Convert 435.5 lbs to kilograms

If an astronaut has a mass of 16 Kg on Earth, what would be his mass on the moon and on the space station

nlexa [21]

Answer:

The astronaut's mass is 16 kg.

Explanation:

Mass can be defined as a measure of the amount of matter an object or a body comprises of. The standard unit of measurement of the mass of an object or a body is kilograms.

Irrespective of the location of an object or a body at a given moment in time, the mass (amount of matter that they're made up of) is constant. This ultimately implies that, whether you're in the moon, space, earth or any other place, your mass remains the same (constant).

Therefore, if an astronaut has a mass of 16 Kg on Earth, his mass on the moon and on the space station would remain the same, as his original mass of 16 Kg because mass is indestructible.

3 0

3 years ago

An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion

anyanavicka [17]

Answer:

attached below is the free body diagram of the missing illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

$E_{e} = E_{3} - E_{1}$

E $_{e}$ = initial kinetic energy of the electron

E $_{1}$ = -4 eV

E $_{3}$ = -1 eV

insert the values into the equation above

$E_{e}$ = -1 -(-4) eV

= -1 + 4 = 3 eV

6 0

3 years ago

________ states that the number of transistors per square inch on an integrated chip doubles every 18 months.

Andrew [12]

Answer:

Moore's Law

Explanation:

5 0

3 years ago

You are at the controls of a particle accelerator, sending a beam of 3.60 x10^7 m/s protons (mass m) at a gas target of an unkno

matrenka [14]

Answer:

a) mass of unknown nucleus = 0.04245 mp, where mp is the proton mass

b) Speed of the unknown nucleus = (7.067 x 10^7) m/s

Explanation:

Considering the initial conditions, the observed collisions are ellastic, i.e, the total kinetic energy are conserved. The proton's mass will refer as $m_{p}$ .

(a)

Total kinetic energy conservation

$\frac{1}{2}m_{p}v_{p_0}^{2}+\frac{1}{2}m_{u}v_{u_o}^{2}=\frac{1}{2}m_{p}v_{p_f}^{2}+\frac{1}{2}m_{u}v_{u_f}^{2}$

where $v_{u_o}$ represents the initial velocity of the unknown element, $m_{u}$ the mass of the unknown element, and $v_{u_f}$ the final velocity of the unknown element

Linear momentum conservation

$m_{p}v_{p_0}+m_{u}v_{u_o}=m_{p}v_{p_f}+m_{u}v_{u_f}$

Using the initial speed of the target nucleus (unknown) is negligible, i.e, its speed is zero. Thereby, using the relation of linear momentum conservation given above, it is possible to find an expression of the final speed of the unknown nucleus in terms of its mass, which can be inserted in the relation of the kinetic energy conservation to obtain the value of the mass of the unknown elements, as follows;

$m_{u}v_{u_f}=m_{p}v_{p_0}-m_{p}v_{p_f}\\\\v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}$

Substituting this expression in the relation of total kinetic energy conservation,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}v^{2}_{u}_{f}$

Then,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}\frac{m^{2}_{p}(v_{p_0}-v_{p_f})^{2}}{m^{2}_{u}}\\\\m_{u}= \frac{m_{p}(v_{p_0}-v_{p_f})^{2}}{(v^{2}_{p_0}-v^{2}_{p_f})}$