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3 years ago

If you dribble a basketball with a frequency of 1.77 Hz, how long does it take for you to complete 12 dribbles?

Physics

1 answer:

Licemer1 [7]3 years ago

8 0

<h2>It takes 6.78 seconds to complete 12 dribbles.</h2>

Explanation:

Frequency of dribble = 1.77 Hz

That is

Number of dribbles in 1 second = 1.77

$\texttt{Time taken for 1 dribble = }\frac{1}{1.77}=0.565s$

Now we need to find how long does it take for you to complete 12 dribbles.

Time taken for 12 dribbles = 12 x Time taken for 1 dribble

Time taken for 12 dribbles = 12 x 0.565

Time taken for 12 dribbles = 6.78 seconds

It takes 6.78 seconds to complete 12 dribbles.

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What would be new resistance if length of conductor is doubled and thickness is halved

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A playground slide is in the form of an arc of a circle with a maximum height of 3.0 m, with a radius of 8.5 m, and with the gro

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Answer:

a) $s \approx 6.676\,m$

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The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:

$U_{g,A} = K_{B} + W_{fr}$

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Read 2 more answers

Two black holes (the remains of exploded stars), separated by a distance of

jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

distance between the two black holes, r = 10 AU = 1.5 x 10¹² m
gravitational force between the two black holes, F = 6.9 x 10²⁵ N.
combined mass of the two black holes = 5.20 x 10³⁰ kg

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0

solve the quadratic equation using formula method;

a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰

$m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20}) \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg$

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg - 4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

m₂ = 5.2 x 10³⁰ kg - 4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

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