Answer:
The runner's acceleration was
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
Solving for a:
The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:
The runner's acceleration was
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