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weeeeeb [17]
3 years ago
15

wnat happens to the sound intensity level (expressed in dB) if we doubled the intensity of the voice of a specific sound?​

Physics
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

You hear a sound at 80 dB.

Explanation:

You might be interested in
Which of these will be the correct relationship between work input and work output?
dsp73

<u>Answer:</u>

Work input = Work output * Work against friction is your answer so C

<u>Explanation:</u>

I hope this helps you :)

8 0
2 years ago
If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now
Andru [333]

Answer:

The new period will be √6 *T

Explanation:

period ,T=2π√(L/g)       ................equation 1

where T is the period on earth

gravitational acceleration on the moon is g/6

T1 = 2π√[L/(g/6)]

T1=2π√(6L/g)               ...............equation 2

divide equation 2 by 1

T1/T =2π√(6L/g)÷2π√(L/g)

T1/T =√(6L/L)

T1/T =√6

T1 = √6 *T

5 0
3 years ago
Un prisma de cemento pesa 2500 N y ejerce una presión de 125 Pa, ¿cuál es el valor del área en la cual se apoya?
Eduardwww [97]

Answer:

Area = 20 m²

Explanation:

Given the following data;

Force = 2500 N

Pressure = 125 Pa

To find the area on which it rest;

Mathematically, pressure is given by the formula;

Pressure = \frac {Force}{area}

Making area the subject of formula, we have;

Area = \frac {Force}{pressure}

Substituting into the formula, we have;

Area = \frac {2500}{125}

Area = 20 m²

3 0
3 years ago
A cart travels 4.00 meters east and then4.00 meters north. determine the magnitude ofthe cart’s resultant displacement.
gogolik [260]
^
|
|
|
----------->|

Square root of (4^2 + 4^2) = 4*squareRoot(2)
7 0
3 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

5 0
3 years ago
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