A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω
1 answer:
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We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
which means
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (
):
Therefore, the kinetic energy gained by the proton is
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>
which means
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (
Therefore, the kinetic energy gained by the proton is
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, , is given by the sum of the rock,
, and the water, V, is given as follows;
=
+ V
= A × h₂
∴ = 10 cm² × 12 cm = 120 cm³
The total volume, = 120 cm³
The volume of the rock, =
- V
∴ = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³