The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank
1 answer:
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Answer:
g' = g/4
Explanation:
- The value of the free-fall acceleration at the surface of the earth, can be obtained applying Newton's 2nd law, assuming that the only force acting on an object at the surface of the earth, is the one produced by the mass of the Earth, i.e. gravity.
- This force can be expressed according the Newton's Universal Law of Gravitation , as follows:
- From Newton's 2nd Law, we have:
- Since the left sides in (1) and (2) are equal each other, both right sides must be equal each other also.
- Simplifying the mass m, we can write the acceleration a in (2) as the acceleration due to gravity, g, as follows:
- Since G is an universal constant, and the mass mE remains constant, if we double the radius of the Earth, the new value for the acceleration due to gravity (let's call it g'), is as follows:
Answer:
a) E = k Ze (1- r³ / R³) 1/r², b) E=0, c) E = -6.62 10¹⁰ N / C
Explanation:
a) For this we can use the law of Gauus
Ф = E- dA = / ε₀
where we take a sphere as a Gaussian surface, so that the electric field lines and the radii of the sphere are parallel, consequently the dot product is reduced to the algebraic product
E A =q_{int} / ε₀
the area of a sphere
A = 4π r²
E 4π r² = q_{int} / ε₀
E = 1 / 4πε₀ q_{int} / r²
k = 1 /4π ε₀
E = k q_{int} / r² (1)
let's analyze the charge inside the gaussian sphere,
let's use the concept of density for electrons, since they indicate that the charge is evenly distributed
ρ = Q / V
where the volume of the sphere is
V = 4/3 πr³
Qe = ρ V
Qe = ρ 4 / 3π r³
the density of the electrons is
ρ = Ze 3 / (4π R³)
where R is the atomic radius
we substitute
Qe = Ze r³/ R³
for protons they are in a very small space, the atomic nucleus, so we can superno that they are a point charge.
The net charge inside our Gaussian surface, the charge of the protoens plus the charge of the electroens (Qe)
q_{int} = q_proton + Q_electron
q_{int} = + Ze - Qe
q_{int} = + Ze - Ze r³ / R³
q_{int} = Ze (1- r³ / R³)
we substitute in equation 1
E = k Ze (1- r³ / R³) 1/r²
b) on the surface of the atom r = R
therefore the electric field is zero
E = 0
c) Calculate the electric field for the Uranium for
r = R / 2 = 0.10 10⁻⁹ / 2 = 0.05 10⁻⁹ = 5 10⁻¹¹ m
E = 8.99 10⁹ 92 1.6 10⁻¹⁹ (1-1/2)³ 1/ (5 10⁻¹¹)²
E = -6.62 10¹⁰ N / C