Answer:
length
<em>Your</em><em> </em><em>well</em><em> </em><em>wisher</em><em> </em><em>:-)</em>
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).
Answer:
E.two angles are vertical angles if, and only if they are not adjacent angles
Answer:
I think C
Explanation:
Since the bus is moving away from John.
{C - V}.
Answer:
The net friction force is 8.01 N
Explanation:
Net friction force = mass of hockey puck × acceleration
From the equations of motion
v^2 = u^2 + 2as
v = 40 m/s
u = 0 m/s (puck was initially at rest)
s = 30 m
40^2 = 0^2 + 2×a×30
60a = 1600
a = 1600/60 = 26.7 m/s^2
The acceleration of the puck is 26.7 m/s^2
Net friction force = 0.3 × 26.7 = 8.01 N