Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
![v^{2}=u^{2}+2\cdot a \cdot s](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D%2B2%5Ccdot%20a%20%5Ccdot%20s)
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/![Sec^{2}](https://tex.z-dn.net/?f=Sec%5E%7B2%7D)
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
![0^{2} = 13^{2} + 2 \cdot a \cdot57.46](https://tex.z-dn.net/?f=0%5E%7B2%7D%20%3D%2013%5E%7B2%7D%20%20%2B%202%20%5Ccdot%20a%20%5Ccdot57.46)
a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)
![v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%2029%5E%7B2%7D%20%20-%202%20%5Ccdot%201.47%20%5Ccdot%20S)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
It doesn't mean that the climate is changing, it is probably just the morning dew.
Answer:
The value is ![F_f = 46.935 \ N](https://tex.z-dn.net/?f=F_f%20%3D%20%2046.935%20%5C%20%20N)
Explanation:
From the question we are told that
The magnitude of the horizontal force is ![F = 92.7 \ N](https://tex.z-dn.net/?f=F%20%20%3D%20%2092.7%20%5C%20%20N)
The mass of the crate is ![m = 40.5 \ kg](https://tex.z-dn.net/?f=m%20%20%3D%20%2040.5%20%5C%20%20kg)
The acceleration of the crate is ![a = 1.13 \ m/s](https://tex.z-dn.net/?f=a%20%3D%20%201.13%20%5C%20m%2Fs)
Generally the net force acting on the crate is mathematically represented as
![F_{net} = F - F_f = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20%20F%20-%20%20F_f%20%3D%20%20ma)
Here
is force of kinetic friction (in N) acting on the crate
So
![92.7 - F_f = 40.5 * 1.13](https://tex.z-dn.net/?f=92.7%20%20-%20%20F_f%20%3D%20%2040.5%20%2A%201.13)
=> ![F_f = 46.935 \ N](https://tex.z-dn.net/?f=F_f%20%3D%20%2046.935%20%5C%20%20N)
Answer:
Therefore the correct statement is B.
Explanation:
In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.
The sum of all these waves results in the diffraction curve of the slit that has the shape
I = Io sin² θ /θ²
Where the angle is a function of the wavelength and the width of the slit.
From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)
Therefore the correct statement is B