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Makovka662 [10]
3 years ago
9

¿Que es un átomo???????

Physics
1 answer:
disa [49]3 years ago
5 0
Un átomo es una porción material menor de un elemento químico que interviene en las reacciones químicas y posee las propiedades características de dicho elemento.
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A cat lifting a barbell of 2.5 kg a distance of 3 meters. How much work is done lifting the barbell? How much work is done if th
Degger [83]

Work = (force) x (distance)

Force = weight of the barbell = (mass) x (gravity) = (2.5 kg) x (9.8 m/s²)

Work = (2.5 x 9.8 newtons) x (3 meters) = 74 newton-meters = 74 joules.

It takes 74 joules to lift the 2.5-kg load 3 meters up off the floor.

After that, as long as the load is held motionless, no more work is done.

7 0
3 years ago
12. The red light emitted by a He-Ne laser has a wavelength of 6.33 x 10^-7 m in the air and travels 3.0 x 10^8 m/s. Find the fr
Ymorist [56]

Answer:

f = 4.76 × 10^14 Hz

Frequency of the laser light f = 4.76 × 10^14 Hz

Explanation:

Given;

wavelength = 6.33 x 10^-7 m

Speed of light v = 3.0 x 10^8 m/s.

Frequency of the laser light f

= speed of light/wavelength

Substituting the given values;

f = 3.0 x 10^8 m/s ÷ 6.33 x 10^-7 m

f = 4.76 × 10^14 Hz

Frequency of the laser light f = 4.76 × 10^14 Hz

4 0
4 years ago
A planet orbits the Sun<br>every 4.5 years. uniform or non uniform velocity?​
Charra [1.4K]
Non uniform velocity
5 0
3 years ago
What is the relationship between the valence electrons of an atom and the chemical bonds the atom can form?​
stellarik [79]

Answer:

Valence electrons are outer shell electrons with an atom and can participate in the formation of chemical bonds. In single covalent bonds, typically both atoms in the bond contribute one valence electron in order to form a shared pair. The ground state of an atom is the lowest energy state of the atom.

8 0
3 years ago
AuniformsphericalshellofmassM=4.5kgandradiusR=8.5cmcan rotate about a fixed vertical axis on frictionless bearings. A massless,
Kazeer [188]

Answer:

v=1.42\frac{m}{s}

Explanation:

There is no friction in the physical system. Thus, according to the law of conservation of energy, recall that the object is released from rest:

\Delta E=0\\U=K_R+K_T\\mgh=\frac{I_p\omega_p^2}{2}+\frac{I_s\omega_s^2}{2}+\frac{mv^2}{2}

Recall that the moment of inertia of a sphere is I_s=\frac{2MR^2}{3}. The angular speed of the pulley is \omega_p=\frac{v}{r} and the angular speed of the sphere is \omega_s=\frac{v}{R}. So, we replace:

mgh=\frac{I_pv^2}{2r^2}+\frac{2MR^2}{3}\frac{v^2}{2R^2}+\frac{mv^2}{2}\\mgh=v^2(\frac{I_p}{2r^2}+\frac{M}{3}+\frac{m}{2})\\v^2=\frac{mgh}{\frac{I_p}{2r^2}+\frac{M}{3}+\frac{m}{2}}\\\\v=\sqrt{\frac{(0.6kg)(9.8\frac{m}{s^2})(0.82m)}{\frac{3*10^{-3}kg\cdot m^2}{2(0.05m)^2}+\frac{4.5kg}{3}+\frac{0.6kg}{2}}}\\v=1.42\frac{m}{s}

5 0
3 years ago
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