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Lapatulllka [165]
3 years ago
15

A train travels 55 km south along a straight track in 34 minutes. What is the train's average velocity in kilometers per hour?​

Physics
2 answers:
AURORKA [14]3 years ago
4 0
Hopefully I’m not late and I apologize if I am, but the answer to your question would be 95.6 km/hr. You know you can look up your question as well to see if they already have a answer to that so you won’t waste your points.
g100num [7]3 years ago
3 0

Answer: v = 96.5 km/h

Explanation: Solve this problem using the following equation:

v= d/t

First convert t in minutes into hours

34 mins x 1 hour / 60 mins

= 0.57 h

Substitute the values

v = 55 km / 0.57 h

= 96.5 km/h

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The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally
dalvyx [7]
(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
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- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
K= \frac{1}{2}mv^2
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 2.20 \cdot 10^{-15} J}{9.1 \cdot 10^{-31} kg} }=6.95 \cdot 10^7 m/s

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
qvB = m \frac{v^2}{r}
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
r= \frac{mv}{qB}= \frac{(9.1 \cdot 10^{-31} kg)(6.95 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(3.00 \cdot 10^{-5} T)}=13.18 m

And finally we can calculate the centripetal acceleration, given by:
a_c =  \frac{v^2}{r}= \frac{(6.95 \cdot 10^7 m/s)^2}{13.18 m}=3.66 \cdot 10^{14} m/s^2
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