Answer:
ρ/ρ2 = 3 / R₀ the two densities are different
Explanation:
Density is defined as
ρ = M / V
As the nucleus is spherical
V = 4/3 π r³
Let's replace
ρ = A / (4/3 π R₀³)
ρ = ¾ A / π R₀³
b)
ρ2 = F / area
The area of a sphere is
A = 4π R₀²
ρ2 = F / 4π R₀²
ρ2 = F / 4π R₀²
Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.
Let's look for the relationship of the two densities
ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)
ρ /ρ2 = 3 (A / F) (1 / R₀)
In this case it does not say that the nucleon number is A (F = A), the relationship is
ρ/ρ2 = 3 / R₀
I see that the two densities are different
I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s
Answer:
2.96 × 10^4 N
Explanation:
1 atm = 101325 N/m², pressure inside the airtight room = 1.02 atm, pressure outside due to hurricane = 0.91 atm
net pressure directed outward = P inside - P outside
net pressure = 1.02 - 0.91 = 0.11 atm
where 1 atm = 101325N/m²
0.11 atm = 0.11 × 101325 N/m² = 11145.75 N/m²
area of the square wall = l × l where l is the length of the wall in meters = 1.63 × 1.63 = 2.6569
net pressure = net force / area
make net force subject of the formula
net force = net pressure × area = 11145.75 × 2.6569 = 2.96 × 10 ^4 N
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Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
