We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p
Now that we have the value of p, we can plug it into the magnification equation.
M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'
So the height of the image produced by the mirror is 9.6cm.
Answer:
The correct answer is the Convex lens. An image is formed when a ray of light coming from a point intersects at another point. The image is formed by the real intersection of light. The image is formed by the virtual intersection of Light.
here is the site : textbook.com
Answer:
A) the maximum acceleration the boulder can have and still get out of the quarry
B) how long does it take to be lifted out at maximum acceleration if it started from rest
Explanation:
A)
let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.
the weight of the chain is:
and maximum tension is ![T=2.50 m_{c} g=1.41*10^4N](https://tex.z-dn.net/?f=T%3D2.50%20m_%7Bc%7D%20g%3D1.41%2A10%5E4N)
total mass and weight is :
![M =m_{c}+ m_{b} =740kg+550kg=1290 kg](https://tex.z-dn.net/?f=M%20%3Dm_%7Bc%7D%2B%20m_%7Bb%7D%20%3D740kg%2B550kg%3D1290%20kg)
![w_{M} =1.2650*10^4N](https://tex.z-dn.net/?f=w_%7BM%7D%20%3D1.2650%2A10%5E4N)
∑![F_{y} =ma_{y}](https://tex.z-dn.net/?f=F_%7By%7D%20%3Dma_%7By%7D)
![T-M_{g} =Ma_{y}](https://tex.z-dn.net/?f=T-M_%7Bg%7D%20%3DMa_%7By%7D)
![a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg](https://tex.z-dn.net/?f=a_%7By%7D%20%3D%28t-M_%7Bg%7D%20%29%2FM%3D%282.50m_%7Bc%7D%20-M_%7Bg%7D%20%29%2FM%3D%282.50.550kg-1290kg%29%289.8m%2Fs%5E2%29%2F1290kg)
![=0.645m/s^2](https://tex.z-dn.net/?f=%3D0.645m%2Fs%5E2)
B)
maximum acceleration
![a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0](https://tex.z-dn.net/?f=a_%7By%7D%20%3D0.645m%2Fs%5E2%5C%5C%5C%5Cy-y_%7B0%7D%20%3D119m%5C%5Cv_%7B0y%7D%20%3D0)
using ![y-y_{0} =v_{oy} t+1/2(a_{y} )t^2](https://tex.z-dn.net/?f=y-y_%7B0%7D%20%3Dv_%7Boy%7D%20t%2B1%2F2%28a_%7By%7D%20%29t%5E2)
to solve for t
![t=\sqrt{2(y-y_{0} )/a_{y} }](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B2%28y-y_%7B0%7D%20%29%2Fa_%7By%7D%20%7D)
![t=\sqrt{2(119m)/0.645m/s^2} =19.20s](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B2%28119m%29%2F0.645m%2Fs%5E2%7D%20%3D19.20s)