Answer:
4 electrons
Explanation:
Carbon, is the group 14 element, with four electrons in its outer shell. Carbon always shares electrons to reach a complete valence shell, making bonds with other atoms.
Answer:
Chemical composition, Temperature, Radial velocity, Size or diameter of the star, Rotation.
Explanation:
Elemental abundances are determined by analyzing the relative strengths of the absorption lines in the spectrum of a star.
The Spectral class to which the star belongs gives the information related to the temperature of the star. It is the spectral lines that determine the spectral class O B A F G K M are the spectral classes.
By measuring the wavelengths of the lines in the star's spectrum gives the radial velocity. Doppler shift is the method used to find the radial velocity.
A star can be classified as a giant or a dwarf . A giant star will have narrow width spectral lines whereas a dwarf star has wider spectral lines.
Broadening of the spectral lines will determine the star's rotation.
Answer:
Explanation:
From the question we are told that:
Mass 
Angle 
Coefficient of static friction 
Generally, the equation for Newtons second Law is mathematically given by
For
for
Where
Therefore
This is possible due to self-discharge. Very small internal currents inevitably occur in these cells over time and they will eventually exhaust the chemistry.
For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as
F= 618.9 N
<h3>What is the centripetal
force?</h3>
Generally, the equation for the angular speed is mathematically given as
w = v/R
Therefore
w= 4.7/1.8
w= 2.611 rad/s
Where total momentum
Tm= 642.96 + 272.32
Tm= 915.28
and total inertia
Ti= 184 + 246.24
Ti= 430.24
In conclusion, centripetal force
F= mrw^2
F = m*R*w2^2
F = 76*1.8*2.127^2
F= 618.9 N
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CQ
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a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?
