Answer:
The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)
Explanation:
Fundamental frequency = wave velocity/2L
where;
L is the length of the stretched rubber
Wave velocity = 
Frequency (F₁) = 
To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.
Given:
L₂ =2L₁ = 2L
T₂ = 2T₁ = 2T
(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)
F₂ = ![\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B2T%7D%7B0.5%28%5Cfrac%7BM%7D%7BL%7D%29%7D%7D%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B4%28%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%29%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%5B%5Cfrac%7B%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%7D%7B2%2AL%7D%5D%20%3D%20F_1)
Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).
The region marked X in the diagram shows that the objects have the same charge.
<h3>
What is conduction?</h3>
The term conduction has to do with the manner of charging in which charge is passed from one object to another. Induction involves charging objects without the objects touching each other.
The region marked X in the diagram shows that the objects have the same charge.
Learn ore about conduction:brainly.com/question/15306642
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Answer: The correct option is (c.).
Explanation:
Mass of the cart A= 1.5 kg
Velocity of Cart A = 1.4 m/s towards right
Mass of the cart B = 1.0 kg
Velocity of Cart B = 1.4 m/s towards left
Momentum (P)= Mass × Velocity

(Negative sign means velocity of the cart is in opposite direction of that of the cart A)
Total Momentum =
Hence, the correct option is (c.).
Answer:

Explanation:
Hello,
In this case, since the acceleration in terms of position is defined as its second derivative:

The purpose here is derive x(t) twice as follows:

Thus, the acceleration turns out 4.8 meters per squared seconds.
Best regards.