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).
As we know that the formula of kinetic energy will be
now here we know that
m = 2 kg
v = 1 m/s
so from the above equation we have
Answer: Remember speed is distance divided by time, so if he travels 1000 m in 7.045 s, his speed is
(1000 m)/(7.045 s) = 141.9 m/s.
Note there are 1609 metres in a mile, or 1 mi = 1609 m, so m = 1/1609 mi, or
141.9/1609 mi/s = 0.08822 mi/s. Now, note that 1 h = 3600 s, so the speed is
0.08822*3600 mi/h = 317.6 mi/h.
Answer:
A) R = (200 i ^ + 100 j ^ + 30k ^) m
, B) L = 223.61 m
, C) R = 225.61 m
Explanation:
Part A
This is a vector summing exercise, let's take a Reference System where the z axis corresponds to the height (flights), the x axis is the East - West and the y axis corresponds to the North - South.
Let's write the displacements
Descending from the apartment
10 flights of 3 m each, the total descent is 30 m
Z = 30 k ^ m
Offset at street level
L1 = 0.2 i ^ km
L2 = 0.1 j ^ km
Let's reduce everything to the SI system
L1 = 0.2 * 1000 = 200 i ^ m
L2 = 100 j ^ m
The distance traveled is
R = (200 i ^ + 100 j ^ + 30k ^) m
Part B
The horizontal distance traveled can be found with the Pythagorean theorem for the coordinates in the plane
L² = x² + y²
L = √ (200² + 100²)
L = 223.61 m
Part C
The magnitude of travel, let's use the Pythagorean theorem for the sum
R² = x² + y² + z²
R = √ (30² + 200² + 100²)
R = 225.61 m
