Answer:
The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"
answers:
a.
b. 
Explanation:
First let state the mathematical expression for the tangential acceleration and the radial acceleration.
a. tangential acceleration is express as

since the distance is expressed as

the derivative is the velocity, hence

hence when we take the drivative of the velocity we arrive at
b. the expression for the radial acceleration is expressed as

Answer:
<em>The flux through the sphere will remain the same, and the magnitude of the electric field will increase by four times.</em>
Explanation:
The electric flux is the number of electric field, passing through a given area. It is proportional to the electric field strength and the area through which this field passes.
If the radius of the sphere is halved, the area of the sphere will reduce by square of the reduction, which will be four times. The electric field lines will become closer together, or technically increase by a fourth of its initial value. The resultant effect is that the electric flux will remain the same.
If originally,
Φ = EA cos∅
where Φ is the electric flux through the sphere
E is the electric field on the sphere
A is the area of the sphere.
If the area of the sphere is reduced to half, then,
the area reduces to A/4,
and the electric field increases to be 4E on the sphere.
The flux now becomes
Φ = 4E x A/4 cos∅
which reduces to
Φ = EA cos∅
which is the initial electric flux on the sphere.
Answer:
rmax/rmin = √1.127
Explanation:
F = GmM / r²
As the masses can be assumed to be constant, the force between the two is proportional to the inverse of the square of the distance between them
(Fmax - Fmin) / Fmin = 0.127
(Fmax - Fmin) = 0.127Fmin
1/rmin² - 1/rmax² = 0.127(1/rmax²)
1/rmin² = 0.127(1/rmax²) + 1/rmax²
1/rmin² = 1.127(1/rmax²)
rmax²/rmin² = 1.127
rmax/rmin = √1.127 ≈ 1.06160256...
mech advantage = load/effort = 275/63=4 and a bit
mech adventure ???
The equation of the energy of a photon is E=h*f.
If we increase the Planck's constant h, the energy would increase.
For example, lets double the value of Planck's constant and name it H:
H=2*h. Now lets put that into the equation for energy that we will call E₂:
E₂=H*f=2*h*f=2*E.
So we can clearly see that E₂=2*E or that if we double Planck's constant, the energy also doubles.