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Masja [62]
3 years ago
14

An empty semi-truck sits on the scale of a weighing station. It is then loaded with 100 irritated pigeons. These two events then

happen in sequence. a) First, the super irritated pigeons just stand on the floor of the truck. b) Second, they take flight inside the truck. They all fly in the enclosed space such that they do not collide with each other or the walls/ceiling/floor. Is the reading on the scale different in a) versus b)? Does the reading on the scale change during the transition of these events?
Physics
1 answer:
mina [271]3 years ago
4 0

Answer:

The reading in the scale is going to be the same, or if it experiences some change it would be minimum. The reason is because the truck will act as a big closed cage, therefore, when the pigeons fly, the air they move with their wings in order to keep flying,  exerts the same force on the closed cage, that if they were standing on the ground.

If the truck however, allows the air flow, the weight might change, because under this scenario, the air flowing could represent less force exerted on the balance.

Explanation:

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
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Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

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.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<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.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

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What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s
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The net force is 270 N

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We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

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