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daser333 [38]
3 years ago
8

A ball is dropped from rest from the top of a cliff that is 15.0 m high. From ground level, a second ball is thrown straight upw

ard at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths.
Physics
1 answer:
lesya [120]3 years ago
7 0

Answer:

3.75 meters below the top of the cliff.

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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m
Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

F = \dfrac{kq_1q_2}{r^2}

F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}

  F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}

    a = 9 x 10²² m/s²

c) speed of the electron

 a =\dfrac{v^2}{r}

 9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}

   v² = 4.77 x 10¹²

  v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

    T=\dfrac{2\pi}{\omega}

    T=\dfrac{2\pi r}{v}

    T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}

          T = 1.53 x 10⁻¹⁶ s

8 0
3 years ago
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Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable,
Charra [1.4K]

To develop this problem it is necessary to apply the concepts related to the kinematic equations of motion. And from the speed found the relationships between wavelength, frequency and last of the period (which is inversely proportional to the frequency)

PART A) We know that the velocity of a body or a wave is equivalent to the distance traveled over a time interval. So,

V = \frac{x}{t}

Where

x = Distance

t = time

V = \frac{3650*10^{3}}{4.59h(\frac{3600s}{1h})}

V = 215.44m/s

PART B) The frequency would then be defined as

f = \frac{V}{\lambda}

Where

\lambda = Wavelength

f = \frac{215.44}{732*10^{3}}

f = 2.943*10^{-4}Hz

PART C) Finally the period is defined as

T = \frac{1}{f}

T = \frac{1}{2.943*10^{-4}}

T = \frac{1}{2.943*10^{-4}}

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5 0
3 years ago
A 23.7 kg kid slides down a
anyanavicka [17]

Answer:

\text { The acceleration of the kid is } 7.18 \mathrm{m} / \mathrm{s}^{2}

Explanation:

Mass of the kid 23.7 kg.

\text { The kid is accelerating down at an angle is } 47.2^{\circ} .

^{\prime \prime} \mathrm{g}^{\prime \prime} \text { acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}

We need to find the acceleration of the kid,

We know that, Parallel force acted on the kid at an angle is

F = m × g × sinθ (F = ma)

m × a = m × g × sinθ

Now, substitute the given values in the above formula to find acceleration of the kid,

23.7 \times a=23.7 \times 9.8 \times \sin 47.2^{\circ}

23.7 × a = 232.26 × 0.733

23.7 × a = 170.24

a=\frac{170.24}{23.7}

a=7.18 \mathrm{m} / \mathrm{s}^{2}

\text { Therefore, acceleration of the kid is } 7.18 \mathrm{m} / \mathrm{s}^{2}

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This is my physics homework
raketka [301]

Answer:

Option B is the correct answer

Explanation:

Atomic number of Americium is 95 however in option B it is given as 94.

Thus, option B is the correct answer.

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How to use proton, neutron, and isotope in a sentence
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An isotope of an element contains the same number of protons as the original element however, the number of neutrons depends on the different isotopes of the element.
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