Explanation:
It is given that,
Length of rod, l = 14 cm = 0.14 m
Total charge, 
We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m
Electric field at the axis of rod is given by :
Where
is the linear charge density, 
So, 
E = −7849988.22 N/C
or
Negative sign shows the direction of electric field is inward in all direction. Hence, this is the required solution.
Explanation:
angular velocity is given by
w = 0.626
now tangential velocity is
V = rw
= 25 x 0.626
= 15.65 m/s
Answer:
Yes . Frictions was not included for the calculation for acceleration .
Explanation:
This is because we're considering the weight of an object which isn't acceleration i:e frictionless.
If friction is included, the equation for acceleration becomes.
F= ma,(Newton second law of motion)
where F=frictional force , m =mass of the object and a=acceleration..
So therefore, a = F/m.
" 30% " means " 0.30 "
30% ÷ 75 means (0.30 / 75) = <em>0.004</em>
Answer:
(a) 1462.38 m/s
(b) 2068.13 m/s
Explanation:
(a)
The Kinetic energy of the atom can be given as:
K.E = (3/2)KT
where,
K = Boltzman's Constant = 1.38 x 10⁻²³ J/k
K.E = Kinetic Energy of atoms = 343 K
T = absolute temperature of atoms
The K.E is also given as:
K.E = (1/2)mv²
Comparing both equations:
(1/2)mv² = (3/2)KT
v² = 3KT/m
v = √[3KT/m]
where,
m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg
v = RMS Speed of Helium Atoms = ?
Therefore,
v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 1462.38 m/s</u>
(b)
For double temperature:
T = 2 x 343 K = 686 K
all other data remains same:
v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 2068.13 m/s</u>
