Answer:
A. 181.24 N
Explanation:
The magnitude of hte electrostatic force between two charged objects is given by the equation
where
k is the Coulomb's constant
q1, q2 are the magnitudes of the two charges
r is the separation between the charges
In this problem, we have:
is the magnitude of the 1st charge
is the magnitude of the 2nd charge
r = 2.5 cm = 0.025 m is the separation between the charges
Therefore, the magnitude of the electric force is:
So, the closest answer is
A) 181.24 N
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
Answer:
6.5e-4 m
Explanation:
We need to solve this question using law of conservation of energy
Energy at the bottom of the incline= energy at the point where the block will stop
Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=
Energy at the point where the block will stop consists of only gravitational potential energy=
Hence from Energy at the bottom of the incline= energy at the point where the block will stop
⇒
⇒
Also
where is the mass of block
is acceleration due to gravity=9.8 m/s
is the difference in height between two positions
⇒
Given m=2100kg
k=22N/cm=2200N/m
x=11cm=0.11 m
∴
⇒
⇒
⇒h=0.0006467m=