Answer:
The net torque is 0.0372 N m.
Explanation:
A rotational body with constant angular acceleration satisfies the kinematic equation:
(1)
with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:
Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:
The negative sign indicates the sphere is slowing down as we expected.
Now with the angular acceleration we can use Newton's second law:
(2)
with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:
With M the mass of the sphere an R its radius, then:
Then (2) is:
Answer: apparent weighlessness.
Explanation:
1) Balance of forces on a person falling:
i) To answer this question we will deal with the assumption of non-drag force (abscence of air).
ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).
iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).
2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.
3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.
Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.
The correct answer is <span>3)
.
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In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
</span>
<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
</span>
<span>
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
</span>
<span>
</span>
Answer:
160 Hz , 240 Hz , 400 Hz
Explanation:
Given that
Frequency of forth harmonic is 320 Hz.
Lets take fundamental frequency = f₁
f₁=80 Hz
Frequency of first harmonic = f₂
f₂=2 f₁
f₂ =2 x 80 = 160 Hz
Frequency of second harmonic = f₃
f₃= 3 f₁=3 x 80 = 240 Hz
Frequency of fifth harmonic = f₅
f₅= 5 f₁= 5 x 80 = 400 Hz
Three frequencies are as follows
160 Hz , 240 Hz , 400 Hz
Answer:
The change in the centripetal acceleration of the brother,
Δa = V₂²/R - V₁²/R
Explanation:
Given data,
A sister spins her brother in a circle of radius, R
The angular velocity of the brother, ω₁ = V₁/R
The angular velocity of the brother, ω₂ = V₂/R
The centripetal acceleration is given by the relation
a = V²/R
Therefore change in the centripetal acceleration of the brother,
Δa = V₂²/R - V₁²/R
