Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards
The electric field due to charge QB at P is EB leftwards
The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
The answer to this question is false
True
The electromagnet will become stronger if we add more coils because there are more field lines in a loop then there is in a straight piece of wire. In a solenoid there are a lot of loops and they are concentrated in the middle, as more loops are added the field lines get larger, therefore making the electromagnet stronger.
Answer:
1.69 T
Explanation:
Applying,
F = BvqsinФ.................. Equation 1
Where F = Force, B = magnetic field, v = velocity, q = charge on an electron, Ф = angle between the electron and the field.
make B the subject of the equation,
B = F/(vqsinФ)............. Equation 2
From the question,
Given: F = 2.0×10⁻¹³ N, v = 7.4×10⁵ m/s, Ф = 90°
Constant: q = 1.60×10⁻¹⁹ C
Substitute into equation 2
B = 2.0×10⁻¹³/(7.4×10⁵×1.60×10⁻¹⁹×sin90°)
B = 0.169×10
B = 1.69 T
