Explanation:
Given that,
Electric field = 5750 N/C
Charge 
Distance = 5.50 cm
(a). When the charge is moved in the positive x- direction
We need to calculate the change in electric potential energy
Using formula of electric potential energy



Put the value into the formula


The change in electric potential energy is 
(b). When the charge is moved in the negative x- direction
We need to calculate the change in electric potential energy
Using formula of electric potential energy



Put the value into the formula


The change in electric potential energy is 
Hence, This is the required solution.
Answer:
176,000 N
Explanation:
Newton's second law:
∑F = ma
F = (4 × 40,000 kg) (1.1 m/s²)
F = 176,000 N
Answer: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4 - Chemical Equation Balancer
Equation is already balanced.
Explanation: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4
The solution to the questions are given as


- the direction of induced current will be Counterclock vise.
<h3>What is the direction of the
current induced in the loop, as viewed from above the loop.?</h3>
Given, $B(t)=(1.4 T) e^{-0.057 t}$




(b) 

c)
In conclusion, the direction of the induced current will be Counterclockwise.
Read more about current
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Answer:
a) E = -4 10² N / C
, b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
-
= m a
Axis y
- W = 0
Initially the system is in equilibrium, so zero acceleration
Fe =
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ =
/ T
T_{y} = cos θ
= T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
= q E
E =
/ q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18
⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
= m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes