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elena55 [62]
3 years ago
10

You are asked to explain what happens to individual atoms when a chemical change takes place in lab you are doing in class. What

is the best explanation?
A. Some are created.
B. They are rearranged.
C. Some are destroyed.
D. Some are destroyed and some are created.
Physics
2 answers:
seraphim [82]3 years ago
7 0

Answer:

whle they cannot be destroyed or created they can be moved around or REARRANGED

Explanation:

B

ss7ja [257]3 years ago
3 0

B. They are rearranged.

The First Law of Thermodynamics states that matter and energy can not be created or destroyed.

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Evgesh-ka [11]
No I don't believe so
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what is the main reason that attitudes are more often revealed in spoken rather than written language
pishuonlain [190]
It's a combination of all those things. probably because we are taught from an early age to write in an academic fashion, giving balanced arguments and a conclusion. When speaking from the heart, there is no opposing argument nor is there a conclusion, just emotion.
5 0
3 years ago
What is the final position of the object if its initial position is x = 0.40 m and the work done on it is equal to 0.21 J? What
r-ruslan [8.4K]

Answer:

a) Final position is x = 0.90 m

b) Final position is x = 0.133 m

Explanation:

The workdone between two points is usually approximated as the area under the force-distance curve between those two points.

From the graph,

As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,

The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J

The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m

Work done under this bar = 0.4 × 0.25 = 0.1 J

Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J

So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N

0.21 = 0.18 + 0.2 (x - 0.75)

0.03 = 0.2x - 0.15

0.2x = 0.18

x = 0.9 m

Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.

b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.

From the starting point where the initial position is 0.40 m, the force here is 0.80 N

The workdone under this bar to the left is

The workdone = 0.8 (0.25 - 0.4) = - 0.12 J

Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N

- 0.19 = -0.12 + 0.6 (x - 0.25)

-0.07 = 0.6x - 0.15

0.6x = 0.08

x = (0.08/0.6) = 0.133 m

Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.

Hope this Helps!!!

4 0
3 years ago
A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

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