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
Answer:
it’s transparent to all visible light
step-by-step explanation:
translucent objects allow some light to travel through them
Answer:
<h2>
m/s ^2</h2><h2 />
Explanation:
Solution,
When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s
Initial velocity ( u ) = 0 m/s
Final velocity ( v ) = 72 km/h ( Given)
We have to convert 72 km /h in m/s
m/s
Final velocity ( v ) = 20 m/s
Time taken ( t ) = 2 seconds
Acceleration (a) = ?
Now,
we have,
m/s ^2
Hope this helps...
Good luck on your assignment..
Answer:
0.0979 N/c
Explanation:
Electric field, E is given as a product of resistivity and current density
E=jP where P is resistivity and j is current density
But the current density is given as
where I is current and A is area and 
Substituting this into the first equation then 
Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
