Answer:
Explanation:
Given
Launch angle =u
Initial Speed is 
Horizontal acceleration is 
At maximum height velocity is zero therefore
Total time of flight 
During this time horizontal range is
For maximum range 
![\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20R%7D%7B%5Cmathrm%7Bd%7D%20u%7D%3D%5Cfrac%7B2v_0%5E2%7D%7Bg%7D%5Cleft%20%5B%20%5Ccos%202u-%5Cfrac%7Ba%7D%7Bg%7D%5Csin%202u%5Cright%20%5D%3D0)
(b)If a =10% g
thus 
The first part of the question is 3,100 V.
The second part of the question is 200 V.
<span>b. The coefficient of static friction for all contacting surfaces is μs=0.35. neglect friction at the rollers.
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The focal length of given concave lens will be -26.85 cm
The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.
Given concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.
We have to find focal length
The focal length can be found out by following way:
Magnification = m = +1.75
m = hi/h
hi = -47 cm
1.75 = -47/h
h = -26.85 cm
So the focal length of given concave lens will be -26.85 cm
Learn more about magnification factor here:
brainly.com/question/6947486
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Answer:
3150
Explanation:
if if you were two times 45 times 70 it would give you that answer
