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Alexxx [7]
3 years ago
15

Using the rules for significant figures, what do you get when you multiply 67.6 by 1.2?

Physics
1 answer:
Anika [276]3 years ago
5 0

Answer:

81

Explanation:

The problem here is to use the rules for significant figures in the multiplication below.

       67.6 x 1.2:

 This product will yield an answer that will produce the least number of significant figures.

   67.6 has 3 significant figures; 6, 7 and 6

    1.2 has 2 significant figures; 1 and 2

The product must give us an answer of two significant figures:

        67.6 x 1.2  = 81.12 to 2significant figures gives 81

The solution to this problem is 81

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A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1)
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total before
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5 0
4 years ago
Diana observed that the plants in her garden were not growing well due to poor soil conditions. She tested the soil and used the
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Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
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