Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = 
v²
v = √( 2K / )
lets relate the cross-sectional area A of the beam to its diameter D;
A = 
πD²
now, we substitute for v and A
n = I / πeD² ×√( 2K / )
n = 4I/π eD² × √( / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Answer:
2683.3N
Explanation:
According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;
F is the force of attraction between the charges
q1 and q2 are the charges
d is the distance between them
k is the coulombs constant
Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m
k = 8.98755 × 109 Nm² /C²
Substituting the given data's in the equation we have;
|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²
|F| = 9.66/0.06²
|F| = 9.66/0.0036
|F| = 2683.3N
The magnitude of the force will be 2683.3N
Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.
Answer:
1.37 seconds before.... splat !
Explanation:
y = y0 + v0 t + 1/2 a t^2 y0 = 30 y = 0
v0 = 0 ( from rest)
a = - 32.2 ft/s^2
0 = 30 - 1/2 * 32.2 * t^2
t=1.366 seconds <u> Note that mass is irrelevant.</u>
D.4
hope it helps have a great day!
