Answer:
a.
b. 
Explanation:
<u>Given:</u>
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
<h2>
(a):</h2>
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.
At time t = 3 seconds,
<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>
(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.
For the time interval of 2 seconds,
The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,
It is the displacement of the particle in 2 seconds.
Answer:
Kilogram(kg) is the SI unit for mass while kilometre(km) is a unit for length. They are both similar in that they are 10^3 of a unit, thus kilo. As kilogram represents mass, it is a measure of how much matter is present in an object. While kilometre is a measure of distance/how long or short an object is.
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
