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2 years ago

The noise level coming from a pig pen with

Physics

1 answer:

sweet [91]2 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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Batteries involve the flow of electricity from one metal to another. What happens to these metals to cause the batteries to, eve

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The metals will start to rust lol. i think. because this messes up how the metals conduct the flow of the electricity.

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2 years ago

A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8

fomenos

Answer:

$v = 186.90\,\frac{m}{s}$

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

$\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h$

The final velocity of the system formed by the ballistic pendulum and the bullet is:

$v = \sqrt{2\cdot g\cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}$

$v\approx 0.78\,\frac{m}{s}$

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

$(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )$

$v = 186.90\,\frac{m}{s}$

5 0

2 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

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3 years ago

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1 year ago

Read 2 more answers

A current of 2 A flows through a resistor. The voltage across the resistor is 18 V.

stepan [7]

Answer:

$R=9\ \Omega$

Explanation:

Given that,

Current, I = 2 A

Voltage across the resistor, V = 18 V

We need to find the value of resistance of the resistor. Let the resistance be R. We can find it using Ohm's law i.e.

V = IR

Where

R is the resistance of the resistor

$R=\dfrac{V}{I}\\\\R=\dfrac{18}{2}\\\\R=9\ \Omega$

So, the resistance of the resistor is equal to $9\ \Omega$ .

3 0

2 years ago