What is the minimum total energy released when an electron and its antiparticle (positron) <span>annihilate each other?
The </span>minimum total energy released when an electron and its antiparticle (positron) annihilate each other is <span>2.73 × 10^–22 J. The answer is number 4.</span>
Answer:
Explanation:
A = 3m
B = 4 m
let the angle between the two vectors is θ.
the resultant of two vectors is given by
(a) R = 7 m
So, 
49 = 9 + 16 + 24 Cosθ
Cosθ = 1
θ = 0°
Thus, the two vectors are inclined at 0°.
(b) R = 1 m
So, 
1 = 9 + 16 + 24 Cosθ
Cosθ = - 1
θ = 180°
Thus, the two vectors are inclined at 180°.
(c) R = 5 m
So, 
25 = 9 + 16 + 24 Cosθ
Cosθ = 0
θ = 90°
Thus, the two vectors are inclined at 90°.
Answer:
Torque, 
Explanation:
Given that,
The number of turns in the coil, N = 47.5
Radius of the coil, r = 5.25 cm
Uniform magnetic field, B = 0.535 T
Current in the coil, I = 26.9 mA
The magnitude of the maximum possible torque exerted on the coil isg given by :
So, the magnitude of the maximum possible torque exerted on the coil is .
Answer:
D. When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.
Explanation:
The density of the metal cube is 2.703 g/cm³.
The mass of a metal cube is m =50.3 grams.
The edge length of the metal cube is l = 2.65 cm.
Now, the density (ρ) of a cube can be given as:
ρ = m/(a)³
Where (a)³ is the volume of the cube, m is the mass of the cube, ρ is the density of the cube, and a is the length of one side of the cube.
Since the sides of a cube are equal therefore the value of a is the same for each edge length of the cube.
Now, the density of the metal cube in g/cm³ is:
ρ = m / a³
ρ = m / l³
ρ = 50.3 g / (2.65 cm)³
ρ = 50.3 g / (2.65 cm)(2.65 cm)(2.65 cm)
ρ = 50.3 g / 18.609 cm³
ρ = 2.703 g/cm³
Learn more about density here:
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