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Marianna [84]
3 years ago
12

Bearing in mind that a sound reflects off an object if the object is larger than the wavelength of the sound, calculate the wave

length of a sound in the air for a sound in the middle of the human hearing range. Then calculate the wavelength for a sound in the middle of the bat hearing range. If you had to use echolocation to detect an insect, which sound has an advantage
Physics
1 answer:
grandymaker [24]3 years ago
4 0

Answer:

mucilage can detect objects the size of millimeters, whereas humans can only detect objects larger than centimeters

Explanation:

The human ear can detect sound in the range of 20 to 20000hz, we are asked to find the mid-range wavelength of the ear at f = 10000 Hz

let's use the relationship between wave speed, wavelength and frequency

           v = λ f

           λ= v / f

           λ = 343/10000

           λ = 3.43 10⁻² m

the sound range of a bat is between 25 10³ Hz and 100 10³ Hz, they also ask for the wavelength in the middle range

         

          f = (100 10³ + 25 10³) / 2

          f = 62.5 10³ Hz

          λ = 343 / 62.5 10³

          λ = 5.5 10⁻³ m

Therefore the size of the object that a bat can detect is much smaller than the size that a human can detect, therefore the bat is capable of detecting insects, but humans.

The mucilage can detect objects the size of millimeters, whereas humans can only detect objects larger than centimeters

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How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

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Julli [10]

Answer:

5s

Explanation:

because m/s=s/t so,

0.5m/s=10m/t

t=10m/0.5m/s

t=5sec

5 0
1 year ago
ECONOMICS !!!!
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Answer:

B

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Answer:

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Explanation:

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Therefore, for this case, since g decreases, the weight decreases but mass remains constant.

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