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3 years ago

13

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at

a location 26 m away from the sound source is 3.0 × 10-4 W/m2. What is the intensity at a spot that is 71 m away?

1 answer:

dybincka [34]3 years ago

5 0

Answer:

The intensity at a spot 71 m away is $4.02*10^{-5} Wm^{-2}$

Explanation:

Given:

Initial intensity, $I_{1}=3.0*10^{-4} Wm^{-2}$ at a distance, $d_{1} = 26 m$

Required:

New intensity, $I_{2} =?$ at a distance, $d_{2} = 71 m$

Using the inverse square law,

I ∝ $\frac{1}{d^{2} }$

⇒ $I_{1}$ $I_{1}d_{1}^{2} =I_{2}d_{2}^{2}$

$I_{2} =\frac{I_{1} d_{1}^{2} }{d_{2}^{2}} =\frac{3.0*10^{-4}*26^{2} }{71^{2} } \\$ ×

$I_{2}=4.02*10^{-5} Wm^{-2}$

Thus, the intensity at a spot that is 71 m away is $4.02*10^{-5} Wm^{-2}$

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Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the

balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

$a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)$

Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

$v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)$

Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

$\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)$

b)

We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

$v_{f} = v_{o} + a*\Delta t (4)$

Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

$\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)$

7 0

2 years ago

You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,

telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

$t_{1}=2.19\ sec$

$t_{2}=2.30\ sec$

$t_{3}=2.26\ sec$

$t_{4}=2.29\ sec$

$t_{5}=2.27\ sec$

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

$t_{1}=2.19\approx 2.2\ sec$

$t_{2}=2.30\approx 2.3\ sec$

$t_{3}=2.26\approx 2.3\ sec$

$t_{4}=2.29\approx 2.3\ sec$

$t_{5}=2.27\approx 2.3\ sec$

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0

3 years ago

Determine the frequency of light whose wavelength is 4.257 x10-7 cm

Bas_tet [7]

<span>Frequency x Wavelength = Speed of light Now, speed of light = 3 x 10^5 km/s = 3 x 10^8 m/s = 3 x 10^10 cm/s Frequency = speed/Wavelength = (3 x 10^10)/(4.257 x 10^-7) =7 x 10^16 Hz</span>

6 0

3 years ago

An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel

Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

S = ut + ½ gt²

∴ g = 2S/t²

Substituting the given values,

g = 2 x 144 /6²

= 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

7 0

3 years ago

Paola can flex her legs from a bent position through a distance of 20.1 cm. Paola leaves the ground when her legs are straight,

makkiz [27]

The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/ $s^{2}$ .

Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.

To determine Paola's acceleration, the third equation of free fall is appropriate.

i.e $V^{2}$ = $U^{2}$ ± 2as

where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.

From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.

So that since Poala flies against gravity, then we have:

$V^{2}$ = $U^{2}$ - 2as

0 = $(4)^{2}$ - 2(a x 0.201)

= 16 - 0.402a

0.402a = 16

a = $\frac{16}{0.402}$

= 39.801

a = 39.80 m/ $s^{2}$

Therefore Paola's acceleration is 39.80 m/ $s^{2}$ .

Visit: brainly.com/question/17493533

7 0

2 years ago

Read 2 more answers