Answer:
Δx = 1.2 m
Explanation:
The CHANGE of spring length) (Δx) can be found using PS = ½kΔx²
Δx = √(2PS/k) = √(2(450)/650) = 1.17669... ≈ 1.2 m
The actual length of the spring is unknown as it varies with material type, construction method, extension or compression, and other variables we have no clue about.
Given Information:
Wavelength = λ = 39.1 cm = 0.391 m
speed of sound = v = 344 m/s
linear density = μ = 0.660 g/m = 0.00066 kg/m
tension = T = 160 N
Required Information:
Length of the vibrating string = L = ?
Answer:
Length of the vibrating string = 0.28 m
Explanation:
The frequency of beautiful note is
f = v/λ
f = 344/0.391
f = 879.79 Hz
As we know, the speed of the wave is
v = √T/μ
v = √160/0.00066
v = 492.36 m/s
The wavelength of the string is
λ = v/f
λ = 492.36/879.79
λ = 0.5596 m
and finally the length of the vibrating string is
λ = 2L
L = λ/2
L = 0.5596/2
L = 0.28 m
Therefore, the vibrating section of the violin string is 0.28 m long.
we know
T=2.pie sqr root(l/g)
l= (sqrof 7/22)*9.8
L= .99 or approx 1m
Volume of tea V = 2.0L = 2000 mL density d = 1.01 g/ mL mass of tea m = V * d = 2000mL * 1.01g/mL = 2020 gWhen we assume that the tea was initially at 72, the final temperature of the tea in F is 91.
The answer in this question is B. 91
Answer:
Option D: 1.5in in front of the target
Explanation:
The object distance is 
.
Because the surface is flat, the radius of curvature is infinity .
The incident index is 
and the transmitted index is 
.
The single interface equation is 
Substituting the quantities given in the problem,
The image distance is then 
Therefore, the coin falls 
in front of the target
