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adell [148]
3 years ago
5

The bird is held in level flight due to the force exerted on it by the air as the bird beats its wings. What is the maximum valu

e of this force due to the air? Note that the bird's weight, the force of gravity acting on it, is calculated using w=mg, a result that will be justified in Chapter 5.
Physics
1 answer:
Cerrena [4.2K]3 years ago
3 0

Answer:

 maximumforce is F = mg

Explanation:

For this case we must use Newton's second law,

     Σ F = m a

bold indicate vectors, so we will write it in its components x and y

 X axis

       Fₓ = maₓ

 Axis y

      Fy - W = m aa_{y}

Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components

      Cos θ = Fₓ / F

      Fₓ = F cos θ

      sin θ = Fy / F

      Fy = F sin θ

Let's replace and calculate

      F sin θ -w = m a

 

As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)

       F sin θ = w = mg

The maximum value of this equation occurs when the sin=1, in this case

      F = mg

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Answer:

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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
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To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

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\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

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Recall that the speed of light is equivalent to

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B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

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3 0
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The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud
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<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = \frac{Gm_1m_2}{r^2}

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

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Thus r = \sqrt{\frac{Gm_1m_2}{F} }

If we substitute the values in the above equation

r = \sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }

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Answer:

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Explanation:

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