Assignment on Physics- Answers

Running head: PHYSICS 1
Physics
Name
Institution

PHYSICS 2
Section A
Question 1
a) Pressure in interstellar space
Temperature= 3K
Density = 1atom/cubic centimeter
P= nRT
V
Where V=volume, n= is the number of atoms, R=gas constant¿ 8.314 J /mol / K , T= absolute
temperature
Eavg =1.5 kb T
V =10−6 m3
PV =nRT
P= nRT
V =1.66 ×10−24 × 8.314 ×3
10−6
¿ 4.14 ×10−17 Pa
b) Velocity of atoms
i. We have to use the V rmsbecause the temperature is constant and thus the it has a
differential magnitude of zero. The V rmstherefore helps in finding the root mean
square speed that has no magnitude as opposed to the average velocity that has
magnitude (Kauzmann, 2012).

PHYSICS 3
ii. The V rms is more appropriate measure of velocity in gas samples as opposed to the
average speed because gases have a net velocity of zero. The net velocity is zero
because gas particles move randomly in all directions thus making it inappropriate to
use the median velocity (Kauzmann, 2012). It is important to note that velocity is a
vector quantity and particles gas particles have particles moving in different
directions at different speed. The resultant velocity, therefore, becomes zero. The
appropriate assessment of velocity, therefore, is done by averaging the squares of the
velocities and calculating the root of the result.
c) Speed of gas molecules near Mars
Near the surface of mass is almost entirely made up of CO2 that has a molecular mass of 44
mass of 1 molecule=44
Therefore, mass of one atom ¿ 44
6.02× 1023 =7.31 ×10−23 g=7.31×10−26 kg
Since v= √ 3 k B T
m = √ 3× 230 K × 1.38× 10−23 J / K
7.31× 10−26 kg ( kg m2
Js2 )
¿ 360 m/s
Question 2
a) Magnetic field
i. The circular path of the charged particle will be perpendicular to the field lines. The
angle is, therefore, ¿ 90 °
ii. The charged particle moves in a circular path because the magnetic force of the field
lines is perpendicular to the particle’s direction of travel. The magnetic force does not

PHYSICS 4
therefore work on the charged particle. As a result, the kinetic energy and the speed
of the particle do not change. Since the speed of the particle is not affected, it will
remain in a circular motion as long as the magnetic field remains constant.
iii. Field strength
F=qvB sin θ
Where q=is the charge, v=velocity, B= magnetic field, θ=angle
Since angle is 90, sin θ=1
But M ( mass )= R2 e B2
2 V (velocity)
2.8 ×1 0−5 =5. 52 × 14.5 ×1 0−6 × B2
2 ×11.4
B2=1.507 , B=1.22777
F=14.5 ×1 0−6 ×11.4 × 1.22777× 1
¿ 2.02 95 ×10−4 N
b) The field due to I1 at a distance r can be expressed as ;
B1= μ° I 1
2 πr ------------(i)
It is however important to note that along wire 2, the field is uniform. Therefore the force F2
exerted by the field on wire 2 can be expressed as
F2=I2 l B1 sin θ
But since θ=90 , sin θ=1