Answer:
B
Explanation:
The net force is the force between action and reaction and when this forces are not the same an acceleration is spurred.
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Answer:

☯ Question :
- How fast is a wave travelling if it has a wavelength of 7 meters and a frequency of 11 Hz?
☯ 
☥ Given :
- Wavelength ( λ ) = 7 meters
- Frequency ( f ) = 11 Hz
☥ To find :
☄ We know ,

where ,
- v = speed of sound
- f = frequency
- λ = wavelength
Now, substitute the values and solve for v.
➺ 
➺ 
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✑ Additional Info :
- Frequency : The number of complete vibrations made by a particle of a body in one second is called it's frequency. It is denoted by the letter f . The SI unit of frequency is hertz ( Hz ).
- Wavelength : The distance between two consecutive compressions or rarefactions of a sound wave is called wavelength of that wave. It is denoted by λ ( lambda ) and it's SI unit is m.
- Speed of a sound wave : The distance covered by a sound wave in one second is called speed of sound wave. It depends on the product of wavelength and frequency of the wave.
Hope I helped!
Have a wonderful time! ツ
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Explanation:
At point B, the velocity speed of the train is as follows.

= 
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y = 

Now, second derivative of the train is calculated as follows.
Radius of curvature of the train is as follows.
![\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B%5B1%20%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D%7D)
= ![\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B1%20%2B%200.2e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B0.2%2810%5E%7B-3%7D%29e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%7D)
= 3808.96 m
Now, we will calculate the normal component of the train as follows.

= 
= 0.1822 
The magnitude of acceleration of train is calculated as follows.
a = 
= 
= 
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is
.
Answer:
frequency of the sound = f = 1,030.3 Hz
phase difference = Φ = 229.09°
Explanation:
Step 1: Given data:
Xini = 0.540m
Xfin = 0.870m
v = 340m/s
Step 2: frequency of the sound (f)
f = v / λ
λ = Xfin - Xini = 0.870 - 0.540 = 0.33
f = 340 / 0.33
f = 1,030.3 Hz
Step 3: phase difference
phase difference = Φ
Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984
Φ = 3.9984 rad * (360°/2π rad)
Φ = 229.09°
Hope this helps!