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makkiz [27]
3 years ago
5

What else did you change in the Levers lab when you changed the position of the fulcrum?

Physics
2 answers:
geniusboy [140]3 years ago
8 0
<h2>Answer:</h2>

<u>By changing the position of the fulcrum, the amount of effort Force also changes</u>

<h2>Explanation:</h2>

The mechanical advantage of a lever is changed when the distance between the pivot point (fulcrum) and the weight is changed. The way levers operate is by an effort applied at a point, which moves a load at another point through a balance point called the fulcrum. If the distance between the fulcrum and the effort is increased, the effort needed to lift a weight decreases proportionally.

konstantin123 [22]3 years ago
6 0

Answer:

the effort force

Explanation:

i just took the test

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Who is known as the "father of modern chemistry" because he first organized all known elements into four different
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D. Henry Moseley

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Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref
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The field lines around one end of a bar magnet are shown
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Answer:

Explanation:

Field lines are lines of forces around a bar magnet. The show the direction of force field in a magnet.

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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
4 years ago
How far does a car travel in 90 seconds if it’s traveling 55 m/s? Show equation
podryga [215]

You just said the car is traveling at the speed of 55 m/s.  If I understand this  correctly, that means the car will cover:

55 meters in the first second,

55 meters in the 2nd second,

55 meters in the 3rd second,

55 meters in the 4th second,

55 meters in the 5th second,

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.

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55 meters in the 87th second,

55 meters in the 88th second,

55 meters in the 89th second, and

55 meters in the 90th second.

That's 55 meters 90 times.  If you just move these words around a little bit, it says "90 times 55 meters" . . . a pretty simple arithmetic problem.

The equation is . . . <em>Distance = (55 m/s) times (time, in seconds)</em> .

I get <em>4,953 meters</em>.  You should check me on this.

8 0
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