Answer:
It would be the brick since the mass and the weight is greater than the peanut I might be wrong on this but thats the best answer i can give
Explanation:
Answer:
The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m
Explanation:
Data
F = 2 N
d = ?
q = 1.6 x 10 ⁻¹⁹ C
k = 8.987 Nm²/C²
Formula
Solve for r
Substitution
Simplification
r = 
r = 
Result
r = 1.07 x 10⁻¹⁴ m
A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.
<h3>What is the right-hand thumb rule?</h3>
Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.
A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.
The proton will be subject to a magnetic pull that is directed into the page.
Hence, option B is correct.
To learn more about the right-hand thumb rule refer to the link;
brainly.com/question/11521829
#SPJ1
Answer:
The inside Pressure of the tank is 
Solution:
As per the question:
Volume of tank, 
The capacity of tank, 
Temperature, T' = 
= 299.8 K
Temperature, T = 
= 288.2 K
Now, from the eqn:
PV = nRT (1)
Volume of the gas in the container is constant.
V = V'
Similarly,
P'V' = n'RT' (2)
Also,
The amount of gas is double of the first case in the cylinder then:
n' = 2n
![\]frac{n'}{n} = 2](https://tex.z-dn.net/?f=%5C%5Dfrac%7Bn%27%7D%7Bn%7D%20%3D%202)
where
n and n' are the no. of moles
Now, from eqn (1) and (2):
These are the most common type of faults not just inductors but also with other elements too like resistors,transformers, generators etc.
open circuit fault means the flow of current is disrupted some how in the circuit and the circuit stops operating. and for short circuit fault the current in the system will be pretty high and this short circuit current or fault current will always run back to the fault location, if the inductor got short circuited somehow then the fault current will only run through it because it will then provide a very low impedence path
