If you are referring to science, then you are correct. A period is also known as a family.
Answer:
11.28 N toward the center of the track
Explanation:
Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.
The formula for centripetal force is given as,
F = mv²/r................................ Equation 1
Where F = force, m = mass of the toy car, v = velocity, r = radius
Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m
Substitute into equation 1
F = 0.108(7.75²)/0.575
F = 11.28 N
Hence the magnitude and direction of the force = 11.28 N toward the center of the track
True
Most photographs will be made using a shutter speed of 1/60 or faster.
Becomes older
Explanation:
As sea floor spreading occurs at divergent margins, the oceanic plate becomes older. Younger plate margin are the closest to the margin whereas the older plates bushes backward away from the spreading centers.
- The idea that the sea floor spreads was postulated by Harry Hess shortly after the second world war around the 1960's.
- At divergent margins new crust materials from the mantle are brought to the surface.
- They crystallize and settle at the flanks of plate margins.
- Older ones are pushed backward away from the margin into far away subduction zones.
Learn more:
Sea floor spreading brainly.com/question/9912731
#learnwithBrainly
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
