Answer:
Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 
Explanation:
The potential energy is given by:
U=Q*V
where:
Q is the charge
V is the potential difference
Potential Difference=V=
So,
Where:
k is Coulomb Constant=
q is the charge on electron=
r is the distance=
For 3 Electrons Potential Energy or work Done is:
Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is
The acceleration of the electron is larger than the acceleration of the proton.
The reason for this is that the mass of the electron is smaller (about 1000 times smaller) than the mass of the proton. The two particles have same charge (e), so they experience the same force under the same electric field E:
However, according to Newton's second law, the force is the product between the mass particle, m, and its acceleration, a:
which can be rewritten as
we said that the force exerted on the two particles, F, is the same, while the mass of the electron is smaller: therefore, from the last formula we see that the acceleration of the electron will be larger than that of the proton.
Athletes of any kind can use technology to monitor things like their BMI (body mass index) or weight or anything of the sort to target what they need to work on to improve their health.
For the purpose we will use the following equation for potential energy:
U = m * g * h
In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth).
When we plug values into the equation, we get following:
U= 65.7kg * 9.8 N/kg *135m = 86921.1 J = 86.92 kJ
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v = 
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t = 
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
