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8 days ago

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Now Abel and Kato use what they learned to answer the following problem. The initial speed of a tennis ball is 54 m/s and the la

unch angle is θi = 21°. Neglect air resistance. What is the maximum height, h, of the tennis ball? m What is the range, R, of the tennis ball?

1 answer:

kicyunya [2.9K]8 days ago

6 0

The answer is h = 19.4 m and R = 199.07 m. Explanation: To address this problem, we utilize parabolic motion equations. Initial speed is defined as 54 m/s with a launch angle θi = 21°. The vertical initial speed component is calculated, leading to a maximum height of 19.1 m and a range of 199.07 m upon substituting values into the relevant formulas.

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Answer:

$n_s = 920 \turns$

Explanation:

Given,

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number of turns in the primary coil (n_p) = 80 turns

number of turns in the secondary coil (n_s) =?

the ratio of turns between primary and secondary coils

$\dfrac{n_p}{n_s} = \dfrac{V_p}{V_s}$

$\dfrac{n_s}{n_p} = \dfrac{V_s}{V_p}$

$n_s = n_p \dfrac{V_s}{V_p}$

$n_s = 80\times \dfrac{345}{30}$

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The number of turns in the secondary coil is equal to $n_s = 920 \turns$

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1 month ago

Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, su

inna [2740]

The solutions and methods related to the exercise are included in the attached files. A detailed explanation of each step is provided below. In the archive below, you will find the required procedures, formulas, or explanations. Feel free to ask if you have any questions, and I will gladly resolve your uncertainties.

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22 days ago

A 6000 kg lorry is reversing into a parking space at a speed of 0.5 m/s but collides with a car. The crumple zone of the car sto

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Answer:

3000 kg.m/s

Explanation:

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14 days ago

A 1.47-newton baseball is dropped from a height of 10.0 meters and falls through the air to the ground. The kinetic energy of th

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Answer:

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1 month ago

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Response:

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Clarification:

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$v=4\times 10^7 m/s$

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Knowing that 1m=100 cm

Our goal is to determine the electric field intensity.

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Applying the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

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$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Electron mass, m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Plugging in the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

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18 days ago