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Evgen [1.6K]
4 years ago
11

Can you access an instance variable from a static method? explain why or why not.

Physics
1 answer:
marusya05 [52]4 years ago
5 0
<span>The reason a static method can't access instance variable is because static references the class not a specific instance of the class so there is no instance variable to access.</span>
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Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4
Zepler [3.9K]

Answer: 5.214(10)^{11} N

Explanation:

According to <u>Coulomb's Law:</u>  

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".</em>

<em />

Mathematically this law is written as:

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}  

Where:

F_{E}  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1}=5.8 C and q_{2}=6.4 C are the electric charges

d=80 cm \frac{1 m}{100 cm}=0.8 m is the separation distance between the charges

Solving:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}  

F_{E}=5.214(10)^{11} N    

7 0
4 years ago
3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a
34kurt

Answer:

The friction force and the x component for the weight should be the reaction forces that are opposite and equal to the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.  

Explanation:

<u>When the locomotive starts to pull the train up, appears two reaction forces opposed to the action force in the direction of the move. </u>

The first one is due to the friction between the wheels and the ground, it will be the friction force (Fr):

Fr = μ*Pₓ =μmg*sin(φ)        

<em>where μ: friction dynamic coefficient, Pₓ: is the weight component in the x-axis, m: total mass = train's mass + locomotive's mass, g: gravity, and sin(φ): is the angle respect to the x-axis.</em>              

And the second one is the x component for the weight (Wₓ):

Wₓ = mg*cos(φ)  

<em>where cos(φ): is the angle respect to the y-axis.    </em>

<em> </em>

These two forces should be the same as the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.          

3 0
4 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
3 years ago
The products of a chemical reaction can never have greater mass than the reactants
jek_recluse [69]
The answer to this is true
6 0
3 years ago
Read 2 more answers
can someone change the words in this (the change in position of one object compared to the position of another ​
adell [148]

change of an object relative to the position of another object.

3 0
4 years ago
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