:
that is the expression or technique to find the intensity of sound.
Did you ever figure it out, bc now I need it lol.
Answer:
Explanation:
Work done = ∫Fdx
= ∫(cx-3.00x²) dx
[ c x² / 2 - 3 x³ / 3 ]₀²
= change in kinetic energy
= 11-20
= - 9 J
[ c x² / 2 - x³ ]₀² = - 9
c x 2² / 2 - 2³ = -9
2c - 8 = -9
2c = -1
c = - 1/2
