The atom's most outer shell is full.
Answer:
R = 1.79*10^6 Ω
C = 2.46*10^-6 F
Explanation:
Given
emf of the source, ε = 120 V
Current passing through the resistor, I = 6.7*10^-5 A
Time constant for the circuit, τ = 4.4 s
From the information above, we can say that RC = 4.4 s
Also, on applying the loop rule, we get
ε - IR = 0
ε = IR
R = ε / I
R = 120 / 6.7*10^-5
R = 1.79*10^6 Ω
Using the first equation, we can thus solve for C
RC = 4.4 s
C = 4.4 / R
C = 4.4 / 1.79*10^6
C = 2.46*10^-6 F
C = 2.46 μF
Therefore, the resistance and capacitance of the capacitor is respectively, 1.79 MΩ and 2.46 μF
Answer:
Force, ![F=6\times 10^9\ N](https://tex.z-dn.net/?f=F%3D6%5Ctimes%2010%5E9%5C%20N)
Explanation:
Given that,
Charge 1, ![q_1=3\ C](https://tex.z-dn.net/?f=q_1%3D3%5C%20C)
Charge 2, ![q_2=2\ C](https://tex.z-dn.net/?f=q_2%3D2%5C%20C)
Distance between charges, d = 3 m
We need to find the force between charges. The electric force between charges is given by :
![F=\dfrac{kq_1q_2}{d^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bkq_1q_2%7D%7Bd%5E2%7D)
k is electrostatic constant
![F=\dfrac{9\times 10^9\times 3\times 2}{(3)^2}\\\\F=6\times 10^9\ N](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%203%5Ctimes%202%7D%7B%283%29%5E2%7D%5C%5C%5C%5CF%3D6%5Ctimes%2010%5E9%5C%20N)
So, the force between charges is
. Hence, this is the required solution.
About 2 pounds... Search kg to lbs to be more sure... Have a great day
Answer:
Yes, it is easier to climb a slanted slope than a vertical or more steep slope.
Explanation:
On a vertical slope, you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope