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erma4kov [3.2K]
3 years ago
12

The noise level coming from a pig pen with

Physics
1 answer:
sweet [91]3 years ago
6 0

Answer:

The decibel of the remaining pigs is 51.5 dB.

Explanation:

Decibel (dB) is a unit of measure of the intensity of a given sound.

Number of pigs = 199, noise level = 74.3 dB.

Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:

                       I    \alpha  N

                       I = kN

where k is the constant of proportionality.

⇒                    k = \frac{I}{N}

                         = \frac{74.3}{199}

                      k = 0.3734

When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.

Thus, the becibel level (I) of the remaining pigs can be determined by:

                  I = kN

                    = 0.3734 × 138

                   = 51.53 dB

The becibel level (I) of the remaining pigs is 51.53 dB.

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Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v (1)

Where:

m_{S}, m_{C} - Masses of the semi truck and the car, measured in kilograms.

v_{S}, v_{C} - Initial velocities of the semi truck and the car, measured in meters per second.

v - Final speed of the system after collision, measured in meters per second.

If we know that m_{S} = 2200\,kg, m_{C} = 2000\,kg, v_{C} = 45\,\frac{m}{s} and v = -15\,\frac{m}{s}, then the initial velocity of the semi truck is:

m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}

v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}

v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}

v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})

v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s}  \right)

v_{S} = -69.545\,\frac{m}{s}  

The semi truck travels at an initial speed of 69.545 meters per second downwards.

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2 years ago
The force required to stretch a Hooke’s-law
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I think this is correct, but I am not entirely certain.

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Find the work done in stretching the spring:

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