Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage
.
The balance point is not on the potentiometer wire - this statement means that
. In that case ,
l > L
V > E
Answer:
2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.
Explanation:
The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:
![2Mg(s)+O_2(g)\rightarrow 2MgO(s)](https://tex.z-dn.net/?f=2Mg%28s%29%2BO_2%28g%29%5Crightarrow%202MgO%28s%29)
Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.
![[Mg]=1s^22s^22p^63s^2](https://tex.z-dn.net/?f=%5BMg%5D%3D1s%5E22s%5E22p%5E63s%5E2)
In order to attain noble gas configuration it will loose two electrons.
![[Mg]^{2+}=1s^22s^22p^6](https://tex.z-dn.net/?f=%5BMg%5D%5E%7B2%2B%7D%3D1s%5E22s%5E22p%5E6)
...[1]
Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.
![[O]=1s^22s^22p^4](https://tex.z-dn.net/?f=%5BO%5D%3D1s%5E22s%5E22p%5E4)
In order to attain noble gas configuration it will gain two electrons.
![[O]^{2-}=1s^22s^22p^6](https://tex.z-dn.net/?f=%5BO%5D%5E%7B2-%7D%3D1s%5E22s%5E22p%5E6)
..[2]
2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.
A lean body mass that the cholesterol level in the body is low and that the excess fats in the body are getting rid off as much as one can. In this case, the exercises A.) cardiorespiratory fitness such as jogging and <span>C.) muscular strength and endurance such as body building, abs routines are employed. Answer is D. both A and C.</span>
Answer:
The answer is 2,416 m/s. Let's jump in.
Explanation:
We do work with the amount of energy we can transfer to objects. According to energy theory:
W = ΔE
Also as we know W = F.x
We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>
Under the force block gains;
W = F.x → ![W=2,55.0,71=1,8105\frac{N}{m}](https://tex.z-dn.net/?f=W%3D2%2C55.0%2C71%3D1%2C8105%5Cfrac%7BN%7D%7Bm%7D)
In the second position block has both kinetic and potential energy. Following the law of conservation of energy;
W = ΔE = Kinetic energy + Potantial Energy
W = ΔE = ![\frac{1}{2} mV^{2} + mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mV%5E%7B2%7D%20%2B%20mgh)
Here we can find h in the triangle i draw in the picture using sine theorem;
In a triangle ![\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7BsinA%7D%3D%5Cfrac%7Bb%7D%7BsinB%7D%3D%5Cfrac%7Bc%7D%7BsinC%7D)
In our situation
→ ![h=0,376](https://tex.z-dn.net/?f=h%3D0%2C376)
Therefore
![1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376](https://tex.z-dn.net/?f=1%2C8105%3D%5Cfrac%7B1%7D%7B2%7D%200%2C274V%5E%7B2%7D%20%2B0%2C274.9%2C81.0%2C376)
→ ![V=2,416](https://tex.z-dn.net/?f=V%3D2%2C416)
A magnifying glass is a simple microscope