Answer:
28 cm and 32 cm
Explanation:
1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of
a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.
As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.
So, the minimum length is 28 cm and the maximum length is 32 cm.
Answer:
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Answers
Related Questions
How many non-square numbers lie between the squares of 12 and 13?
Answer
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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.
Complete step-by-step answer:
In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.
Square of 12 = 122=144 and square of 12 = 132=169
As 12 and 13 are consecutive numbers so all numbers between their squares will be non-square numbers.
Therefore, 169 – 144 = 25
Total number of numbers between 169 and 144 (i.e., excluding 144 and 169) = 25 – 1 = 24.
Explanation:
Brian least po please
Answer:
v₁ = 3.5 m/s
v₂ = 6.4 m/s
Explanation:
We have the following data:
m₁ = mass of trailing car = 400 kg
m₂ = mass of leading car = 400 kg
u₁ = initial speed of trailing car = 6.4 m/s
u₂ = initial speed of leading car = 3.5 m/s
v₁ = final speed of trailing car = ?
v₂ = final speed of leading car = ?
The final speed of the leading car is given by the following formula:
![v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B2m_1%7D%7Bm_1%2Bm_2%7Du_1-%5Cfrac%7Bm_1-m_2%7D%7Bm_1%2Bm_2%7Du_2%5C%5C%5C%5Cv_2%3D%5Cfrac%7B%282%29%28400%5C%20kg%29%7D%7B400%5C%20kg%2B400%5C%20kg%7D%286.4%5C%20m%2Fs%29-%5Cfrac%7B400%5C%20kg-400%5C%20kg%7D%7B400%5C%20kg%20%2B%20400%5C%20kg%7D%283.5%5C%20m%2Fs%29)
<u>v₂ = 6.4 m/s</u>
The final speed of the leading car is given by the following formula:
![v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7Bm_1-m_2%7D%7Bm_1%2Bm_2%7Du_1%2B%5Cfrac%7B2m_2%7D%7Bm_1%2Bm_2%7Du_2%5C%5C%5C%5Cv_1%3D%5Cfrac%7B400%5C%20kg-400%5C%20kg%7D%7B400%5C%20kg%20%2B%20400%5C%20kg%7D%286.4%5C%20m%2Fs%29%2B%5Cfrac%7B%282%29%28400%5C%20kg%29%7D%7B400%5C%20kg%2B400%5C%20kg%7D%283.5%5C%20m%2Fs%29)
<u>v₁ = 3.5 m/s</u>
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Answer: D. The elements have the same number of valence electrons
Explanation: The chemical reactivity of elements is governed by the valence electrons present in the element.
The elements present in the same group or vertical column have similar valence configurations and thus behave similarly in chemical reactions or have similar bonding properties.
For Example: Both fluorine and chlorine belong to same family or group and both have 7 electrons in their valence shell and thus accept single electron to attain noble gas configuration.
![F:9:1s^22s^22p^5](https://tex.z-dn.net/?f=F%3A9%3A1s%5E22s%5E22p%5E5)
![F^-:10: 1s^22s^22p^6](https://tex.z-dn.net/?f=F%5E-%3A10%3A%201s%5E22s%5E22p%5E6)
![Cl:17:1s^22s^22p^63s^23p^5](https://tex.z-dn.net/?f=Cl%3A17%3A1s%5E22s%5E22p%5E63s%5E23p%5E5)
![Cl^-:18:1s^22s^22p^63s^23p^6](https://tex.z-dn.net/?f=Cl%5E-%3A18%3A1s%5E22s%5E22p%5E63s%5E23p%5E6)
thus both would bond with a cation bearing a single positive charge.