Answer:
a=3.53 m/s^2
Explanation:
Vo=0 m/s (because he is not moving at the start)
V1=15 m/s
t= 4.25 s
a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
a. metallic bond
b. the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “cloud” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.
c. due to the presence of free electrons in its outer energy levels
Answer:
yes
Explanation:
objects with constant velocity also have zero net external force. this means the forces on the object are balanced. this mean they are in equilibrium
<h2>
Time taken is 0.459 seconds</h2>
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 81 km/hr = 22.5 m/s
Time, t = ?
Acceleration, a = 49 m/s²
Substituting
v = u + at
22.5 = 0 + 49 x t
t = 0.459 seconds
Time taken is 0.459 seconds