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uranmaximum [27]
3 years ago
13

A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes.

What is the smallest insect a bat can detect? Assume that bats emit a chirp at a frequency of 33.3 kHz, and that the speed of sound in air is 536 m/s. Answer in units of mm.
Physics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

size of insect = wavelength = 16.1 mm

Explanation:

Since the bat can detect the size of small insects which are of size equal to the wavelength of sound in air

So here we can say that the wavelength of sound is the ratio of speed of sound and its frequency

here we know that

speed of sound in air is 536 m/s

frequency of sound is 33.3 kHz

so here we can say

\lambda = \frac{v}{f}

\lambda = \frac{536}{33.3 \times 10^3}

\lambda = 0.0161 m

\lambda = 16.1 mm

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A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme
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Answer: The density of this piece of jewelry is 8.90g/cm^3

Explanation:

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\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

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Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = 14.7cm^3   1cm^3=1ml

Putting values in above equation, we get:

\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3

Thus density of this piece of jewelry is 8.90g/cm^3

8 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
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