Answer:
A= 61.35
B= -44.40
Explanation:
1. Using the components method we have:

Considering that the vector sum
, then:

Then:

It means the value of x and y component is 0.
2. Determinate the equations that describe each component:

Form Eq. (1):

Replacing A in Eq. (2):

Replacing values of C, α and β in (4):

Replacing value of B in (3)

Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz

velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
Answer:
1626.4 N
Explanation:
Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?
The parameters to be considered are:
Distance S = 3m
Time t = 0.55s
Since the man started from rest, initial velocity u = 0
Using second equation of motion
S = Ut + 1/2at^2
3 = 1/2 × a × 0.55^2
3 = 1/2 × a × 0.3025
a = 3/ 0.15125
a = 19.83 m/s^2
Force = mass × acceleration
Force = 82 × 19.83
Force = 1626.4 N
Therefore, the force that water exerted on him is 1626.4 N
Answer:
x(t) = -8sin2t
Explanation:
See the attachment for solution
From my solving, we can deduce that w² = 4, and thus, w = 2
Therefore, the general solution is
x(t) = c1 cos2t + c2 sin2t
Considering the final variable, we can conclude that
x(0) = 0
x'(0) = -8 m/s
The final solution, thus
x(t) = -8sin2t
Answer:
or 0.32 μm.
Explanation:
Given:
The radiations are UV radiation.
The frequency of the radiations absorbed (f) = 
The wavelength of the radiations absorbed (λ) = ?
We know that, the speed of ultraviolet radiations is same as speed of light.
So, speed of UV radiation (v) = 
Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

Now, expressing the above equation in terms of wavelength 'λ', we have:

Now, plug in the given values and solve for 'λ'. This gives,
![\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B3%5Ctimes%2010%5E8%5C%20m%2Fs%7D%7B9.38%5Ctimes%2010%5E%7B14%7D%5C%20Hz%7D%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-7%7D%5C%20m%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2010%5E%7B6%7D%5C%20%5Cmu%20m%5C%20%5B1%5C%20m%3D10%5E6%5C%20%5Cmu%20m%5D%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-1%7D%3D0.32%5C%20%5Cmu%20m)
Therefore, the wavelength of the radiations absorbed by the ozone is nearly
or 0.32 μm.