Answer:
0.68 kg-m²
Explanation:
F = Force applied by the muscle = 2615 N
r = effective perpendicular lever arm = 2.85 cm = 0.0285 m
α = Angular acceleration of the forearm = 110.0 rad/s²
I = moment of inertia of the boxer's forearm = ?
Torque is given as
τ = I α eq-1
Torque is also given as
τ = r F eq-2
using eq-1 and eq-2
r F = I α
(0.0285)(2615) = (110.0) I
I = 0.68 kg-m²
Answer:
230.4 N
Explanation:
From the question given above, the following data were obtained:
Charge (q) of each protons = 1.6×10¯¹⁹ C
Distance apart (r) = 1×10¯¹⁵ m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The force exerted can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²
F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰
F = 2.304×10¯²⁸ / 1×10¯³⁰
F = 230.4 N
Therefore, the force exerted is 230.4 N
<span>Describe the relationship of attractive forces between molecules and the ability of a solvent to dissolve a substance. Solvents can dissolve a substance only if the attraction of the solvent molecules is greater than the attraction between the molecules of the substance.</span>
5.1 m
Explanation:
Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by
(1)
where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground
is given by
(2)
At the instant the two balls collide, they will have the same displacement, therefore

or

Solving for t, we get

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):


Answer: Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively
Explanation: