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Paraphin [41]
3 years ago
8

A sound from a source has an intensity of 270 db when it is 1 m from the source. what is the intensity of the sound when it is 3

m from the source?
Physics
1 answer:
Lina20 [59]3 years ago
4 0
Sound intensity is inversely proportional to the square of the distance between the source and the receiver.
That is 
I = k/r^2
 where
 k = constant
 r = radius

When r=1,  the intensity is I₁ = k/1 = k
When r=3, the intensity I₂ = k/3² = k/9
Therefore
 I₂ = I₁ /9

In decibels,
I = 10 log₁₀(I/I₀)
where I₀ = reference intensity

When r=1,
10 log₁₀ (I₁/I₀) = 270

When r =3,
10 log₁₀ (I₂/I₀) = 10 log₁₀ [(I₂/I₁)*(I₁/I₀)]
                     = 10 log₁₀ [(1/9)*(I₁/I₀)]
                     = 10 log₁₀(1/9) + 270
                     = 260.5

Answer: 260.5 dB (nearest tenth)
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Answer:

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2 years ago
an object is dropped from a height of 25 meters. at what velocity will it hit the ground? a 7.0 m/s b 11 m/s c 22 m/s d 49 m/s e
kipiarov [429]
Assuming that the object starts at rest, we know the following values:

distance = 25m
acceleration = 9.81m/s^2 [down]
initial velocity = 0m/s

we want to find final velocity and we don't know the time it took, so we will use the kinematics equation without time in it:

Velocity final^2 = velocity initial^2 + 2 × acceleration × distance

Filling everythint in, we have:

Vf^2 = 0^2 + (2)(-9.81)(-25)
The reason why the values are negative is because they are going in the negative direction

Vf^2 = 490.5

Take the square root of that

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3 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

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