Answer: the options to the questions are
a. 1.0 moles of N2
b.0.5 moles of New
c.0.2 moles of CO2
d.2 moles of He
Answer D
Explanation:
The average molecular speed v of gas is given by =√(8RT,/πM)
From the equation it can be seen that substance with lowest molar mass has the highest velocity has He is the answer
Simple reaction time involves selecting a
specific and correct response from several choices when presented with several
different stimuli. This is very important because historically, this was the
first indicators of intelligence pioneered by Francis Galton. To measure one’s
intelligence is to know how he quick a person could respond to the stimulus
with an already expected response wherein the stimulus is given unknown to the
receiver. In other terms, the intelligence is measured on how quick a person
could grasp certain concepts and how he could think fast and answer them
correctly.
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Answer:
Classification of the Elements. These three groups are: metals, nonmetals, and inert gases.
Explanation:
Given data:
* The mass of the baseball is 0.31 kg.
* The length of the string is 0.51 m.
* The maximum tension in the string is 7.5 N.
Solution:
The centripetal force acting on the ball at the top of the loop is,
![\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%2Bmg%3D%5Cfrac%7Bmv%5E2%7D%7BL%7D_%7B%7D%20%5C%5C%20v%5E2%3D%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%7D%20%5Cend%7Bgathered%7D)
For the maximum velocity of the ball at the top of the vertical circular motion,
![v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}](https://tex.z-dn.net/?f=v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T_%7B%5Cmax%20%7D%2Bmg%29%7D%7Bm%7D%7D)
where g is the acceleration due to gravity,
Substituting the known values,
![\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%287.5_%7B%7D%2B0.31%5Ctimes9.8%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%2810.538%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B17.34%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D4.16%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.
C. Eastward. Acceleration is the change in speed so it can be a positive (speeding up) or negative (slowing down) acceleration
