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Elenna [48]
3 years ago
8

How fast should a rocket ship move for its mass to be dilated to 167% of its restmass?​

Physics
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.

Hopefully, this answer helps you! :)

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3 years ago
Need help ASAP!!!! here are the options: Amplitude, Compression, Rarefaction, and Wavelength
Strike441 [17]
Amplitude is the pair of vertical buttons, so to speak. Compressions are the bunched up vertical lines with the purple arrows pointing left and right. Rarefactions are purple arrows pointing down. Wavelength is crest to crest purple buttons. Associated LH and RH pointing arrows.
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3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)
nlexa [21]
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.

so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f

3 0
3 years ago
In this type of bond, electrons are lost or gained by atoms, and the atoms are held together by electrical attraction.
expeople1 [14]
<span>In Ionic type of bonding, electrons are lost (more protons than electrons and positive charge) or gained (more electrons than protons, still a negative charge) by atoms, and the atoms are held together by electrical attraction in the process. Covalent bondings are the sharing of electrons as well as partial bondings. Covalent bondings’ electrons have the same charges thus, there is no gaining or losing electrons in the process of sharing. Strong bondings are applicable only to Hydrogen (H) atoms. </span>
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