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Kipish [7]
4 years ago
10

While out hiking, you and your buddy decide to have some fun with the rocks laying all around. While sitting on a high cliff, ov

erlooking a lake, you each drop a rock into the water.
Describe the change in motion of the rock as it enters the water.
Physics
1 answer:
Ulleksa [173]4 years ago
8 0

When we drop the stone in air then during its motion in air the net force on the stone is due to gravity

so its net force is given as

F = mg

here we know that

ma = mg

so we have

a = g

now as soon as it will enter into the water then it will have resistive force due to water as well as buoyant force due to water on it in opposite direction

so the net force on the stone will decrease

so as soon as it will enter into the water the acceleration of stone will decrease due to which the rate of increase in speed of the stone will decrease as it will enter into the water

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You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio
Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

E = \dfrac{\sigma }{2 \varepsilon_o}

E = \dfrac{8 \times 10^{-6} }{2 \times  8.85 \times 10^{-12}}

E = 4.51977 \times 10^5 \ V/m

On the square plate; The electric force F = Eq

F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)

F = 1.3559 \times 10^{-2} \ N

The acceleration a =\dfrac{ F}{m}{

a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}

a = 2.25988 \times 10^6 \ m/s^2

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

(\dfrac{1}{2}) mv^2 = \Delta Vq

v^2 = \dfrac{2 Eq}{dm}

v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }

v^2 = 64568142.86  \ m/s

v =\sqrt{ 64568142.86  \ m/s}

\mathbf{v = 8.035 \times 10^3 \ m/s}

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