D oxygen i believe is your answer
Answer:
see explanation
Explanation:
Given that,
velocity of 1.50 km/s = 1.50 × 10³m/s
acceleration of 2.00 ✕ 1012 m/s2
electric field has a magnitude of strength of 18.0 N/C
![\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]](https://tex.z-dn.net/?f=%5Cbar%20F%3D%20q%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%5Cbar%20B%5D%5C%5C%5C%5C%5Cbar%20F%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D%5C%5C%5C%5C%5C%5Cm%20%5Cbar%20a%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D)
![9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]](https://tex.z-dn.net/?f=9.1%20%5Ctimes%2010%5E-%5E3%5E1%20%5Ctimes%202%5Ctimes%2010%5E1%5E2%20%5Chat%20k%3D-1.6%5Ctimes10%5E-%5E1%5E9%20%5Chat%20k%20%5B18%5Chat%20k%2B%201.5%5Ctimes%2010%5E3%20%5Chat%20i%20%5Ctimes%20%28B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20k%29%5D)




Answer:
A. 5 m/s
Explanation:
From the graph, for the first 2 seconds, the graph is a straight line meaning that the slope is a constant.
Average speed of an object is the rate of change of position. Here, the position of the object changes from 0 m to 10 m for a time interval of 2 seconds.
The change in position (
) and time interval (
) are given as:

Therefore, the average speed (
) is given as the ratio of the total change in position and the time interval for the change.

Hence, the average speed is 5 m/s.
Answer:
Explanation:
The forces compare together as a result of the fact that the force exerted by that of the ball and the force exerted by that of the wall both have the same magnitude.