What type of wave carry sound energy forward?

What is the Difference between accuracy and precision ?<br>

Pie

Answer:

Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.

6 0

2 years ago

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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

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An electron is confined to a one dimensional infinite potential well 150 pm. How much energy must it absorb if it is to jump to

igor_vitrenko [27]

Answer:

\Delta E=1.22\times 10^{-22}J

Explanation:

The energy of electron in any state is given by $E=\frac{n^2h^2}{8mL^2}$ here h is planck's constant n is state of electron L is the infinte potential well m is the mass of electron

We know that $h=6.67\times 10^{-34}$

Potential well dimension = $150pm=150\times 10^{-12}m$

Mass of electron $=9.1\times 10^{-31}kg$

So energy required to electron to jump from ground state to 3rd state

$\Delta E=\frac{h^2}{8mL^2}\left ( 3^2-1^2 \right )$

$\Delta E=\frac{\left ( 6.67\times 10^{-34} \right )^2}{8\times 9.1\times 10^{-31}(150\times 10^{-12})^2}\left ( 9-1 \right )$

$\Delta E=1.22\times 10^{-22}J$

7 0

3 years ago

It is important to organize your data because

serg [7]

I would say B. hope it helps

5 0

3 years ago

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What are the steps in order to get the correct outcome?

In-s [12.5K]

1 Purpose/Question. Ask a question. 2 Research. Conduct background research. ... 3 Hypothesis. Propose a hypothesis. ... 4 Experiment. Design and perform an experiment to test your hypothesis. ... 5 Data/Analysis. Record observations and analyze the meaning of the data. ... 6 Conclusion. Conclude whether to accept or reject your hypothesis.

Sure hope this helps you. Pls mark me as brainiest

7 0

2 years ago