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Bond [772]
2 years ago
8

The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of th

em are talking at the same time as loudly as she is, what sound level are you being exposed to?
Physics
1 answer:
Lady_Fox [76]2 years ago
4 0

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

=10^{-5.9}

intensity of sound of 5 persons

I=5\times 10^{-5.9}

dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

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In the shadow of a tree with a dense, leafy canopy, one sees a number of light spots. Surprisingly, they all appear to be circul
Bad White [126]

The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.

In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.

Diffraction can be classified depending on the distance to the observer:

  • Near field or fresnel. In this case the distance from the observer is small and we can see the shape of the object that creates the diffraction.
  • Far field or Fraunhoger. In this case the distance between the obstacle (leaves) and the person is great, here the information on the shape of things is lost and we have two observable forms. Lines for the case of slits and circles for the case of objects with a closed shape.

In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.

In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Learn more about diffraction here:  brainly.com/question/20140459

8 0
2 years ago
Help quick
Yanka [14]

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8 0
2 years ago
Why does the current splits in parallel circuit and why does the voltage remains equal?​
Xelga [282]

<u>Voltage:</u>

It is basically the difference between the charges of the materials on the ends of the Wire

<em>also known as potential difference</em>

It is very similar to the movement of air, it moves from higher density to lower density. in this case, the change in density is the potential difference

So, since voltage is the difference between the charge available on the ends of a wire. Even if the wire splits in parallel circuit, the difference of the charges remains the same

<em>the more the potential difference, the faster electrons will move to the material with lower charge</em>

<u>Current:</u>

Current is the amount of electrons moving through a cross-section of a wire in a period of time

So basically, it is the amount of electrons that move across a given point on a wire in a period of time

If the wire splits, we will have the same amount of electrons moving through as they would if the wire was not split but now, the electrons passing are divided and hence, if we measure the current after the split, we will find that we have a lower current

that's because we have less charge moving through the cross-section of the wire since some of those electrons are moving through a different wire

That's why the current splits in a parallel circuit

3 0
2 years ago
You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of
solong [7]

Answer:

The velocity of the man from the frame of  reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

                                V₂₁ = V₂ -V₁

∴                               V₂ =  V₂₁ + V₁

On substitution

                                 V₂ = 3 + 2

                                      = 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0
3 years ago
n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
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