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torisob [31]
3 years ago
15

De que trata la primera ley de newton

Physics
1 answer:
Tasya [4]3 years ago
4 0
uhm i think i don’t understand what it says but i can answer if it was English ! i also tried to translate it but it didn’t help !
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Going around a ferris wheel without stopping is an example of _____ circular motion<br>(apex)
victus00 [196]

going around a ferris wheel without stopping is an example of uniform circular motion (apex).

3 0
3 years ago
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Object that is relatively close to mirror ray diagrams for convex mirrors
Anit [1.1K]

Answer: Such objects are closer than they look.

Explanation:

Image formed by convex mirror are relatively virtual, smaller than the image and upright, because it forms behind the mirror. Thank you.

6 0
3 years ago
Which hypothenical scenerio would you result in the moon not having dofferent phases? A.The moon takes twice as long as it does
maw [93]
<span>The hypothetical scenario that would result in the moon not having different phases would be t</span>he moon always stays in one position relative to the earth and sun. Why? If the moon didn't have an phases it would stay in one position. Phases mean change, and if you observe the moon every night, you'll see how it changes; it repeats, so thus it's a pattern. 

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5 0
3 years ago
Help please ASAP !! Thanks
SVEN [57.7K]

Answer:

I think its C

Explanation:

8 0
3 years ago
answer A vertical spring stretches 3.4 cm when a 8-g object is hung from it. The object is replaced with a block of mass 26 g th
victus00 [196]

Answer:

0.695s

Explanation:

From Hooke's law, the restoring force is given has

F = -ky .......1

Where F is the force, y is the spring displacement and k force constant of the spring.

Also recall,

F=mg ............ 2

Where m is the mass of object, g is the acceleration due to gravity.

Equating 1 and 2

Ky = mg

Given that g=9.8m/s2 , y is 3.4cm and g is 8g

K×3.4/100m =8/1000kg × 9.8m/s2

K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034

K= 0.0784÷0.035

K=2.24N/m

Mass ofvthe second object is 25g =0.025kg

Period of oscillation T

T=2π√m/k

T=2×3.142√0.025/2.24

T=6.284√0.0111

T=0.659seconds

8 0
3 years ago
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