Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
Let d is the distance moved in 2.25 s. Using second equation of motion,
So, it will move 6.32 m from rest in 2.25 seconds.
Answer:
a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N g) 16.9 N
h) 24.3 N θ = 44.2º
Explanation:
a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.
So, Fcax = 0
b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.
Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²
Fyca = 29. 9 N
c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:
rbc² = (3.00 m)² + (4.00m)² = 25.0 m²
⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²
⇒ Fbc = 21.7 N
d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:
Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:
cos θ = x/r = 4.00 / 5.00 m =
Fcbx = 21.7*(-0.8) = -17.4 N
e) The y component can be calculated in the same way, projecting the force over the y-axis, as follows:
Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N
f) The sum of both x components gives :
Fcx = 0 + (-17.4 N) = -17.4 N
g) The sum of both y components gives :
Fcy = 29.9 N + (-13.0 N) = 16.9 N
h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:
Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)² = 24.3 N
The angle from the horizontal can be found as follows:
Ф = arc tg (16.9 / 17.4) = 44.2º
Answer:
100N
Explanation:
Newton's third law states that whenever an object exerts a force on a second object, it exerts a force of equal magnitude and direction but in the opposite direction on the first. It is often stated as follows: Each action always opposes an equal but opposite reaction.
The subject 1 of 100kg is making a force F, to move an object from 50Kg to 2m / s ^ 2. This Force the object of 50Kg will reflect it in the opposite direction by Newton's third law.
Once the parameter of the force that both are experiencing is clarified, Newton's second law is applied to their respective calculation.
That is the force the boy exert on the man during the shove.