Answer:
See the answers below
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
<u>First case</u>
Vf = 6 [m/s]
Vo = 2 [m/s]
t = 2 [s]
![6=2+a*2\\4=2*a\\a=2[m/s^{2} ]](https://tex.z-dn.net/?f=6%3D2%2Ba%2A2%5C%5C4%3D2%2Aa%5C%5Ca%3D2%5Bm%2Fs%5E%7B2%7D%20%5D)
<u>Second case</u>
Vf = 25 [m/s]
Vo = 5 [m/s]
a = 2 [m/s²]
![25=5+2*t\\t = 10 [s]](https://tex.z-dn.net/?f=25%3D5%2B2%2At%5C%5Ct%20%3D%2010%20%5Bs%5D)
<u>Third case</u>
Vo =4 [m/s]
a = 10 [m/s²]
t = 2 [s]
![v_{f}=4+10*2\\v_{f}=24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D4%2B10%2A2%5C%5Cv_%7Bf%7D%3D24%20%5Bm%2Fs%5D)
<u>Fourth Case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
![v_{f}=5+8*10\\v_{f}=85 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D5%2B8%2A10%5C%5Cv_%7Bf%7D%3D85%20%5Bm%2Fs%5D)
<u>Fifth case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]

Answer:
I don't know about this, this is which class question
Answer:
The intensity level of sound 2 is 93.3\ W/m².
Explanation:
Given that,
Intensity of sound 1 = 45.0 W/m²
Intensity level of sound 2 = 3.2 dB

We need to calculate the intensity
Using equation of the sound level intensity



The intensity of sound 2 is greater than 3.2 dB.
Therefore,


Calculate the intensity of sound 2



Hence, The intensity level of sound 2 is 93.3\ W/m².
Answer:
Hooke's law can be written as:
F = -k*Δx
Where k is a constant that depends on the spring, called the spring constant.
And Δx = (x - x₀) is the displacement of the spring with respect to the rest position x₀
And a proportional equation is something like:
y = c*x
Where c is a constant of proportionality.
In the case of Hooke's law, the constant is K, then the force is proportional to the displacement of the spring with respect to the rest position Δx.