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Answer:
= 4.3 × 10 ⁻¹⁴ m
Explanation:
The alpha particle will be deflected when its kinetic energy is equal to the potential energy
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C
Kinetic energy of the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)
= 8.459 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus
r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³
= 4.3 × 10 ⁻¹⁴ m
Answer:
Q = 47.06 degrees
Explanation:
Given:
- The transmitted intensity I = 0.464 I_o
- Incident Intensity I = I_o
Find:
What angle should the principle axis make with respect to the incident polarization
Solution:
- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:
I = I_o * cos^2 (Q)
- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:
Q = cos ^-1 (sqrt (I / I_o))
- Plug the values in:
Q = cos^-1 ( sqrt (0.464))
Q = cos^-1 (0.6811754546)
Q = 47.06 degrees
37° Celsius is equal to 98.6° Fahrenheit
Answer:
Explanation:
Since the wires attract each other , the direction of current will be same in both the wires .
Let I be current in wire which is along x - axis
force of attraction per unit length between the two current carrying wire is given by
x 
where I₁ and I₂ are currents in the wires and d is distance between the two
Putting the given values
285 x 10⁻⁶ = 10⁻⁷ x 
I₂ = 16.76 A
Current in the wire along x axis is 16.76 A
To find point where magnetic field is zero due the these wires
The point will lie between the two wires as current is in the same direction.
Let at y = y , the neutral point lies
k 2 x 
= k 2 x 
25.5y = 16.76 x .3 - 16.76y
42.26 y = 5.028
y = .119
= .12 m
