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IrinaVladis
1 month ago
5

8) A soccer goal is 2.44 m high. A player kicks the ball at a distance 10.0 m from the goal at an angle of 25.0°. The ball hits

the crossbar at the top of the goal. What is the initial speed of the soccer ball?
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
6 0
The soccer ball's initial speed stands at 16.38 m/s. Given that the vertical distance is y = 2.44 m, the horizontal span x = 10.0 m, and the angle of launch θ = 25.0°. The initial velocity comprises two components, Vₓ and V. The calculations are as follows: Vₓ = V cosθ and V = V sinθ. The formula for horizontal distance becomes x = Vₓt. Since g is deemed 0, we can state that: x = Vₓt or 10 = V cos 25 * t. Solving for V gives us 10 = 0.906V * t, thus V * t = 10 / 0.906 = 11.038 m. Regarding the vertical distance (with g being negative due to the upward movement opposing gravity), we use y = V sinθ * t - 1/2 * g * t². Following through with the calculations leads us to determine that the soccer ball's initial speed is indeed 16.38 m/s.
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Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
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v = 9.037 m/s

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0 = (9.037 m/s)² + 2 × a × 0.5 m
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A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a
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If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

C=\frac{\epsilon_0A}{d}

In this equation, \epsilon_0 refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

C \alpha \frac{1}{d}.......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

\frac{C_1}{C_2} =\frac{d_2}{d_1}......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

E_1=\frac{Q^2}{2C_1}......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

E_2=\frac{Q^2}{2C_2}......(4)

By dividing equation (4) by (3),

\frac{E_2}{E_1} =\frac{C_1}{C_2}

According to equation (2),

\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}

This results in a 3.5 fold increase in energy.

\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

E_1=\frac{1}{2} C_1V^2......(5)

The final energy when the plate separation transitions to d₂ can be written as:

E_2=\frac{1}{2} C_2V^2.....(6)

Referencing equations (5) and (6)

\frac{E_2}{E_1} =\frac{C_2}{C_1}

From equation (2),

\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}

Thus, in this particular scenario,

\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm





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