Answer:
E(x,t) = Emaxcos(kx - ωt + φ),
B(x,t) = Bmaxcos(kx - ωt + φ).
Explanation:
E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves the electric field E and the magnetic field B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E x B.
<u>Answer:</u>
The stopping car has negative acceleration.
<u>Explanation:</u>
We know acceleration is the rate of change of velocity. That is
Acceleration(a) = ( Final velocity - Initial velocity )/ Time taken.
Here Final velocity = 0 km/h( car stops) and initial velocity = 30 km/h
So acceleration = (0-30)/time = -30/ time
Time is always positive , so -30/time is negative, so the stopping car has negative acceleration.
Hydrogen and oxygen bond because when more molecules are present, as is the case with liquid water, more bonds are possible because the oxygen of one water molecule has two lone pairs of electrons, each of which can form a hydrogen bond with a hydrogen on another water molecule.
Acceleration is any change in speed or direction of motion.
Speeding up, slowing down, or moving along a curve are all accelerations.
Answer:
0.785 m/s
Explanation:
Hi!
To solve this problem we will use the equation of motion of the harmonic oscillator, <em>i.e.</em>
- (1)
- (1)
The problem say us that the spring is released from rest when the spring is stretched by 0.100 m, this condition is given as:
![x(0) = 0.100](https://tex.z-dn.net/?f=x%280%29%20%3D%200.100)
![\frac{dx}{dt}(0) = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%280%29%20%3D%200)
Since cos(0)=1 and sin(0) = 0:
![x(0)=A](https://tex.z-dn.net/?f=x%280%29%3DA)
![\frac{dx}{dt}(0) = -B\omega](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%280%29%20%3D%20-B%5Comega)
We get
![A =0.100\\B = 0](https://tex.z-dn.net/?f=A%20%3D0.100%5C%5CB%20%3D%200)
Now it say that after 0.4s the weigth reaches zero speed. This will happen when the sping shrinks by 0.100. This condition is written as:
![x(0.4) = - 0.100](https://tex.z-dn.net/?f=x%280.4%29%20%3D%20-%200.100)
Since
![x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)](https://tex.z-dn.net/?f=x%28t%29%20%3D%200.100%20cos%28%5Comega%20t%29%5C%5C%20-0.100%3Dx%280.4%29%3D0.100cos%28%5Comega%200.4%29)
This is the same as:
![-1 = cos(0.4\omega)](https://tex.z-dn.net/?f=-1%20%3D%20cos%280.4%5Comega%29)
We know that cosine equals to -1 when its argument is equal to:
(2n+1)π
With n an integer
The first time should happen when n=0
Therefore:
π = 0.4ω
or
ω = π/0.4 -- (2)
Now, the maximum speed will be reached when the potential energy is zero, <em>i.e. </em>when the sping is not stretched, that is when x = 0
With this info we will know at what time it happens:
![0 = x(t) = 0.100cos(\omega t)](https://tex.z-dn.net/?f=0%20%3D%20x%28t%29%20%3D%200.100cos%28%5Comega%20t%29)
The first time that the cosine is equal to zero is when its argument is equal to π/2
<em>i.e.</em>
![t_{maxV}=\pi /(2\omega)](https://tex.z-dn.net/?f=t_%7BmaxV%7D%3D%5Cpi%20%2F%282%5Comega%29)
And the velocity at that time is:
![\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%28t_%7BmaxV%7D%20%29%20%3D-%200.100%5Comega%20sin%28%5Comega%20t_%7BmaxV%7D%29%5C%5C%5Cfrac%7Bdx%7D%7Bdt%7D%28t_%7BmaxV%7D%20%29%20%3D-%200.100%5Comega%20sin%28%5Cpi%2F2%29%5C%5C)
But sin(π/2) = 1.
Therefore, using eq(2):
![\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%28t_%7BmaxV%7D%20%29%20%3D%200.100%2A%5Comega%20%3D%200.100%5Cfrac%7B%5Cpi%7D%7B0.400%7D%20%3D%20%5Cpi%2F4)
And so:
![V_{max} = \pi / 4 =0.785](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%20%5Cpi%20%2F%204%20%3D0.785)