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3 years ago

Define personal health.

Physics

2 answers:

nekit [7.7K]3 years ago

7 0

Answer:

Personal Health is the ability to take charge of your health by making conscious decisions to be healthy.

mafiozo [28]3 years ago

4 0

Answer:Personal Health is the ability to take charge of your health by making conscious decisions to be healthy. It not only refers to the physical well being of an individual but it also comprises the wellness of emotional, intellect, social, economical, spiritual and other areas of life.

Explanation:

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What is the correct term for the flow of charged particles?

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Answer: An electric charge

Explanation:

Because it is carried by charged particles

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3 years ago

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Which pair of sentences is describing the same velocity? A car is parked. A car is moving in circles. A bus drives 40 miles per

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The bus and the truck have the same velocity.

Also, Valerie and Owen have the same velocity.

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Two objects of mass m move in opposite directions toward each other. The green object moves at velocity v, and the blue object m

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. The velocity of a mass attached to a spring is given by v = (1.5 cm/s) sin(ωt + π/2), ..... Which of the following is the motion of objects moving in two dimensions

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4 years ago

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The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 9.00×1010 atoms of zinc e

KengaRu [80]

Answer:

$8.4\cdot 10^{-8}J$

Explanation:

The energy emitted by a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

For the photons emitted by the zinc atoms,

$\lambda = 214 nm = 214 \cdot 10^{-9} m$

So the energy of a single photon emitted is

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{214\cdot 10^{-9}}=9.3\cdot 10^{-19}J$

And since the number of atoms is

$N=9.0\cdot 10^{10}$

The total energy emitted will be

$E=NE_1 = (9.0\cdot 10^{10})(9.3\cdot 10^{-19})=8.4\cdot 10^{-8}J$

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4 years ago

Two hypothetical planets of masses m1 and m2 and radii r1 and r 2 , respectively, are nearly at rest when they are an infinite d

Leto [7]

Answer:

(a) $v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) Kinetic Energy of planet with mass m₁, is KE₁ = 1.068×10³² J

Kinetic Energy of planet with mass m₂, KE₂ = 2.6696×10³¹ J

Explanation:

Here we have when their distance is d apart

$F_{1} = F_{2} =G\frac{m_{1}m_{2}}{d^{2}}$

Energy is given by

$Energy \,of \,attraction = -G\frac{m_{1}m_{2}}{d}}+\frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$

Conservation of linear momentum gives

m₁·v₁ = m₂·v₂

From which

v₂ = m₁·v₁/m₂

At equilibrium, we have;

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$ which gives

$2G{m_{1}m_{2}}= d m_{1} v^2_1+ dm_{2} (\frac{m_1}{m_2}v_1)^2= dv^2_1(m_1+(\frac{m_1}{m_2} )^2)$

multiplying both sides by m₂/m₁, we have

$2Gm^2_{2}}= dv^2_1 m_2+dm_1v^2_1 =dv^2_1( m_2+m_1)$

Such that v₁ = $\sqrt{\frac{2Gm^2_2}{d(m_1+m_2)} }$

$v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

Similarly, with v₁ = m₂·v₂/m₁, we have

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2\Rightarrow 2G{m_{1}m_{2}}= dm_{1} (\frac{m_2}{m_1}v_1)^2 +d m_{2} v^2_2= dv^2_2(m_2+(\frac{m_2}{m_1} )^2)$

From which we have;

$2G{m^2_{1}}= dm_{2} v_2^2 +d m_{1} v^2_2$ and

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

The relative velocity = v₁ + v₂ = $v_1+v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} } + m_2\sqrt{\frac{2G}{d(m_1+m_2)} } = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$