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12. [2/4 Points] DETAILS PREVIOUS ANSWERS SERPSE10 22.6.P.025.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A proton moves at $5.40 \times 10^5$ m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of $9.40 \times 10^3$ N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.50 cm horizontally. 83.3 ns (b) Find its vertical displacement during the time interval in which it travels 4.50 cm horizontally. (Indicate direction with the sign of your answer.) 3.158 ?mm (c) Find the horizontal and vertical components of its velocity after it has traveled 4.50 cm horizontally. $\vec{v} = ($ X Your response is off by a multiple of ten. $\hat{i} + 7.5e5$ X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. $\hat{j}$) km/s Need Help? Read It Submit Answer

          12. [2/4 Points]
DETAILS
PREVIOUS ANSWERS
SERPSE10 22.6.P.025.MI.
MY NOTES
ASK YOUR TEACHER
PRACTICE ANOTHER
A proton moves at $5.40 \times 10^5$ m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of $9.40 \times 10^3$ N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 4.50 cm horizontally.
83.3
ns
(b) Find its vertical displacement during the time interval in which it travels 4.50 cm horizontally. (Indicate direction with the sign of your answer.)
3.158
?mm
(c) Find the horizontal and vertical components of its velocity after it has traveled 4.50 cm horizontally.
$\vec{v} = ($
X
Your response is off by a multiple of ten. $\hat{i} + 7.5e5$
X
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. $\hat{j}$) km/s
Need Help? Read It
Submit Answer

Show more…

12. [2/4 Points]
DETAILS
PREVIOUS ANSWERS
SERPSE10 22.6.P.025.MI.
MY NOTES
ASK YOUR TEACHER
PRACTICE ANOTHER
A proton moves at 5.40 × 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.40 × 10^3 N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 4.50 cm horizontally.
83.3
ns
(b) Find its vertical displacement during the time interval in which it travels 4.50 cm horizontally. (Indicate direction with the sign of your answer.)
3.158
?mm
(c) Find the horizontal and vertical components of its velocity after it has traveled 4.50 cm horizontally.
v⃗ = (
X
Your response is off by a multiple of ten. î + 7.5e5
X
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. ĵ) km/s
Need Help? Read It
Submit Answer

Added by Cheryl K.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sam Stansfield

08/28/2023

Step 1

50 cm horizontally. Given that the speed of the proton is 7.5 km/s, we can convert this to cm/ns: 7.5 km/s = 7.5 * 10^5 cm/s = 7.5 * 10^5 cm/s * (1 s / 10^9 ns) = 7.5 * 10^-4 cm/ns To find the time interval required to travel 4.50 cm horizontally: Time = Distance Show more…

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(a) Find the time interval required for the proton to travel 4.50 cm horizontally. 83 ns (b) Find its vertical displacement during the time interval in which it travels 4.50 cm horizontally. (Indicate direction with the sign of your answer.) 3.158 mm (c) Find the horizontal and vertical components of its velocity after it has traveled 4.50 cm horizontally. Your response is off by a multiple of ten. î + 7.5 km/s

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Transcript

00:01 So this one, the problem that's written out in the question is not the same as the thing that was copied in.

00:08 So i'm going to do the one that was written out.

00:12 So we have a proton that's moving horizontally with a velocity of 3 .8 times 10 to the 5th meters per second.

00:20 It enters a vertical electric field with a magnitude of 7 .6 times 10 to the 3rd newtons per coulomb.

00:28 And here's the mass of the proton and it's a charge.

00:31 And it's positively charged.

00:33 Okay? so in part a, we want to know, we're given that it's going to travel a distance horizontally of 0 .04 meters.

00:48 That's 4 centimeters.

00:52 Okay? we want to find the time it takes.

00:55 So delta t is delta x over v, which really, that's one of our handy dandy formulas from physics one.

01:07 0 .04 meters divided by 3 .8 times 10 to the 5th meters per second gives us a time of 1 .05 times 10 to the minus 7 seconds.

01:29 Okay.

01:29 So it's going to also move upwards because the electric field, we know that the acceleration in the y direction is going to be qe over m.

01:40 And so the y, y minus y zero, there is no initial velocity in the y direction.

01:53 Comes out to be one half a sub y t squared.

02:00 So that's half...

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