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2 years ago

The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?

Physics

2 answers:

SIZIF [17.4K]2 years ago

8 0

The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
$I(r)= \frac{1}{r^2}$
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put
$I= \frac{1}{r^2}= \frac{1}{(1)^2}=1$
The intensity at r=3 m will be
$I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}$
Therefore, the sound intensity has decreased by a factor $1/9$ .

RSB [31]2 years ago

3 0

its a on edge just finished my exam

heres the diagram

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How does fabric softener help prevent static electricity?

Olin [163]

Answer:

Dryer sheets stop static cling by absorbing static electricity formed by different fabrics during your dryer cycle. The fabric softeners that coat dryer sheets are positively charged ions to balance the electrons and ions that cause static cling, leaving you with soft clothes without the static.

Explanation:

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1 year ago

The work done in lifting a brick of mass 2kg through a height of 5m above the ground will be

antoniya [11.8K]

Answer:

D. 100 J

Explanation:

Given

Mass (m) = 2 kg

Height (h) = 5 m

Acceleration due to gravity (g) = 10 m/s²

Now

Work done(W)

= m * g * h

= 2 * 10 * 5

= 100 Joule

Hope it will help :)❤

8 0

2 years ago

I what is the resistance of

Lynna [10]

The resistance of a conductor is given by:
$R= \frac{\rho L}{A}$
where
$\rho$ is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area

We can use this formula to solve both parts of the problem.

a) The length of the copper wire is L=1.0 m. Its diameter is d=0.50 mm, so its radius is
$r= \frac{d}{2}=0.25 mm=0.25 \cdot 10^{-3} m$
And its cross-sectional area is
$A=\pi r^2 = \pi (0.25 \cdot 10^{-3}m)^2 = 1.96 \cdot 10^{-7} m^2$
The copper resistivity is $\rho=1.68 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
$R= \frac{\rho L}{A}= \frac{(1.68 \cdot 10^{-8} \Omega m)(1.0 m)}{1.96 \cdot 10^{-7} m^2}= 8.57 \cdot 10^{-2} \Omega$

b) The length of this piece of iron is L=10 cm=0.10 m. Its cross-sectional size is L=1.0 mm=0.001 m, so its cross-sectional area is
$A=L^2 = (0.001 m)^2 =1 \cdot 10^{-6}m^2$
The iron resistivity is $\rho = 9.71 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
$R= \frac{\rho L}{A}= \frac{(9.71 \cdot 10^{-8} \Omega m)(0.10 m)}{1.0 \cdot 10^{-6} m^2}=9.71 \cdot 10^{-3} \Omega$

3 0

3 years ago

Two identical satellites orbit the earth in stable orbits. one satellite orbits with a speed v at a distance r from the center o

wariber [46]

The available options are: (found the complete text on internet)
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r

Solution:
The correct answer is C) at a distance greater than r.

In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
$G \frac{Mm}{r^2}=m \frac{v^2}{r}$
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed

Re-arranging the equation, we get
$r= \frac{GM}{v^2}$
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.

7 0

2 years ago

Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc

slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

$F=55\ kg\times 1.4\ m/s^2$

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0

3 years ago