Answer:
GPS consist of the space segment, the control segment, and the user segments
When a charged object is brought near to but does not touch a neutral object, it causes the side of the neutral object that the charged object is near to become the other charge. It causes charge migration within the neutral object so the two charges (positive and negative) move to opposite sides of the object. Because the two objects do not touch, they do not repel each other, but rather have a slight attraction because of charge migration. If the two object were to touch then they would repel.
Explanation:
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 30 m/s
- Mass (m) = 2400 kg
- Force (F) = 12000 N
Let us find the time taken first.
→ F = ma
- Acceleration (a) = (v – u)/t
→ 12000 = 2400 × (30 – 10)/t
→ 12000 ÷ 2400 = (20)/t
→ 5 = 20/t
→ 5t = 20
→ t = 20 ÷ 5
→ <u>t</u><u> </u><u>=</u><u> </u><u>4</u><u> </u><u>seconds</u>
Now, find the acceleration.
→ a = (v – u)/t
→ a = (30 – 10)/4
→ a = 20/4
→ <u>a</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>m</u><u>/</u><u>s²</u>
Now, by using the third equation of motion,
→ v² – u² = 2as
→ (30)² – (10)² = 2 × 5 × s
→ 900 – 100 = 10s
→ 800 = 10s
→ 800 ÷ 10 = s
→ <u>8</u><u>0</u><u> </u><u>m</u><u> </u><u>=</u><u> </u><u>s</u>
Therefore, distance travelled is 80 m.
A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.
b) The acceleration will be the horizontal force - the kinetic friction force (since they act in opposite directions) divided by the mass. Kinetic friction force = (coefficient of kinetic friction)*(normal force or weight).
F(net) = (2207.25 N)-(0.30)(4905 N) = 735.75 N
a = (735.75 N)/(500kg)= 1.4715 m/s^2
