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:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²
Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m
(a) Angular frequency w= 2πf
w= 2π × 760 = 4775.2Hz
(b) Maximum speed v is given as the product of angular frequency and maximum displacement
V=wx
V= 4775.2 × 0.00085
V= 4.06m/s
(c) The maximum acceleration a
= w²x
= (4775.2)² × (0.00085)
a= 19382.15m/s².
Equilibrium force is the force that will keep the small
mass in place, hence no movement must be made. So we know that 32 N of force is
acted towards the positive direction so +32 N. Which is counteracted by 26 N
force so:
32 N – 26 N = 6 N (positive)
Since positive 6 is left, therefore this must be acted by
an equilibrant negative 6 N.
Answer:
<span>- 6 N </span>
500 N is the answer, you just tell that the moon is attracted towards the person because of the Earth's huge mass.
The force result in stretching the spring 10.0 centimeters is 2.5N.
<h3>
What is Hooke's law?</h3>
If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.
F = kx
where k is the proportionality constant called the spring constant or force constant.
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.
Let the spring constant be very low 0.04N/m
The force applied is
F = 10 cm / 0.04
F = 0.1 m / 0.04
F = 2.5 N
Thus, the force result in stretching the spring 10cm is 2.5 N.
Learn more about hooke's law.
brainly.com/question/13348278
#SPJ1
