The fine horizontal scratches etched onto a bullet after it has been fired are called Striations.
<h3>What is Striations?</h3>
- Striations are the minute differences in the curve of the bullet's surface. The bullet and the gun barrel are the harder and softer materials, respectively, in firearms evidence.
- The striations left on the fired bullet are used as a comparison by firearm examiners.
- When a gun is discharged, the bullet blasts down the barrel where it strikes ridges and grooves, spinning and improving shot accuracy.
- These ridges cause striations in the bullet's soft metal by digging into it.
- An examiner analyses these distinctive markings to verify whether a given bullet was shot from a specific firearm.
- A barrel will produce individual markings in addition to a bullet's land and groove impressions as the projectile passes through.
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Answer:
150000000

49050000 N/C
Explanation:
q = Charge = 24 pC
m = Mass of honeybee = 0.12 g
E = Electric field = 100 N/C
g = Acceleration due to gravity = 9.81 m/s²

Number electrons is

The number of electrons added or removed was 150000000
Force is given by

The ratio is

The ratio is 
Balancing the forces we get

The electric field required is 49050000 N/C
It can help with measurements and when you want to add measurements to a cylinder or a beaker so ya
Answer:
8F_i = 3F_f
Explanation:
When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.
Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.
Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.
The electrostatic force, Fi, in the initial configuration can be calculated as follows.
![F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f](https://tex.z-dn.net/?f=F_i%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7BQ%5E2%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3EThe%20electrostatic%20force%2C%20Ff%2C%20in%20the%20final%20configuration%20is%20%3C%2Fp%3E%3Cp%3E%5Btex%5DF_f%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B3Q%5E2%2F8%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3ETherefore%2C%20the%20relation%20between%20Fi%20and%20Ff%20is%20as%20follows%3C%2Fp%3E%3Cp%3E%5Btex%5DF_i%20%3D%20F_f%5Cfrac%7B3%7D%7B8%7D%5C%5C8F_i%20%3D%203F_f)
<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>
➷ Earth's gravity is approximately 9.81
weight = mass x gravity
weight = 4.6 x 9.81
weight = 45.126
Answer is B. 45N
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