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andreyandreev [35.5K]
3 years ago
9

An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-dia

meter eardrum so exposed?
Physics
2 answers:
Dafna11 [192]3 years ago
7 0

Answer:

1.44x10⁻³J

Explanation:

Given :

Time = 8hrs *(3600secs/1hr)= 28.8*10³seconds

Intensity= 90dB

D= 0.008m

Radius=0.008m/2

=0.004m

the sound level in decibel can be expressed below as

I=10dBlog[I/I₀]

where I₀=10⁻¹²/m² which is the refrence intensity

90=10log[I/10⁻¹²]

I=0.001W/m²

we know that intensity of the wave which id the average rate per unit area which energy is transfered can be calculated using below formular

I = P/A

where P= powerr which is renergy transfer at a time

A= area= πr²

making P subject of formular we have

P= I πr²

= 5.02 x10⁻⁸W

Energy =power x time

E=28800*5*10⁻⁸

=0.001443J

therefore,the energy in joules is 1.44x10⁻³J

11111nata11111 [884]3 years ago
5 0

Answer:

1.4E-3J

Explanation:

Given that

Time = 8hrs = 28.8E3 seconds

Intensity= 90dB

D= 0.008m

Radius= 0.004m

So intensity is sound level Bis

10dBlog(I/Io)

I= 10 (B/10dB)Io

= 10( 90/10) x 10^-12

=0.001W/m²

But we know that

I = P/A

P= I πr²

= 5.02 x10^-8W

But energy is power x time

So E= 5.02E-8 x 28.8E3

= 1.4E-3J

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Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
Which of the following statements is true about the scientific process?
KiRa [710]
The best and most correct answer among the choices provided by your question is the third choice or letter C.
<span>The statement "Your hypothesis must be testable." is true about the scientific process.


</span>I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
4 0
3 years ago
How does the current in a circuit change if the resistance is doubled? The current is halved. The current is doubled. The curren
Inessa05 [86]

Answer:

The current is halved

Explanation:

The relationship between the current and the resistance is given by Ohm's Law, as follows:

V = IR\\I = \frac{V}{R}

where,

V = Voltage

I = Current

R = Resistance

Therefore, if we double the resistance:

I' = \frac{V}{2R}\\\\I' = \frac{1}{2}I

Hence the correct option is:

<u>The current is halved</u>

7 0
3 years ago
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Car A starts in Sacramento at 11am. It travels along 400 mile route to Los Angeles at 60 mph. Car B starts from Los Angeles at n
damaskus [11]

Answer:

  • 38.89 miles

Explanation:

from the question we have the following:

distance between Sacramento and los angles = 400 miles

speed of car A = 60 mph

start time of car A = 11 am

speed of car B = 75 mph

start time of car B = 12 pm

distance of Fresno from Los Angeles = 150 miles

  • To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
  • Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
  • From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
  • Distance covered by car A = speed x time(t) = 60 x t = 60t
  • Distance covered by car B = speed x time(t) = 75 x t = 75t
  • 60t + 75t = 340 miles
  • 135t = 340
  • t = 2.51 hours
  • Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
  • Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
  • 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.

4 0
3 years ago
How much will the kinetic energy of a school bus increase if its velocity is tripled?
son4ous [18]

Answer:

Explanation:

Given a school bus.

Let say initially the school bus is traveling with speed "v"

Let assume mass of school bus is "m"

Then, the initial kinetic energy is

K.E_initial = ½mv²

Now, if the initial velocity is tripled,

Then, the new velocity is

v_new = 3v.

Note: the mass of the school does not change it is constant

Then, new kinetic energy is

K.E_new = ½m(v_new)²

v_new = 3v

Then,

K.E_new = ½m(3v)²

K.E_new = ½m × 9v²

K.E_new = 9 × ½mv²

Since K.E = ½mv²

Then,

K.E_new = 9 × K.E

So, the new kinetic energy will be 9 times the initial kinetic energy.

So, option D is correct

D. It will be nine times greater.

4 0
3 years ago
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