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Likurg_2 [28]
2 years ago
11

An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal

culate the displacement of the ant.
Physics
2 answers:
borishaifa [10]2 years ago
8 0

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be <u>80cm</u>.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is <u>20 cm.</u>

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u>I hope this helps!</u></em>

docker41 [41]2 years ago
3 0

Answer:

Distance = 80cm

Displacement = 20cm

Explanation:

Distance does not depend on direction, its the total amount traveled.

Displacement has a direction. Displacement is how far your stopping point is from the beginning point.

The ant could have walked 50cm to the east, stopped and turned around, walked 50cm to the west, walking a total distance of 100cm, but the displacement would be 0 because he ended up at the same spot that he started. It can get much more complicated than this, but this is a good start.

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Answer:

Do neither of these things ( c )

Explanation:

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7 0
3 years ago
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

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7 0
3 years ago
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Explanation:

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3 years ago
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C.
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8 0
4 years ago
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Which value represents accuracy: the mean, median, mode, range from center, or distance between farthest beanbags?
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Answer:

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