Using the equation v(avg)=distance/time
and the equation v=v(original)+a(t)
solve for acceleration
2600=0+a(12)
a=216.66666 m/s^2
Then, you use the equation
v^2=v(original)+2a*(change in x)
2600^2=2(216.666666)*change in x
6760000/2/216.666666 = 15600 meters which is the length of the race
Then using v(avg)=x/t
15600/12= 1300 m/s
Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie
Answer:

and

Explanation:
Given:
- first charge,

- second charge,

- position of first charge,

- position of second charge,

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.
<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

- since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.





Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

and

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.
<u>Explanation</u>:
We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the
and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

Better understood from numerical example as given:
If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?
This can be solved as follows:


It shows that man A will have more K.E.
Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.
Answer:
45.3°C
Explanation:
Heat gained = mass × specific heat × increase in temperature
q = mC (T − T₀)
Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:
305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)
T = 45.3°C