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azamat
3 years ago
14

The atomic mass of an element is defined as the weighted average mass of that element’s

Physics
1 answer:
marishachu [46]3 years ago
4 0

Answer:

False

Explanation:

Atomic mass (Also called Atomic Weight, although this denomination is incorrect, since the mass is property of the body and the weight depends on the gravity) Mass of an atom corresponding to a certain chemical element). The uma (u) is usually used as a unit of measure. Where u.m.a are acronyms that mean "unit of atomic mass". This unit is also usually called Dalton (Da) in honor of the English chemist John Dalton.

It is equivalent to one twelfth of the mass of the nucleus of the most abundant isotope of carbon, carbon-12. It corresponds roughly to the mass of a proton (or a hydrogen atom). It is abbreviated as "uma", although it can also be found by its English acronym "amu" (Atomic Mass Unit). However, the recommended symbol is simply "u".

<u> The atomic masses of the chemical elements are usually calculated with the weighted average of the masses of the different isotopes of each element taking into account the relative abundance of each of them</u>, which explains the non-correspondence between the atomic mass in umas, of an element, and the number of nucleons that harbors the nucleus of its most common isotope.

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A 20kg bike accelerates at 10 m/s2. With what force was the person padeling?
Romashka-Z-Leto [24]
Formula:
F = ma
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Oceanic crust is much denser than continental crust
Free_Kalibri [48]
Oceanic because it’s denser
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2 years ago
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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T
wlad13 [49]

Explanation:

Given:F=m\ddot{x}=Fe^{-\frac{t}{T}}

Solving for \ddot{x}:

\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}

where:

T=\sqrt{\frac{m}{F}}

Integrating to get \dot{x} with initial conditions \dot{x}(0)=0:

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}

Integrating to get x with initial conditions x(0) = 0:

x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}

When t=T:

x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})

4 0
3 years ago
Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0
3 years ago
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