Answer:
i hardly havent got teach this sorry
Answer:
0.546 
Explanation:
From the given information:
The force on a given current-carrying conductor is:

where the length usually in negative (x) direction can be computed as

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:



![F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%5E3_1%20%5Chat%20k)
![F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B3%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
where;
current I = 7.0 A
![F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B27%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
![F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B26%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
F = 546 × 10⁻³ T/mT 
F = 0.546 
Answer:
Inez thinks it is important to be honest with her friends.
Byron tries to treat all people with respect.
Lydia cares about expressing her creativity.
Explanation:
Just did on Edge
Answer: pick A moves 10.2 metre
Explanation: since they were given a quick push simultaneously the product of their distance moved and velocity would be equal.
Let the distance moved by puck B x then that of puck A would be 18-x.
So therefore we have that,
1.3*(18-x) = 1.7*x
23.4 - 1.3x = 1.7x
23.4=3x
Making x subject of formula we have
x = 23.4/3 = 7.8m
But the distance moved by puck A is 18-x. So therefore,
18-7.8=10.2meter