Answer:
Explanation:
The expression which represent the first diffraction minima by a circular aperture is given by --------eqn 1
The angle through which the first minima is diffracted is given by ---------eqn 2
As is very small so we can write
So from eqn 1 and eqn 2 we can write
--------eqn 3
Here is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture
It is given that diameter of circular aperture is 14.7 cm so
Now putting all these value in eqn 3
F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m
Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved
Explanation:
Let the charges be q1 and q2 and the distance between the charges be 'd'
Mathematical representation of coulombs law will be;
F1=kq1q2/d²...(1)
Where k is the electrostatic constant.
If q1 and q2 is doubled and the distance halved, we will have;
F2 = k(2q1)(2q2)/(d/2)²
F2 = 4kq1q2/(d²/4)
F2 = 16kq1q2/d²...(2)
Dividing equation 1 by 2
F1/F2 = kq1q2/d² ÷ 16kq1q2/d²
F1/F2 = kq1q2/d² × d²/16kq1q2
F1/F2 = 1/16
F1 = 1/16F2
This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved
Density is the ratio of a substance's mass to its volume. On the other hand, according to Archimedes' principle, the volume of water displaced is equal to the volume of the object placed on the water. Thus, the density of the metal is equal to 8.39 mL. So, the density would be
Density = 32.5 g/8.39 mL = 3.87 g/mL