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3 years ago

A person on a merry-go-round is constantly accelerating away from the center?true or false

Physics

2 answers:

netineya [11]3 years ago

7 0

Answer:

False

Explanation:

Lemme know if you need an explanation! :))

Alisiya [41]3 years ago

4 0

That statement is false. It's exactly the opposite direction.

In circular motion, the "centripetal" force ... and therefore the centripetal acceleration ... points TOWARD the center.

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The sun orbits the eth at 2kilogram per sec

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3 years ago

How much work does this force do as the particle moves along the x-axis from x = 0 to x = l? express your answer in terms of the

nydimaria [60]

<h3><u>Answer</u>;</h3>

= F0 L ( 1 - 1/e )

<h3><u>Explanation;</u></h3>

Work done is given as the product of force and distance.

In this case;

Work done = ∫︎ F(x) dx

= F0 ∫︎ e^(-x/L) dx

= F0 [ -L e^(-x/L) ] between 0 and L

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3 years ago

The synodic period for a planet is different from its sidereal period because

IrinaK [193]

<span>Earth (and hence the observer) moves.</span>

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3 years ago

Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?

Ksenya-84 [330]

Answer:

The answer is " $6\ \frac{m}{s}$ "

Explanation:

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$\to \overline{v}={\frac{\Delta x}{\Delta t}}$

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7 0

3 years ago

The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,

madam [21]

Answer:

<u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

E = hf

E = h.v/λ

λ = h/(mv)

E = (1/2)mv²

Where:

E = kinetic energy of the particle
λ = de-Broglie wavelength
m = mass of the particle
v = speed of the particle
h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

$\dfrac{m_\alpha}{m_p}=4$

For the kinetic energies you find:

$\dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}$

$\dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4$

Thus:

$\dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}$

$m_\alpha v_\alpha=m_pv_p$

From de-Broglie equation, λ = h/(mv)

$\dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1$

5 0

3 years ago

A person on a merry-go-round is constantly accelerating away from the center?true or false ​

A person on a merry-go-round is constantly accelerating away from the center?true or false