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)
Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
Ohms law = v= Ir
V= 0.02 x 4000 = 80v
Answer:
a)54L/min
b)0.845
Explanation:
a) A x V=
where suffix 1,2,3 refers to the three pipes.
=27L/min+16L/min+11 L/min
=54L/min
b) A x V=54L/min =>
x v
d= 2 cm
x v = 54
v=
x
->
x
=27L/min =>
x 
= 1.3cm
x
= 27
=
x 
Next is to find the ratio of speed i.e 
x
/
x
=>

= 0.845
To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.
If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

Replacing,


The total speed change
we have that the value is 0.71m/s
If we know that acceleration is the change of speed in a fraction of time,

We have that,


Therefore the Rocket should be fired around to 1403.16s