Answer:
r = 0.114 m
Explanation:
To find the speed of the proton, from conservation of energy, we know that
KE = PE
Thus, we have;
(1/2)mv² = qV
Where;
V is potential difference = 1kv = 1000V
q is charge on proton which has a value of 1.6 x 10^(-19) C
m is mass of proton with a constant value of 1.67 x 10^(-27) kg
Let's make the velocity v the subject;
v =√(2qV/m)
v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))
v = 4.377 x 10^(5) m/s
Now to calculate the radius of the circular motion of charge we know that;
F = mv²/r = qvB
Thus, mv²/r = qvB
Divide both sides by v;
mv/r = qB
Thus, r = mv/qB
Value of B from question is 0.04T
Thus,
r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)
r = 0.114 m
r = 8.76 m
Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ... V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ... s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1; -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ... s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1; -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)
Answer:
Check the explanation
Explanation:
A) There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface).
To find out how they relate, draw a picture with the path the light follows in the plastic and the normal to both surfaces.
Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normal are parallel lines. An important theorem from geometry will give you the relationship between the angles.
Using Snell's Law, θa = asin[(nw/na)*sin(θw)]
B) D = l/tan(θw)
C) D = l/θw
D) d/D = na/nw
Answer:
457.81 Hz
Explanation:
From the question, it is stated that it is a question under Doppler effect.
As a result, we use this form
fo = (c + vo) / (c - vs) × fs
fo = observed frequency by observer =?
c = speed of sound = 332 m/s
vo = velocity of observer relative to source = 45 m/s
vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).
fs = frequency of sound wave by source = 459 Hz
By substituting the the values to the equation, we have
fo = (332 + 45) / (332 - (-46)) × 459
fo = (377/ 332 + 46) × 459
fo = (377/ 378) × 459
fo = 0.9974 × 459
fo = 457.81 Hz
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = - 1.4 m
