Linear expansivity, area expansivity and volume or cubic expansivity are
Answer:
Ф_cube /Ф_sphere = 3 /π
Explanation:
The electrical flow is
Ф = E A
where E is the electric field and A is the surface area
Let's shut down the electric field with Gauss's law
Фi = ∫ E .dA = 
/ ε₀
the Gaussian surface is a sphere so its area is
A = 4 π r²
the charge inside is
q_{int} = Q
we substitute
E 4π r² = Q /ε₀
E = 1 / 4πε₀ Q / r²
To calculate the flow on the two surfaces
* Sphere
Ф = E A
Ф = 1 / 4πε₀ Q / r² (4π r²)
Ф_sphere = Q /ε₀
* Cube
Let's find the side value of the cube inscribed inside the sphere.
In this case the radius of the sphere is half the diagonal of the cube
r = d / 2
We look for the diagonal with the Pythagorean theorem
d² = L² + L² = 2 L²
d = √2 L
we substitute
r = √2 / 2 L
r = L / √2
L = √2 r
now we can calculate the area of the cube that has 6 faces
A = 6 L²
A = 6 (√2 r)²
A = 12 r²
the flow is
Ф = E A
Ф = 1 / 4πε₀ Q/r² (12r²)
Ф_cubo = 3 /πε₀ Q
the relationship of these two flows is
Ф_cube /Ф_sphere = 3 /π
Cover your cough
Wash your hands
Keep a clean space
Avoid smoking
Stay active
Hope this helps :)
1 year = (365 / 121) = 3.02 half-lifes. Let's call it 3 .
The amount of radioactive isotope remaining after 3 half-lifes is
(1/2) x (1/2) x (1/2) = 1/8
A year after the medical lab received the 24 kg of W-181,
there will still be 24 kg of stuff in the container.
But only 3 kg of it will still be W-181. The other 21 kg will be
whatever substances W-181 becomes when it decays.
Sadly, even the 3 kg of good stuff won't be usable anymore ...
it'll be thoroughly mixed with the 21 kg of junk. It would be harder
and more expensive to try and separate them than to buy a new
can of pure W-181, and USE it before 7/8 of it has deteriorated.
Answer:
D. 9 N
Explanation:
The tension on the string is equivalent to the centripetal force.
Centripetal force is the force exerted by an object in circular motion or path towards the center of the circular path.
Centripetal force = mv²/r
where m is the mass of the object, v is the velocity and r is the radius of the circular path.
Centripetal force = (0.25 kg × 6²)/ 1
= 9 N
Thus, the tension on the string is 9 N
