The energy of moving electrical charges is Electrical energy
Hope its the answer you are finding and hope it helps....
Answer:
(a) A = 1 mm
(b) 
(c) ![a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D606.4%20m%2Fs%5E%7B2%7D%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EDistance%20moved%20back%20and%20forth%20%3D%202%20mm%20%3C%2Fp%3E%3Cp%3EFrequency%2C%20f%20%3D%20124%20Hz%3C%2Fp%3E%3Cp%3ESo%2C%20amplitude%20is%20the%20half%20of%20the%20distance%20traveled%20back%20and%20forth.%20%3C%2Fp%3E%3Cp%3E%28a%29%20So%2C%3Cstrong%3E%20amplitude%2C%20A%20%3D%201%20mm%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%28b%29%20Angular%20frequency%2C%20%CF%89%20%3D%202%20%CF%80%20f%20%3D%202%20x%203.14%20x%20124%20%3D%20778.72%20rad%2Fs%20%3C%2Fp%3E%3Cp%3EThe%20formula%20for%20the%20maximum%20speed%20is%20given%20by%20%3C%2Fp%3E%3Cp%3E%5Btex%5DV_%7Bmax%7D%3D%5Comega%20%5Ctimes%20A)


(c) The formula for the maximum acceleration is given by


[tex]a_{max}=606.4 m/s^{2}/tex]
A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be 372.77 m/s
<h3>What is a sound wave?</h3>
It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.
For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula
V= √F/μ
where v is the wave velocity of the wave travel on the string
F is the tension in the string of piano
μ is the mass per unit length of the string
As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.
The μ is the mass per unit length of the string would be
μ = 4.80/(1.150×1000)
μ = 0.0041739 kg/m
By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity
V= √F/μ
V=√(580/0.0041739)
V = 372.77 m/s
Thus, the speed of transverse waves on the wire comes out to be 372.77 m/s
Learn more about sound waves from here
brainly.com/question/11797560
#SPJ1
Answer:
1.85 J/K
Explanation:
The computation of total change in entropy is shown below:-
Change in Entropy = Sum Q ÷ T
= 

= -3.12 + 4.97
= 1.85 J/K
Therefore for computing the total change in entropy we simply applied the above formula.
As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.
Answer:
block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s
Explanation:
We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.
Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)
Before the crash
p₀ = m v₀ + 0
After the crash
= (m + M) v
p₀ = 
m v₀ = (m + M) v (1)
Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring
Initial
Em₀ = K = ½ m v2
Final
E
= Ke = ½ k x2
Emo = E
½ m v² = ½ k x²
v² = k/m x²
Let's look for the spring constant (k), with Hook's law
F = -k x
k = -F / x
k = - 0.75 / -0.25
k = 3 N / m
Let's calculate the speed
v = √(k/m) x
v = √ (3/8.00) 0.15
v = 0.09186 = 9.18 10⁻² m/s
This is the spped of the block plus bullet rsystem right after the crash
We substitute calculate in equation (1)
m v₀ = (m + M) v
v₀ = v (m + M) / m
v₀ = 0.09186 (0.008 + 0.992) /0.008
v₀ = 11.5 m / s