Answer:
electrons
Explanation:
The total charge Q is the sum of the charge of the N electrons contained in the sphere:

charge of a electron
We solve to find N:

Answer:
The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.
The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.
Answer:
Explanation:
In the x direction the force will be
½(-w₀)L/2 = -¼w₀L
acting ⅔(L/2) = L/3 below the x axis.
In the y direction the force will be
½(-w₀)L + ½w₀L/2 = -¼w₀L
the magnitude of the resultant will be
F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛
in the direction
θ = arctan(-¼w₀L / -¼w₀L) = 225°
to find the distance, we balance moments
(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]
(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]
(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]
(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L
(√⅛)[d] = 5L/24
d = 5L/24 / (√⅛)
d = 5√⅛L/3
Answer:
im pretty sure it B but I recommend waiting for another person. I used the workdone formula (Force*Dictance*cos(theta) and got 55 Joules
Explanation:
Explanation:
Given that,
Radius R= 2.00
Charge = 6.88 μC
Inner radius = 4.00 cm
Outer radius = 5.00 cm
Charge = -2.96 μC
We need to calculate the electric field
Using formula of electric field

(a). For, r = 1.00 cm
Here, r<R
So, E = 0
The electric field does not exist inside the sphere.
(b). For, r = 3.00 cm
Here, r >R
The electric field is

Put the value into the formula


The electric field outside the solid conducting sphere and the direction is towards sphere.
(c). For, r = 4.50 cm
Here, r lies between R₁ and R₂.
So, E = 0
The electric field does not exist inside the conducting material
(d). For, r = 7.00 cm
The electric field is

Put the value into the formula


The electric field outside the solid conducting sphere and direction is away of solid sphere.
Hence, This is the required solution.