The energy of the system, tension in the string, and speed of each cube are respectively; U = 0.72 J, T = 14 N and (V₂, V₄) = (-22, 14)
<h3>Calculation of energy and tension in a String</h3>
We are given;
Mass of cube 1; m₁ = 2 g
Mass of cube 2; m₂ = 4 g
Distance between cubes; d = 5 cm = 0.05 m
Charges of cubes; Q = q = +2. 0 μc = 2.0 × 10⁻⁶ C
A) Formula for the energy of the system is;
U = kQq/d
U = 8.99 × 10⁹ × (2.0 × 10⁻⁶ C)²/0.05 m
U = 0.03596/0.05
U = 0.72 J
B) Formula for the Tension is;
T = U/d
T = 0.72/0.05
T ≈ 14 N
C) Momentum is conserved, and as such the initial momentum is zero. Thus;
0 = (0.0020 × V₂) + (0.0040 × V₄)
⇒ V₂ = -2V₄
Energy is also conserved and so;
(½ × 0.0020 × (-2V₄)²) + (½ × 0.004 × (V₄)²) = 0.72 J
-0.0040V₄² + 0.002V₄² = 0.72 J
0.0060V₄² = 0.72 J
V₄² = 0.72/0.0060
V₄² = 120
V₄ = √120
V₄ ≈ 11 m/s
Recall that; V₂ = -2V₄
Thus;
V₂ = -2(11) m/s
V₂ = -22 m/s
Thus;
(V₂, V₄) = (-22, 14)
Read more about Conservation of Momentum at; brainly.com/question/7538238
The tempo of the excerpt can best be described as Adagio.
<h3>What is an adagio tempo?</h3>
This refers to when the tempo of a musical piece is done slowly or rather in slow time.
This enables for powerful music as it feels like every part of the musical piece by Maria Szymanowska is heard individually.
Rest of the question talks about an excerpt from a music piece by Maria Szymanowska.
The music piece is a Mozart Piano Concerto.
Options for this question are:
Find out more on adagio tempo at brainly.com/question/7146515
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Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.
Explanation:
Let a Force pushes the hockey puck of mass m.
Then acceleration, a= \frac{F}{m}a=mF
From the equation of motion,
\begin{gathered}\➪ v=u+at\\ v=0+\frac{F}{m}\Delta t\end{gathered}⇒v=u+atv=0+mFΔt ......(1)
In the second case, when mass is 2m, then acceleration,
a'=\frac{F}{2m}a′=2mF
and t' is the time taken.
The final speed is v,
\begin{gathered}\➪ v=0+ a't'\\ \➪ \frac {F}{m}\Delta t=\frac{F}{2m}t'\\ \➪ t'= 2\Delta t\end{gathered}⇒v=0+a′t′⇒mFΔt=2mFt′⇒t′=2Δt using equation (1)
Hence, it would take two times the previous amount of time to push the pluck of double mass.
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