Answer:
ω = 0.0347 rad/s²
a ≅ 1.07 m/s²
Explanation:
Given that:
mass of the model airplane = 0.741 kg
radius of the wire = 30.9 m
Force = 0.795 N
The torque produced by the net thrust about the center of the circle can be calculated as:
where;
F represent the magnitude of the thrust
r represent the radius of the wire
Since we have our parameters in set, the next thing to do is to replace it into the above formula;
So;
(b)
Find the angular acceleration of the airplane when it is in level flight rad/s²
where;
I = moment of inertia
ω = angular acceleration
The moment of inertia (I) can also be illustrated as:
I = ( 0.741) × (30.9)²
I = 0.741 × 954.81
I = 707.51 Kg.m²
Making angular acceleration the subject of the formula; we have;
ω = 
ω = 0.0347 rad/s²
(c)
Find the linear acceleration of the airplane tangent to its flight path.m/s²
the linear acceleration (a) can be given as:
a = ωr
a = 0.0347 × 30.9
a = 1.07223 m/s²
a ≅ 1.07 m/s²
In order for particles to perform a simple harmonic motion, we must follow the law of force of the form F = -kx, where x is the displacement of the object from the equilibrium position and k is the spring constant. The
force shown in <span>F = -kx is always the restoring force in the sense
that the particles are pulled towards the equilibrium position.
The
repulsive force felt when the charge q1 is pushed into another charge
q2 of the same polarity is given by Coulomb's law
F = </span><span>k *q1* q2 / r^2.
</span>It is clear that Coulomb's law is an inverse-square relationship. It does not have the same mathematical form as the equation <span><span>F = -kx.</span> Thus,
charged particles pushed towards another fixed charged particle of
the same fixed polarity do not show a simple harmonic motion when
released. Coulomb's law does not describe restoring force. When q1 is released, it just fly away from q2 and never returns.</span>
Answer:
The correct option is;
D. There is not enough information to answer this question
Explanation:
The universal gravitational constant = 6.67408 × 10⁻¹¹ 3³/(kg·s²)
For an in between distance of 1 m and equal masses of 60 kg, we have;
The gravitational attraction ≈ 2.403 × 10⁻⁷ N, which does not correspond with the answers, therefore, the best option is that there is not enough information to answer this question.
Answer:
a.
b. 
Explanation:
<u>Given:</u>
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
<h2>
(a):</h2>
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.
At time t = 3 seconds,
<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>
(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.
For the time interval of 2 seconds,
The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,
It is the displacement of the particle in 2 seconds.
