Answer:
Number of neautrons
Explanation:
Atoms from two different elements may have the same number of neutrons, but never the same number of protons. The number of protons is unique to the element and represents the atomic number. (sorcastic.org)
Answer:
1/4 times your earth's weight
Explanation:
assuming the Mass of earth = M
Radius of earth = R
∴ the mass of the planet= 4M
the radius of the planet = 4R
gravitational force of earth is given as = 
where G is the gravitational constant
Gravitational force of the planet = 
=
=
recall, gravitational force of earth is given as = 
∴Gravitational force of planet = 1/4 times the gravitational force of the earth
you would weigh 1/4 times your earth's weight
If F =m*a
and the question says how much force the s needed to accelerate a 68kg skier to a rate of 1.2ms^-2
Then F = 68*1.2
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³