Answer:
Distance covered 28 km
displacement is 22.8 km North-East
Explanation:
Distance shows how far apart objects or points are from each other. The distance he covered is the sum of all the distance travelled. Therefore:
Distance covered = 12 km + 6 km + 10 km = 28 km
Displacement is a vector quantity (has direction). It is the overall change in position.
The total distance traveled north = 12 km + 10 km = 22 km
The distance traveled east = 6 km
The displacement (d) is:
d² = 22² + 6² = 484 + 36
d² = 520
d = √520 = 22.8 km
Therefore the displacement is 22.8 km North-East
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒ 
(b)
For m₃,
⇒ 
Or,
⇒ 
A. The magnitude (in N) of the electric force that one particle exerts on the other is 8.60×10⁻⁸ N
B. The force is repulsive
<h3>A. How to determine the magnitude of the electric force</h3>
From the question given above, the following data were obtained:
- Charge 1 (q₁) = 7.03 nC = 7.03×10¯⁹ C
- Charge 2 (q₂) = 4.02 nC = 4.02×10¯⁹ C
- Electric constant (K) = 9×10⁹ Nm²/C²
- Distance apart (r) = 1.72 m
- Force (F) =?
The magnitude of the electric force can be obtained by using the Coulomb's law equation as shown below:
F = Kq₁q₂ / r²
F = (9×10⁹ × 7.03×10¯⁹ × 4.02×10¯⁹) / (1.72)²
F = 8.60×10⁻⁸ N
<h3>B. How to determine whether the force is attractive or repulsive</h3>
From the question given, we were told that:
- Charge 1 (q₁) = 7.03 nC
- Charge 2 (q₂) = 4.02 nC
Since both charge are positive, then the force attraction between them is repulsive as like charges repels and unlike charges attracts
Learn more about Coulomb's law:
brainly.com/question/506926
#SPJ1
Answer:
Explanation:
Given that:
angular frequency = 11.3 rad/s
Spring constant (k) = 
k = (11.3)² m
k = 127.7 m
where;
= 0.065 m
= 0.048 m
According to the conservation of energies;
∴
