Answer:
19.68 × 10⁻³ m
Explanation:
Given;
Original Length, L₁ = 41.0 m
Temperature Change, ΔT = 40.0°C
Thermal Linear expansion of steel is given to be, ∝
= 12 × 10⁻⁶ /°C
Generally, Linear expansivity is expressed as;
∝ = ΔL / L₁ΔT
Where
∝ is the Linear expansivity
ΔL is the change in length, L₂ - L₁
L₂ is the final length
L₁ is the original length
ΔT is the change in temperature θ₂ - θ₁ (Final Temperature - Initial Temperature)
From equation of linear expansivity
ΔL = ∝L₁ΔT
ΔL = 12 × 10⁻⁶ /°C × 41.0 m × 40.0 °C
ΔL = 19.68 × 10⁻³ m
ΔL = 19.68 mm
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = 
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:
Plug in the given values and solve for 
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
Answer:
Frequency=10 Hz
Period=0.1 s
Explanation:
Frequency of a wave is the number of complete cycles per unit time. Usually, frequency is cycles per second, with unit as Heartz
Given cycles of the wave as 200 and time as 20 seconds
Frequency=200÷20=10 cycles per second
Therefore, frequency is 10 Hz
Period is usually the reviprocal of frequency hence reciprocal of 10 Hz will be 1/10=0.1 s
My teacher said 36m when I asked her
