Aswer:
False, the values of the distance traveled and the displacement only coincide when the trayectorie is a straight line. Otherwise, the distance will always be greater than the offset.
Although these terms are used synonymously in other cases, they are totally different. Since the distance that a mobile travels is the equivalent of the length of its trajectory. Whereas, the displacement will be a vector magnitude.
<u>xXCherryCakeXx</u>.
Utilize the formula: 
= Final Velocity (86 m/s)
= Initial Velocity (0 m/s)
a = acceleration (m/s²)
t = Time (100 seconds)
As a result,
86 m/s = 0 + (a)(100 seconds)
Using algebra, divide 86 m/s by 100 seconds:
86 m/s = 100a
a = 0.86 m/s²
Rounded to one decimal place: 0.9 m/s²
Let me know if you have any questions!
Answer:
Explanation:
General equation of the electromagnetic wave:
![E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%20E_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
where
Phase angle, 0
c = speed of the electromagnetic wave, 3 × 10⁸
wavelength of electromagnetic wave, 698 × 10⁻⁹m
E₀ = 3.5V/m
Electric field equation
![E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%203.5sin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)
Magnetic field Equation
![B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%20B_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
Where B₀= E₀/c

![B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)
Answer:
The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant
The law can be expressed mathematically as follows;

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously
Therefore, the law is verified indirectly, by rearranging the above equation as follows;

From which it can be shown that the following relation holds with the limits of error in the experiment
m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²
Explanation: