Answer:
i answered this question not that long ago....it's D
Explanation:
Force = Mass * Acceleration therefore the red ball with the higher mass will have more force and greater acceleration
Answer:
1)a. It is constant the whole time the ball is in free-fall.
2)b. = 14 m/s
3) e. = 19.6 m/s
Explanation:
1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.
2) speed = distance/time
horizontal distance = 56m
Time = 4 seconds
Speed = 56m/4s = 14m/s
3) acceleration due to gravity g = 9.8m/s^2
Initial vertical velocity = u
Final vertical velocity = v = -u
Using the law of motion;
v = u + at
a = acceleration = -g = -9.8m/s^2
t = time of flight = 4
Substituting the values;
-u = u - 4(9.8)
-2u = -4(9.8)
u = -4(9.8)/-2
u = 2(9.8) = 19.6 m/s
Initial vertical velocity = u = 19.6 m/s
Answer:
1. 8437500 N
2. The force between the two charges is attractive.
Explanation:
1. Determination of the force between the two charges.
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
Distance apart (r) = 80 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 2 × 3 / 80²
F = 5.4×10¹⁰ / 6400
F = 8437500 N
Thus, the force of attraction between the two charges is 8437500 N
2. From the question given, the charges are:
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.
Thus the force between them is attractive.
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