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lubasha [3.4K]
3 years ago
8

A

Physics
1 answer:
Arada [10]3 years ago
4 0

Answer:

i would think c

Explanation:

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Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l
goblinko [34]

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}

where,

Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

\Delta t = \frac{0.18\ m}{340\ m/s}

<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>

4 0
3 years ago
Two boxes are 8 cm apart. Which of the following should Janet do to decrease the gravitational force between the boxes?
OlgaM077 [116]

Answer:

the answer is 2.

Explanation:

4 0
3 years ago
Which statement is the correct representation of these electric field lines?
GuDViN [60]
I would say a. hope this helps
4 0
3 years ago
Read 2 more answers
The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
likoan [24]

Answer:

14869817.395 m

Explanation:

\theta=22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians

From Rayleigh Criterion

\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0
3 years ago
What is the fundamental frequency (in Hz) of a 0.632 m long tube, open at both ends, on a day when the speed of sound is 344 m/s
Greeley [361]

Answer:

f=272.15Hz

Explanation:

Given data

Length of tube L=0.632 m

Speed of sound v=344 m/s

To find

Fundamental frequency f

Solution

The fundamental frequency of the tube can be given as:

f=\frac{v}{2L}\\ f=\frac{344m/s}{2(0.632m)}\\ f=272.15Hz

4 0
4 years ago
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