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goldenfox [79]
2 years ago
6

find the following Distance, Time, and Velocity in the problem; a track runs a 100-m dash in 11 seconds. What is his average vel

ocity during the run? (please include explanation and formula thankyou)
Physics
1 answer:
kow [346]2 years ago
5 0

Answer:

9.09m/s

Explanation:

now velocity of a body is measure of the distance covered by a body in a particular time frame,so simply its the rate at which distance is covered,

thus mathematically V(velocity)=d(distance covered)/t(time it takes to cover the distance)

hence V=100m/11s

which is V=9.09m/s.

so this means the runner covers 9.09m in every second ,<em> </em><em>well</em><em> </em><em>not</em><em> </em><em>bad</em><em> </em><em>lol</em><em> </em><em>!</em>

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The diagram shows a ballistic pendulum. A 200 g bullet is fired into the suspended 4 kg block of wood and remains embedded insid
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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:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

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

3) 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

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

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

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

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

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

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

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.

5 0
3 years ago
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