Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U = 
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ = 
starting point.
Em₀ = K + U₁ + U₂
Em₀ = 
final point
Em_f = 
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( 
)
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ 
) = 9 109 (1.6 10-19) ²( 
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( 
)
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 = 
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
The type of radiation that can penetrate through paper, but not through wood is called beta rays. Beta rays can penetrate paper and air, but a thin piece of alimony can stop it. Gamma can cut through anything except lead and many inches of concrete. Alpha can be stopped by paper and not penetrated. The correct answer is B.
when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as
T=mv²/r
where T = tension force in the string , m = mass of the apple
v = speed of apple , r = radius of circle.
clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.
at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks
The answer is 24N. Since the body is moving with constant velocity all the forces must balance (equal & opposite)
Answer:
h = 90.10 m
Explanation:
Given that,
A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m
The initial speed of the stone, u = 10 m/s
The path followed by the projectile is given by :
....(1)
For maximum height,
Put dh/dt = 0
So,
Put the value of t in equation (1).
So, the maximum height of the stone is equal to 90.10 m.
