Explanation:
Given that,
Initial speed of the sports car, u = 80 km/h = 22.22 m/s
Final speed of the runner, v = 0
Distance covered by the sports car, d = 80 km = 80000 m
Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :
Value of g, 
Hence, this is required solution.
Answer:
a.Distance = 150 m
b. Displacement = 50 m
Time lapsed = 5 seconds
Explanation:
a. Distance is the change in the position of an object.
The distance covered by the car = 100 + 50
= 150 m
b. Since displacement is a vector quantity,
Displacement of the car = 100 - 50
= 50 m due east
c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.
Thus, the time elapsed = 4 + 1
= 5 seconds
Answer:
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Explanation:
At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.
Answer:
The ball will continue to move at a constant speed forever until another force stops it.
Explanation:
Newton's first law of motion can be seen as a law of inertia. It explains that an object at rest or in a state of uniform motion will remain in that state unless it is acted upon by an external force.
Following the above, when the ball is thrown into space, the ball will continue to move with the velocity with which it was thrown until it comes in contact with another object that stops it. If this does not happen, it will continue to drift forever at that velocity.
Answer:
Explanation:
The <em>3N force </em>must be attributed to electrostatic attraction between both charges.
The formula is known as Coulomb's law:
Where:
is the elecrostatic force: 3 N- k is the constant for Coulomb's law ≈ 9.00 × 10⁹ N . m² / C²
- d is the distance that separate the centers of the charges
are the values of the charges: ≈ ± 1.6 × 10⁻¹⁹ C
When you are not interested in the direction of the electrostatic force you just use the magnitudes of the charges.
Substitute in the formula and solve for d:
That can be converted to angstroms with the conversion factor:
