The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854 *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854 *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854 *10 ^-12 equal to -1.908 *10^7.
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As you gave no pic I took them on one lined
- F_1=3N
- F_2=10N
- F_3=3root 3 N
- F_4=6N





Answer:
<em>The total time is: t=451.22 sec</em>
<em>The average speed is: V=34.57 m/s</em>
Explanation:
<u>Average speed</u>
The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.
We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is
x=2*7,800 m=15,600 m
The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

The total time is:


The average speed is:


There's actually potential energy before the kinetic energy came into play, but the sum of potential and kinetic energy is MECHANICAL ENERGY.
Answer:
18 km
Explanation:
Convert km/h to m/s:
120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s
The time it takes the bomb to travel the 2000 meters is:
2000 m / (33.3 m/s) = 60 s
So it takes 60 seconds for the bomb to fall. The distance it fell is therefore:
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²
Δy = 18,000 m
Δy = 18 km