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adell [148]
2 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]2 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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An engineer wants to design an oval racetrack such that 3.20 × 10 3 lb racecars can round the exactly 1000 ft radius turns at 10
Reptile [31]

Answer:

The banking angle necessary for the race cars is 34.84°

Explanation:

For normal reaction the expression is:

\\Nsin\theta = \frac{mv^{2} }{R}  =Fc\\tan\theta =\frac{v^{2} }{Rg}  \\\theta =tan^{-1} (\frac{v^{2} }{Rg} )\\\theta =tan^{-1} (\frac{(102*0.447)^{2} }{1000*0.3048*9.8} )=34.84

4 0
1 year ago
If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these
insens350 [35]

To develop this problem it is necessary to apply the concepts related to Sound Intensity.

By definition the intensity is given by the equation

\beta = 10Log(\frac{I}{I_0})

Where,

I = Intensity of Sound

I_0= Intensity of Reference

At this case we have that 15 engines produces 15 times the reference intensity, that is

I= 15I_0

And the total mutual intensity is 100 dB, so we should

\beta = 100-10*log(\frac{15I_0}{I_0})

\beta = 100-10*log(15)

\beta = 100-11.76

\beta = 88.23dB

Therefore each one of these engines produce D. 88dB.

5 0
1 year ago
A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco
34kurt

Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform  = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

gh = 1/2v²

v² = 2gh

v = √(2gh)

v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s

So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

= -310.1 kgm/s ÷ 0.02 s

= -15,505 N

So, the force transmitted to her bones is 15,505 N

3 0
1 year ago
A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0
2 years ago
The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth i
faltersainse [42]

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

 

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

 

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

<span>v = 18.55 miles / s = 29.85 km / s</span>

5 0
2 years ago
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