The first ariplanr was made December 17, 1903
Answer:
v = 4.18 m/s
Explanation:
given,
frequency of the alarm = 872.10 Hz
after passing car frequency she hear = 851.10 Hz
Speed of sound = 343 m/s
speed of the jogger = ?
speed of the
![v_f = \dfrac{872.10-851.10}{2}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Cdfrac%7B872.10-851.10%7D%7B2%7D)
![v_f =10.5\ Hz](https://tex.z-dn.net/?f=v_f%20%3D10.5%5C%20Hz)
v_o = 872.1 - 10.5
![V_0 = 861.6\ Hz](https://tex.z-dn.net/?f=V_0%20%3D%20861.6%5C%20Hz)
The speed of jogger
![v = \dfrac{v_1 \times 343}{v_0}-343](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7Bv_1%20%5Ctimes%20343%7D%7Bv_0%7D-343)
![v = \dfrac{872.1 \times 343}{861.6}-343](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7B872.1%20%5Ctimes%20343%7D%7B861.6%7D-343)
v = 4.18 m/s
Total displacement along the length of mountain is given as
L = 235 m
angle of mountain with horizontal = 35 degree
now we will have horizontal displacement as
x = L cos35
x = 235 cos35 = 192.5 m
similarly for vertical displacement we can say
y = L sin35
y = 235 sin35 = 134.8 m
Answer:
0.25 m.
Explanation:
We'll begin by calculating the spring constant of the spring.
From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:
Force (F) = 0.1 N
Extention (e) = 0.125 m
Spring constant (K) =?
F = Ke
0.1 = K x 0.125
Divide both side by 0.125
K = 0.1/0.125
K = 0.8 N/m
Therefore, the force constant, K of spring is 0.8 N/m
Now, we can obtain the number in gap 1 in the diagram above as follow:
Force (F) = 0.2 N
Spring constant (K) = 0.8 N/m
Extention (e) =..?
F = Ke
0.2 = 0.8 x e
Divide both side by 0.8
e = 0.2/0.8
e = 0.25 m
Therefore, the number that will complete gap 1is 0.25 m.
Answer: 1.95
Explanation:
You should start off from the decay formula and solve for τ:
![I = I_{0}e^{\frac{t}{\tau\\ } }](https://tex.z-dn.net/?f=I%20%3D%20I_%7B0%7De%5E%7B%5Cfrac%7Bt%7D%7B%5Ctau%5C%5C%20%20%7D%20%7D)
![\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7BI_%7B0%7D%7D%20%3D%20e%5E%7B%5Cfrac%7B-t%7D%7B%5Ctau%7D%20%7D)
Apply inverse logarithmic function:
![ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B0.2%20A%7D%7B1.2%20A%7D%20%29%20%3D%20%5Cfrac%7B-t%7D%7B%5Ctau%7D)
The final form will be:
![\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}](https://tex.z-dn.net/?f=%5Ctau%3D%5Cfrac%7B-3.5s%7D%7Bln%28%5Cfrac%7B0.2A%7D%7B1.2A%7D%20%29%7D)
Inputing values for I, IO, and t: