Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B carries a charge of -q. Sphere C
1 answer:
Answer:
Part a)
Final charge on C : q = 1.875
Part b)
Ratio for A = 6 : 1.25
Ratio for B = -1 : 1.875
Ratio for C = 0
Explanation:
When two identical metal sphere are connected together then the charge on them will get equally divided on both after connecting them by conducting wire
So here we have
Step 1: We connected A and B and then separate them
so we have
Step 2: We connected A and C and then separate them
so we have
Step 3: We connected B and C and then separate them
so we have
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Answer:
Maximum permitted surface crack length is 1.29 mm
Explanation:
As per the question:
Tensile stress,
Surface energy of magnesium oxide, SE =
Modulus of elasticity of the material, E = 225 GPa =
Now,
To calculate the maximum allowable surface crack length:
You have two different phases of the star: 1 the Sun-like phase and 2 the neutron star phase. The given quantities are:
r₁ = r(Sun) = 695700km
m₁ = 8 Msun
f₁ = 1 rev / 12 days
m₂ =
·m₁
First thing, you need to transform the frequency in units of revolution/seconds
f₁ = 1 rev / (12·24·60·60) = 1 rev / 1036800 s
and then into angular velocity through the formula
ω₁ = 2πf = 6.06E-6 rad/s
a) If the angular momentum stays the same: L₁ = L₂
where L = I·ω
and the momentum of inertia I is given by I = m·r²
Therefore, substituting we have:
m₁·r₁²·ω₁ = m₂·r₂²·ω₂
And we can find:
ω₂ = <span>
</span>
(remember m₂=·m₁ so we can cancel out the two m₁)
We obtain:
ω₂ = (<span>695700²·6.06E-6)/(0.25·12²) = 81473.1 rad/s
we can transform it back into frequency:
</span>f₂ = ω₂/2<span>π = 1 rev / 12967 s = 7.7e-5 rev/s
b) If L</span>₂ =·L₁
we do expect an angular velocity 4 times smaller.
Using the same formulas as above:
·m₁·r₁²·ω₁ = ·m₁·r₂²·ω₂
ω₂ = <span>
=
= </span><span>[(695700²·6.06E-6)(12²) = 20368 rad/s
</span>and
f₂ = 1 rev / 3242 s = 3.08e-4 rev/s
r₁ = r(Sun) = 695700km
m₁ = 8 Msun
f₁ = 1 rev / 12 days
m₂ =
First thing, you need to transform the frequency in units of revolution/seconds
f₁ = 1 rev / (12·24·60·60) = 1 rev / 1036800 s
and then into angular velocity through the formula
ω₁ = 2πf = 6.06E-6 rad/s
a) If the angular momentum stays the same: L₁ = L₂
where L = I·ω
and the momentum of inertia I is given by I = m·r²
Therefore, substituting we have:
m₁·r₁²·ω₁ = m₂·r₂²·ω₂
And we can find:
ω₂ = <span>
(remember m₂=
We obtain:
ω₂ = (<span>695700²·6.06E-6)/(0.25·12²) = 81473.1 rad/s
we can transform it back into frequency:
</span>f₂ = ω₂/2<span>π = 1 rev / 12967 s = 7.7e-5 rev/s
b) If L</span>₂ =
we do expect an angular velocity 4 times smaller.
Using the same formulas as above:
ω₂ = <span>
= </span><span>[(695700²·6.06E-6)(12²) = 20368 rad/s
</span>and
f₂ = 1 rev / 3242 s = 3.08e-4 rev/s