Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by
So, 
.... (1)
Now,
Substitute the value of v from equation (1), we get
Let the radius of the alpha particle is r2.
For proton
So, 
... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So, 
... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get
q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values
So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
<h3>Question 1</h3>
Answer
option C) velocity
Explanation
acceleration = Δv ÷ Δt
<h3>Question 2</h3>
Answer
option C) m/s²
Explanation
Δv ÷ Δt
= m/s ÷ s
= m/s x 1/s
= m/s²
<h3>Question 3</h3>
Answer
option B) velocity has both direction and speed.
That is why velocity can be negative but speed can not and velocity is rate of change of displacement where as speed is rate of change of distance.
Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.
Explanation:
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
Answer: 585 J
Explanation:
We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:
where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is
The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:
