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Triss [41]
3 years ago
10

An unknown noise is generated in a building. Sound waves move toward an open doorway. Waves of which wavelength would diffract t

he least through the doorway and spread into the room?
15 centimeters
1 meter
5 meters
12.5 meters
Physics
2 answers:
nadezda [96]3 years ago
6 0
<span>The answer is 15 centimeters. The waves with the smallest wavelengths (also highest frequency) have the least capacity to diffract around objects hence have a shadowing effect, behind the object, especially if the wavelength is small compared to the object. This is also the reason why high-frequency electromagnetic waves do not have a good reach in mountainous regions. </span>




Fofino [41]3 years ago
3 0

The correct choice is

15 centimeters

The diffraction of the sound waves takes place when the wavelength of sound wave is comparable with the size of the object. Waves of wavelengths 1 meter, 5 meter, and 12.5 meters are comparable to the size of the doorway while the wavelength 15 centimeters is very small compared to the size of the doorway. hence the waves of wavelength 15 cm diffracts the least.

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Which statement describes the relationship of resistance and current?
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Choices 'B'; and 'D' both begin with the correct words.
But they should end with the equation

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8 0
3 years ago
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Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke
nirvana33 [79]

Answer:

4 times greater

Explanation:

<u>Step 1:</u> Calculate light-collecting area of a  20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² =\frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 20-m

A_{1} = \frac{\pi d^{2}}{4}

A_{1} = \frac{\pi (20^{2})}{4}

A₁ = 314.2 m²

<u>Step 2:</u> Calculate light-collecting area of a  10-meter Keck telescope (A₂)

A_{2} = \frac{\pi d^{2}}{4}

Where d is the diameter of the circle = 10-m

A_{2} = \frac{\pi (10^{2})}{4}

A₂ = 78.55 m²

<u>Step 3</u>: divide A₁ by A₂  

= \frac{314.2 m^2}{78.55 m^2}

= 4

Therefor,  the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

5 0
4 years ago
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after
frez [133]

Answer:

(a) the deceleration of the player is -80.36 m/s²

(b) the time the collision last is 0.093 s

Explanation:

Given;

Initial velocity of the football player, u = 7.50 m/s

Final velocity of the football player, v = 0

distance traveled = compression of the pad, s = 0.35 m

Part (a) the deceleration of the player

v² = u² + 2as

0 = 7.5² + (2 x 0.35)a

0 = 56.25 + 0.7a

- 56.25 = 0.7a

a = -56.25 / 0.7

a = -80.36 m/s²

Part (b) the time the collision last

v = u + at

t = (v - u)/a

t = (0 - 7.5)/ -80.36

t = - 7.5 / -80.36

t = 0.093 s

7 0
3 years ago
Most distances in the Galaxy are measured in light-years instead of meters. Why do you think this is the case?
lbvjy [14]

Answer and Explanation:

Most of the distances in the galaxy are measured in light years instead of meter because the distances in galaxy are very large and it is very difficult to measure in meters and light year is the largest unit of distance so it is very easy to measure large distances in light year so we prefer light year instead of meters for measuring distances in galaxy.

3 0
4 years ago
Four forces act on bolt A as shown; F1 150N, F2 80N, F3 110N and F4 100N. Determine the magnitude and direction of the resultant
netineya [11]

Complete Question

The  complete question(reference (chegg)) is shown on the first uploaded image

Answer:

The magnitude of the resultant force is  F  =  199.64 \ N

The  direction of the resultant force is  \theta  =  4.1075^o from the horizontal plane

Explanation:

Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be  Fcos(\theta ) while if the force is  moving away from the angle  then the resolved force is  Fsin (\theta )

Now  from the diagram let resolve the forces to their horizontal component

    So

          \sum F_x  =  150 cos(30) + 100cos(15) -80sin (20)

          \sum F_x  =  199.128 \ N

Now  resolving these force into their vertical component can be mathematically evaluated as

         \sum  F_{y}  =  150 sin(30) - 100sin(15) -110 +80 cos(20)

         \sum  F_{y}  =  14.30

Now the resultant force is mathematically evaluated as

        F  =  \sqrt{F_x^2 + F_y^2}

substituting values

        F  =  \sqrt{199.128^2 + 14.3^2}

        F  =  199.64 \ N

The  direction of the resultant force is  evaluated as

       \theta  =  tan^{-1}[\frac{F_y}{F_x} ]

substituting values

       \theta  =  tan^{-1}[\frac{ 14.3}{199.128} ]

       \theta  =  4.1075^o from the horizontal plane

5 0
4 years ago
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