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nikitadnepr [17]
3 years ago
7

"Influence" between bodies is reduced as the bodies are

Physics
2 answers:
Degger [83]3 years ago
5 0

They are moved farther apart.

azamat3 years ago
4 0
Both particles and big objects will be influenced by each other less if they are moved further apart. 
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when in object moves stop moving changes speed or changes direction how do scientist describe that condition?
Kruka [31]

im taking test rn nd this question was on there, im saying unbalanced force, if it not correct i will put right answer, but im pretty sure the answer is unblanced

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3 years ago
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I want ti know how to study​
arlik [135]

Answer:

Make sure everything is organized have a planner it can help

Get rid of all distractions

Listen to music if it helps you concentrate

Have your notes

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4 years ago
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Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile? Answ
Vinil7 [7]

Answer: 272.82 drop/tile

Explanation:

Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?

Tiles/ft^2 × drop/tiles = drop/ft^2

Tiles will cancel out. Leaving the answer to be drop/ ft^2

Substitutes all the magnitude of the above units.

17 × drop/tiles = 4638

Make drop/tiles the subject of formula

Drop/tiles = 4638/17

Drop/tiles = 272.82

Therefore, 272.82 drop/tile drops fall on each tile? 

8 0
3 years ago
a soccer ball accelerates more than a bowling ball when thrown with the same force. which law of newton best supports this?
Vedmedyk [2.9K]

Answer:

Newtons second law of motion known as the law of acceleration

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The second law explains that a greater mass requires a greater force

3 0
2 years ago
Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a
masha68 [24]
Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m
And by using Hook's law, we can find the constant of the spring:
k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m
4 0
3 years ago
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