To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

Here,
k = Coulomb's constant
q = Charge of proton and electron
r = Distance
Replacing we have that,


The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.
The acceleration of the electron is given as



The acceleration of the proton is given as,



Answer:
3.78 m/s
Explanation:
Recall that the formula for average speed is given by
Speed = Distance ÷ Time taken
Where,
Speed = we are asked to find this
Distance = given as 340m
Time taken = 1.5 min = 1.5 x 60 = 90 seconds
Substituting the values into the equation:
Speed = Distance ÷ Time taken
= 340 meters ÷ 90 seconds
= 3.777777 m/s
= 3.78 m/s (round to nearest hundredth)
Complete Question
A football coach walks 18 meters westward, then 12 meters
eastward, then 28 meters westward, and finally 14 meters
eastward.
a
From this motion what is the distance covered
b
What is the magnitude and direction of the displacement
Answer:
a

b
Magnitude
Direction
West
Explanation:
From the question we are told that
The first distance covered westward is ![d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2](https://tex.z-dn.net/?f=d_w_1%20%20%3D%20%2018%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20The%20%20first%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e1%20%3D%20%2012%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20second%20distance%20covered%20westward%20is%20%5Btex%5Dd_w_2%20%20%3D%20%2028%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20%20second%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e2%20%3D%20%2014%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%3C%2Fp%3E%3Cp%3EGenerally%20the%20distance%20covered%20is%20mathematically%20represented%20as%20%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%5Btex%5DD%20%3D%20%20d_w1%20%2B%20d_w2%20%2B%20d_e1%20%2B%20d_e2)
=> 
=> 
For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative
The magnitude of the displacement is

=>
=>
The direction is west
Answer:
Angle: 
Explanation:
<u>Two-Dimension Motion</u>
When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.
Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.
To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

where
is the speed of the boy in still water and
is the angle respect to the shoreline. If the river flows at speed
, we now set



The current will decrease as the resistance has now increased, meaning less current will be 'let through' the resistor. (assuming it's in series, there's no image)