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Ahat [919]
3 years ago
8

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find

the initial speed of the rock.
a. 3m/s
b. 30.3 m/s
c. None of the above
Physics
1 answer:
butalik [34]3 years ago
5 0

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N  

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy    ..............................1

so it can be written as

work = force × distance     ...................2

and

KE is express as

K.E = 0.5 × m × v²  

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)²      ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m  (-9.8)  , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m  (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² =  919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

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Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

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2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

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\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

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Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

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When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

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