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
-- loud sounds
-- bright lights
-- strong radio signals
-- Slinkies that can pinch you painfully
-- a tsunami in the ocean
-- earthquakes above Richter 5 or 6
Answer:
40m/s
Explanation:
The horizontal component of velocity remains constant because there are no external forces in that direction
By applying motion equations, V= U+ at
where ,
- v - final velocity
- u - initial velocity
- a-acceleration
- t - time
v = u +at
As no force act on the ball ( we neglect air resistance here) no acceleration is seen,
So v = u = 40m/s
<span>The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.</span>
Answer and Explanation:
Data provided in the question
Force = 50N
Length = 5mm
diameter = 2.0m = 
Extended by = 0.25mm = 
Based on the above information, the calculation is as follows
a. The Stress of the wire is
here area of circle = perpendicular to the are i.e cross-sectional i.e
= 
= 
Now place these above values to the above formula
= 15.92 MPa
As 1Pa = 1 by N m^2
So,
MPa = 10^6 N m^2
b. Now the strain of the wire is
= 
