a 9.0-kg dog runs at 4.0 m/s and jumps onto a stationary skateboard the mas of the skateboard is 1.0 m/s what speed is the speed
1 answer:
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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:
3.82 Ns
Explanation:
Time varying horizontal Force is given as
F(t) = A t⁴ + B t²
F(t) = 4.50 t⁴ + 8.75 t²
Impulse imparted is given as
Answer:
Part A
The intensity is
Part B
The intensity is
Explanation:
From the question we are told that
The intensity of the light detected by first eye is
Now at initial state according the question the light ray is perpendicular to the eye so it means that it is at 90° the eye
Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light )would detect when the head is rotated by 20° its previous orientation
This is mathematically evaluated as
Now the second question is to obtain the intensity the first eye (the first eye in this case is the one that is not focused on the light )would detect when the head is rotated by 20° its previous orientation
Now in this case the angle between the eye and the light is 90-20 = 70°
So