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2 years ago

Which range is the approximate audible range for humans?

Physics

2 answers:

allsm [11]2 years ago

8 0

Answer:

Explanation:

bcz mre than that they will be affected

nlexa [21]2 years ago

6 0

$\triangleright$ $\overline{\qquad{\qquad{\qquad{\qquad{\qquad{\qquad{\qquad{\qquad}}}}}}}}$ $\triangleleft$

$\boldsymbol{Answer.}$

$\bullet$ Option [A].

$\triangleright$ $\overline{\qquad{\qquad{\qquad{\qquad{\qquad{\qquad{\qquad}}}}}}}$ $\triangleleft$

$\boldsymbol{Explanation.}$

Hi student! Let me help you out on this question.

The approximate audible range for humans is.

$\boldsymbol{20-20,000\:Hertz\,(Hz)}$ .

Humans can't hear sounds lower than 20 Hz or higher than 20,000 Hz.

$\boldsymbol{\therefore,\:Option\:\big[A\big]\,is\,correct.}$

Hope that this helped you out! Best wishes!

$\star\triangleright\boldsymbol{Reach\:far.\,Aim\,high.\,Dream\,big.}\triangleleft\star$

$\ddagger\it{-Greetings!-}\dagger$

_______________________

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A scientist performs an experiment and asks other scientists around the

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Answer: C

Explanation: I think

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It is easier to open the lid of a can using a spoon why?

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It depends on how you use the spoon...

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3 years ago

Read 2 more answers

Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce

Taya2010 [7]

Answer:

So the sound intensity level they would experience without the earplugs is 110.32dB.

Explanation:

Given data

Sound intensity by factor =215

Sound intensity level =87 dB

To find

Sound intensity level they would experience without the earplugs

Solution

First we need to find the new sound intensity level

$I_{n}=215(10^{\frac{87}{10} } )\\I_{n}=1.08*10^{11})$

The dB can be calculated as:

$dB=10log(I_{n})\\$

Substitute the given values

$dB=10log(1.08*10^{11})\\dB=110.32dB$

So the sound intensity level they would experience without the earplugs is 110.32dB.

7 0

2 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

2 years ago

what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?

zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

5 0

3 years ago