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aleksandr82 [10.1K]
3 years ago
6

You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie

nd moves a distance 6 cm from you?
Physics
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance.  i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by =  $\frac{1}{0.06^2}$

                       = $3.6 \times 10^{-3}$ times

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A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high
Lelu [443]

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}

Where:

v_{o} - Initial velocity, measured in meters per second.

y - Final height, measured in meters.

y_{o} - Initial height, measured in meters.

t - Time, measured in seconds.

g - Gravitational constant, measured in meters per square second.

(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}

Given that y = 0\,m, v_{o} = 15\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} and t = 16\,s, the final height of the package is:

y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}

y_{o} = 1015.296\,m

The package was released at a height of 1015.296 meters.

7 0
3 years ago
Need help science the top and bottom! Please
Oksi-84 [34.3K]

errrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

7 0
3 years ago
A thin lens with a focal length of 6.0 cm is used as a simple magnifier by (a) what angular magnification is obtainable with the
Serhud [2]

Answer:

4.167

4.83871 cm

Explanation:

u = Object distance

v = Image distance = 25 cm

f = Focal length = 6 cm

Angular magnification is given by

m=\frac{25}{f}\\\Rightarrow m=\frac{25}{6}\\\Rightarrow m=4.167

The angular magnification of the lens is 4.167

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{6}=\frac{1}{u}+\frac{1}{-25}\\\Rightarrow \frac{1}{6}+\frac{1}{25}=\frac{1}{u}\\\Rightarrow \frac{1}{u}=\frac{31}{150}\\\Rightarrow u=4.83871\ cm

The closest distance by which the object can be examined is 4.83871 cm

5 0
2 years ago
Earth has a surface area of 197 million square miles. Convert this area into each of the following units.
slamgirl [31]

Answers:

a) 510,225,121.4 km^{2}

We know the Earth's surface area in square miles is:

197000000 mi^{2}

On the other hand, we know 1 mi=1.60934 km.

Then:

197000000 mi^{2} \frac{(1.60934 km)^{2}}{1 mi^{2}}=510,225,121.4 km^{2}

b) 510.225 Mm^{2}

In this part, we can work with the obtained value in part a:

510,225,121.4 km^{2}

And knowing 1 km=0.001 Mm

Hence:

510,225,121.4 km^{2} \frac{(0.001 Mm)^{2}}{1 km^{2}}=510.225 Mm^{2}

c) 5.1022(10)^{16} dm^{2}

Knowing 1 Mm=10^{7} dm:

510.225 Mm^{2} \frac{(10^{7} dm)^{2}}{1 Mm^{2}}=5.1022(10)^{16} dm^{2}

7 0
3 years ago
A skier begins skiing straight down a hill witha constant slope, starting from rest. If friction is negligible, as the skier goe
makvit [3.9K]

Answer:

acceleration is constant, with a value less than 10 m/s²

Explanation:

The force pushing her downward are his weight and the at an angle

The weight is given as

W=mg

The component of the weight along x axis is Wsin θ

Forward force is Wsin θ

This forward force is the only force acting on the x-axis since it frictionless

Then, using newton law

ΣF = ma

WSinθ=ma

mgSinθ=ma

Then, a=gSin θ

Since sinθ is always between 1 and -1

-1<=sinθ<=1

The only time a=g is when It is on a straight line and not an incline plane and since we are give an inclined plane

Then, a will be less that g

Where g is constant =10m/s²

So the correct answer is D

4 0
3 years ago
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