<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
Answer:
The decibel of the remaining pigs is 51.5 dB.
Explanation:
Decibel (dB) is a unit of measure of the intensity of a given sound.
Number of pigs = 199, noise level = 74.3 dB.
Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:
I 
N
I = kN
where k is the constant of proportionality.
⇒ k = 
= 
k = 0.3734
When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.
Thus, the becibel level (I) of the remaining pigs can be determined by:
I = kN
= 0.3734 × 138
= 51.53 dB
The becibel level (I) of the remaining pigs is 51.53 dB.
The actual position of the object is <span>at a great distance, effectively infinite. The other options given in the question are not at all correct. The correct option among all the options that are given in the question is the last option or option "D". I hope that this answer has actually come to your great help.</span>
Answer:
The correct option is;
A. Circular
Explanation:
Some of the light that impinges on the surface are reflected and the rest are transmitted to a different medium
At the surface of the next medium also, some of the light are transmitted while the others are reflected and refracted through the first medium
The speed of light (and hence the wavelength and color) refracted through the thin film is changed as the distance the refracted light travels through the thin film is increased as we move away from the point directly in the front view to some distance as the reflected light path from those distance to the eye is increased due to their inclination giving them a different wavelength which are all equal at a radial distance from the eye hence forming a circular fringes.
Answer:
The frequency would not change because the frequency of a simple pendulum does not depend on mass.
Explanation:
