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zhenek [66]
3 years ago
14

Particles (mass of each = 0.40 kg) are placed at the 60-cm and 100-cm marks of a meter stick of negligible mass. This rigid body

is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the magnitude of the initial linear acceleration of the end of the body opposite the pivot?

Physics
1 answer:
hichkok12 [17]3 years ago
7 0

Answer:

12 m/s ∧2

Explanation:

The picture attached explains it all and i hope it helps. Thank you

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Which of Newton's laws explains why satellites need very little fuel to stay in oribit?
Angelina_Jolie [31]

Sattelites don't need any fuel to stay in orbit. The applicable law is...."objects in motion tend to stay in motion". Having reached orbital velocity, any such object is essentially "falling" around the earth. Since there is no (or at least very little) friction in the vacuum of space, the object does not slow.... It simply continues.


Sattelites in "low" earth orbit do encounter some friction from the very thin upper atmosphere, and they will eventually "decay".

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3 years ago
All but one statement could describe a scientific theory. That is:
murzikaleks [220]

D is the wrong answer. New information does often completely change the theory. Its hard to change something and leave the major theory intact.

8 0
3 years ago
Read 2 more answers
. Write the following in the scientific notation of 0.0000002400m ​
aleksandr82 [10.1K]

Explanation:

<h2><u>Steps </u><u>:</u></h2>
  1. <u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
  2. <u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
  3. <u>That's</u><u> </u><u>all </u>

<u>hope</u><u> it</u><u> </u><u>help</u>

<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>
7 0
3 years ago
How to do this, i'm completely lost
vaieri [72.5K]
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
4 years ago
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