Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu

<span>this might be useful
Regarding the field, the two charges placed opposite cancel each other out!
Therefore, E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
given k = 8.99×10^9 N·m²/C²,
E = 1.789×10¹⁰ N·m²/C² * Q / d² </span>

Response:

AB = 100 km; BC = 80 km; AC = 180 km

Time of arrival = 11:30

Reasoning:

1. Distance from A to B

(a) Duration of travel

Duration = 10:00 - 8:00 = 2.00 hours

(b) Distance

Distance = speed × time = 50 km/h × 2.00 h = 100 km

2. Distance from B to C

Distance = 80 km/h × 1 h = 80 km

3. Summary of Distances

AB = 100 km

BC = 80 km

AC = 180 km

4. Time of Arrival

Departure from A = 08:00

Travel duration to B = 2:00

Arrival at B = 10:00

Waiting time at B = 0:30

Departure from B = 10:30

Travel duration to C = 1:00

Arrival at C = 11:30

Numerous items, including bags, toys, and various goods, incorporate plastic. Its remarkable durability and flexibility are key characteristics.:) I hope this information is beneficial to you! c;

In this scenario, the principles of momentum conservation can be applied since there are no external forces acting on the system. Consequently, the conservation of momentum principle is applicable here. After the bird lands on it, both the bird and the bark will have a unified final speed. Thus, this final speed will be 1 m/s.

Answer:

17.35 × 10^(-6) m

Explanation:

Mass; m = 50 kg

Weight; W = 554 N

From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m