A complete description of simple harmonic motion must take into account several physical quantities and various mathematical rel
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A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the edge of the table in the horizontal direction,
- t= 0.45 seconds.
- V=2.22m/s
- VT=4.95 m/s
This is further explained below.
<h3>What is its speed when it hits the floor...?</h3>Generally, the equation for motion is mathematically given as
S= ut + 0.5at²
Therefore
y = Voy t + 0.5gt^2
1 = 0.5x 98 x 6²
1=4.9t^2
t= 0.45 seconds.
b) Horizontal motions are uniform.
V=Horizontal displacement/time
V=1/0.45
V=2.22m/s
C)
Vx: 2.22 m/s At bottom,
Vy² = Voy² + 2as
Vy² = 2x95x1
Vy² = 19.6
Total velocity
VT=4.95 m/s
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1) Focal length
We can find the focal length of the mirror by using the mirror equation:
(1)
where
f is the focal length
is the distance of the object from the mirror
is the distance of the image from the mirror
In this case,
, while 
(the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
from which we find
2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
We can find the focal length of the mirror by using the mirror equation:
where
f is the focal length
In this case,
from which we find
2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is: