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Hitman42 [59]
3 years ago
8

Sam stands on a 20 m high cliff and throws a 45 g rock with an initial velocity of 5 m/s [forward] to the water below. Use the c

onservation of energy to determine the speed of the rock when it has fallen 12 m.
Physics
1 answer:
storchak [24]3 years ago
5 0

Answer:

v = 12.52 [m/s]

Explanation:

To solve this problem we must use the energy conservation theorem. Which tells us that potential energy is transformed into kinetic energy or vice versa. This is more clearly as the potential energy decreases the kinetic energy increases.

Ep = Ek

where:

Ep = potential energy [J] (units of joules]

Ek = kinetic energy [J]

Ep = m*g*h

where:

m = mass of the rock = 45 [g] = 0.045 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = (20 - 12) = 8 [m]

Ek = 0.5*m*v²

where:

v = velocity [m/s]

The reference level of potential energy is taken as the ground level, at this level the potential energy is zero, i.e. all potential energy has been transformed into kinetic energy. In such a way that when the Rock has fallen 12 [m] it is located 8 [m] from the ground level.

m*g*h = 0.5*m*v²

v² = (g*h)/0.5

v = √(9.81*8)/0.5

v = 12.52 [m/s]

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Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

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Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

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Using equation (1), the potential difference between points A and B is:

V_{1}=\frac{W_{1} }{q}

Substitute the suitable values in the above equation.

V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }

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V_{2}=\frac{W_{2} }{q}

Substitute the suitable values in the above equation.

V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

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