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-BARSIC- [3]
4 years ago
11

Under what circumstances does distance traveled equal magnitude of displacement? what is the only case in which magnitude of dis

placement and displacement are exactly the same?
Physics
2 answers:
hodyreva [135]4 years ago
6 0

Explanation:

We know that the total path covered by an object is called its distance covered while the shortest path covered by it is called its displacement. The distance of an object cannot be equal to 0 but the displacement can. When an object moves in a straight line and not come back to its initial position, its magnitude of displacement and distance will always be equal.

attashe74 [19]4 years ago
4 0

if both are in the same straight line.

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Your friend says that for a moving object to continue moving, a force must be continually applied to it. Do you agree with your
Zigmanuir [339]

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I would disagree with my friend because that according to Newton's first law, an object in motion will continue to be in motion until stopped by another object/force. If the object is already moving, it will stay in motion until something else stops it. There is no need for a force to be <em>continually</em> applied to it while it's already moving.

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2 years ago
Match the type of fitness with the correct example.
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cardiovascular fitness: 3, 4, 7

flexibility: 1, 5

muscular fitness: 2, 6

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3 years ago
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When the diaphragm contracts , air pressure in the chest increases . True or False ?
nexus9112 [7]

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When the diaphragm contracts, the muscles will also contract and pull upward and increase the size of the thoracic cavity thus decreases air pressure inside during inspiration. After the diaphragm contracts, it goes to relaxation, the muscles will also relaxed. It gets looser and return to its original position higher up in the chest. This increase the pressure in the chest, which force the air in the lungs out through the nose.

 

 

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3 years ago
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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m&#10;\\d=150rev

d= 155 rev = 155(2\pi ) = 310\pi rad

a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi )  = 4.0\pi \frac{rev}{s^{2} }

d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449

w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}

Now the angular velocity is the blade speed so:

V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s

Now is the same time the piece travel horizontally

L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m

blade piece travels  HORIZONTALLY = (24.5)(0.397) = 9.73 m  ANS

6 0
4 years ago
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