Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C 
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
Your answer is false since science is a subject not a religion
Answer:
s = 23.72 m
v = 21.56 m/s²
Explanation:
given
time to reach the ground (t) = 2.2 second
we know that
a) s = u t + 0.5 g t²
u = 0 m/s
g = 9.8 m/s²
s = 0 + 0.5 × 9.8 × 2.2²
s = 23.72 m
b) impact velocity
v = √(2gh)
v = √(2× 9.8 × 23.72)
v = √464.912
v = 21.56 m/s²
There are two particular cases, the first is when Object A is attracted to the neutral wall. This would indicate that the object is not neutral, as there is an attraction.
At the same time we know that Object A is attracted to an object B. And therefore, the load of A must be opposite to that of B. Remember that opposite charges attract each other. If the charge of object B is positive, then the charge of object A will be negative.
Option B is correct: It has a negative charge.
1st Blank Answer: Balanced
2nd Blank Answer: 0n
I took the test and got it right
I would appreciate a Brainliest...
