An electron in a cathode-ray tube is traveling horizontally at 2.10×109 cm/s when deflection plates give it an upward acceleration of 5.40×1017 cm/s2 .
How long does it take for the electron to cover a horizontal distance of 6.30 cm ?What is its vertical displacement during this time?
To find the time it takes for the electron to cover a horizontal distance of 6.30 cm, we need to use the horizontal velocity and distance formula.
The formula for calculating the time (t) is:
t = d / v
Where:
d = horizontal distance
v = horizontal velocity
Given:
d = 6.30 cm
v = 2.10×10^9 cm/s
Plugging in the values:
t = 6.30 cm / 2.10×10^9 cm/s
Since the units are different, we need to convert cm to m to ensure consistent units:
t = 0.063 m / 2.10×10^9 m/s
Calculating the value:
t = 0.063 x 10^-2 / 2.10×10^9
t ≈ 3.0 x 10^-12 s
Therefore, it takes approximately 3.0 x 10^-12 seconds for the electron to cover a horizontal distance of 6.30 cm.
To find the vertical displacement during this time, we can use the equation of motion:
Δy = v_0 * t + (1/2) * a * t^2
Where:
Δy = vertical displacement
v_0 = initial vertical velocity = 0 (since it starts from rest in the vertical direction)
a = upward acceleration = 5.40×10^17 cm/s^2
t = time = 3.0 x 10^-12 s (from the previous calculation)
Plugging in the values:
Δy = (0 * 3.0 x 10^-12) + (1/2) * (5.40×10^17) * (3.0 x 10^-12)^2
Calculating the value:
Δy = 0 + (1/2) * (5.40×10^17) * 9.0 x 10^-24
Δy = 2.7 x 10^-24 cm
Therefore, the vertical displacement during this time is approximately 2.7 x 10^-24 cm.