In a cathode-ray tube, electrons are projected horizontally at 1.0×10^6 m/s and travel a horizontal distance 11.5 cm across the tube.
How long are the electrons in flight?
Express your answer to two significant figures and include the appropriate units.
Find the vertical distance the electrons fall (under the influence of gravity) during their flight.
Express your answer to two significant figures and include the appropriate units.
t = distance/speed = .115 * 10^-6 seconds
which is
1.2 *10^-7 seconds
h = (1/2) g t^2
h = 4.9 (1.44)(10^-14)
h = 7.1 * 10^-14 meters
supposedly the answer to the second part of the question is y = −6.5×10−14 m
any insight?
I rounded .115 to .12 in part a
(as instructed :)
If they use 4.9 (1.15)^2
they would get 6.48
which rounds to 6.5
To find the time the electrons are in flight, we can use the formula:
Time = Distance / Velocity
Given:
Horizontal distance = 11.5 cm = 0.115 m (converted to meters)
Horizontal velocity = 1.0×10^6 m/s
Using the formula, we can calculate the time:
Time = 0.115 m / 1.0×10^6 m/s = 1.15×10^-7 s
Therefore, the time the electrons are in flight is approximately 1.15×10^-7 seconds.
To find the vertical distance the electrons fall under the influence of gravity, we can use the formula for free fall:
Vertical distance = 0.5 * acceleration due to gravity * time^2
The acceleration due to gravity is approximately 9.8 m/s^2. Plugging in the values:
Vertical distance = 0.5 * 9.8 m/s^2 * (1.15×10^-7 s)^2
Calculating this, we get:
Vertical distance ≈ 6.8×10^-15 m
Expressing the answer to two significant figures, the vertical distance the electrons fall is approximately 6.8×10^-15 meters.