<h3>
Answer:</h3>
800 meters
<h3>
Explanation;</h3>
<u>We are given;</u>
- Speed as 40 m/s
- Time as 20 seconds
We are required to determine the distance traveled
- Speed refers to the rate of change in distance.
- It is given by;
Speed = Distance ÷ time
Rearranging the formula;
Distance = speed × time
In this case;
Distance = 40 m/s × 20 sec
= 800 meters
Thus, the distance traveled by the car is 800 m
Well both ends of the magnet have a diffrent magnetic force. i hope this was the answer you where looking for, and have a nice rest of the day
Answer:
F = 2.26 × 10⁻³ N
Explanation:
given,
length of rod = 11 cm
charge = 19 nC
linear charge density = 3.9 x 10⁻⁷ C/m
electric force at 2 cm away.

F = E q

integrating from 0.02 to 0.02 + L
![F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2K%5Clambda%5C%20q%7D%7BL%7D%5Bln%280.02%2BL%29-ln%280.002%29%5D)
![F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2%5Ctimes%209%20%5Ctimes%2010%5E9%5Ctimes%203.9%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2019%20%5Ctimes%2010%5E%7B-9%7D%7D%7B0.11%7D%5Bln%280.02%2B0.11%29-ln%280.002%29%5D)
F = 2.26 × 10⁻³ N
Answer:
Answer 11
Explanation:
. Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them
For the orbital speed of a spacecraft we know that

here
M = mass of the planet around which satellite is revolving
r = orbital radius
Now when thruster is used by the spacecraft its speed will change due to which orbital speed will change.
Since here while changing the speed mass of the planet will be same
we can say the speed of the spacecraft will changed by thruster due to which its orbital radius will change
so the correct answer must be
<em>b. the spacecraft will change motion and will maintain this new orbit until the thruster is fired again.</em>