New force is 1/4 of force when distance is doubled
Answer:
2*F
Explanation:
If we put an object of a given size exactly at a distance 2*F from the lens, the virtual image (the image generated by the lens) will be generated at a distance 2*F from the lens and the size will be equal to the size of the real object (but the image will be inverted)
Now let's do the math.
The relation between the distance of the object to the lens O, and the distance between the image and the lens I is:
1/O + 1/I = 1/F
solving for O, we get:
1/O = 1/F - 1/I = (I - F)/(F*I)
O = F*I/(I - F)
Such that the relation between the height of the original object, H and the height of the virtual image H' is:
H/H' = -I/O
Replacing by O we get:
H/H' = -I/(F*I/(I - F))
If the sizes are equal, then H/H' = - 1 (remember that the image is inverted, thus the sign)
-1 = -I/(F*I/(I - F))
F*I/(I - F) = I
F*I = (I - F)*I
F = (I - F)
F + F = I = 2*F
The distance between the image and the lens is 2*F
O = F*I/(I - F) = F*2*F/(2*F - F) = 2*F
The object is at a distance 2*F from the lens.
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)
I think the answer is 25J
Answer:
0.333 m/s
Explanation:
avg speed= (total distance)/(total time)
10m/30s
0.333 m/s