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Masja [62]
3 years ago
12

Please help I'll mark brainliest!!!!

Physics
2 answers:
madam [21]3 years ago
7 0

This question is based on the fundamental assumption of  vector direction.

A vector is  a physical quantity which has  magnitude as well direction  for its complete specification.

The magnitude of a physical quantity is simply a  numerical number .Hence it can not be negative.

A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector.  For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.

As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.

The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.

Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.

In a general sense we assume that vertically downward motion  is negative and vertically upward is positive. In case of a falling object the motion is  vertically downward. So the velocity of that object is negative .

So last   option is  partially  correct  as  the vector can be negative depending on our choice of co-ordinate system.





exis [7]3 years ago
4 0

The velocity of a falling object is a vector that should be negative

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0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)

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Answer:

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(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

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Alternatively, you can explicitly solve for the acceleration, then for the distance:

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0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

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(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

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<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

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