The watch hand covers an angular displacement of 2π radians in 60 seconds.
ω = 2π/60
ω = 0.1 rad/s
v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
Answer:
128.21 m
Explanation:
The following data were obtained from the question:
Initial temperature (θ₁) = 4 °C
Final temperature (θ₂) = 43 °C
Change in length (ΔL) = 8.5 cm
Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)
Original length (L₁) =.?
The original length can be obtained as follow:
α = ΔL / L₁(θ₂ – θ₁)
17×10¯⁶ = 8.5 / L₁(43 – 4)
17×10¯⁶ = 8.5 / L₁(39)
17×10¯⁶ = 8.5 / 39L₁
Cross multiply
17×10¯⁶ × 39L₁ = 8.5
6.63×10¯⁴ L₁ = 8.5
Divide both side by 6.63×10¯⁴
L₁ = 8.5 / 6.63×10¯⁴
L₁ = 12820.51 cm
Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
12820.51 cm = 12820.51 cm × 1 m / 100 cm
12820.51 cm = 128.21 m
Thus, the original length of the wire is 128.21 m
Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone, 
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
Answer:
W = 145.8 [N]
Explanation:
To solve this problem we must remember that weight is defined as the product of mass by gravity, in this case lunar gravity.
W = m*g
where:
m = mass = 90 [kg]
g = gravity acceleration = 1.62 [kg/m²]
W = 90*1.62
W = 145.8 [N]
