the gravity force between Mars and an object near Earth's surface as much lower than the force between an object on Earth surfac
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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
To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
Where,
= Mass of each object
= Initial velocity of each object
= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
Therefore the velocity right after catching the ball is 0.0975m/s