Please help me out! This is a big question!

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ( $\Delta t$ ), in seconds:

$\Delta t = t_{A}-t_{C}$ (1)

Where:

$t_{C}$ - Time spent by the sound in concrete, in seconds.

$t_{A}$ - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

$\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}$

$\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)$

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$ (2)

Where:

$v_{C}$ - Speed of the sound in concrete, in meters per second.

$v_{A}$ - Speed of the sound in the air, in meters per second.

$x$ - Distance traveled by the sound, in meters.

If we know that $\Delta t = 6.4\,s$ , $v_{C} = 3000\,\frac{m}{s}$ and $v_{A} = 343\,\frac{m}{s}$ , then the distance travelled by the sound is:

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$

$x = 2478.585\,m$

The impact occured at a distance of 2478.585 meters from the person.

7 0

3 years ago

A generator connected to an RLC circuit has an rms voltage of 140 V and an rms current of 31 mA. Part A

vredina [299]

Answer:

53.06 ohm

Explanation:

We have given the rms voltage V =140 V

And the rms current i=31 mA

So the impedance $Z=\frac{V}{i}=\frac{140}{3\times 10^{-3}}=46.666kohm$

Resistance is given as R = 3 kohm

And capacitive reactance $X_C=6.5kohm$

We know that $Z=\sqrt{R^2+(X_L-X_C)^2}$

$46.666=\sqrt{3^2+(X_L-6.5)^2}$

Squaring both side $2177.1556=9+(X_L-6.5)^2$

$(X_L-6.5)^2=2168.155$

$(X_L-6.5)=46.5634$

$X_L=53.063ohm$

8 0

3 years ago

What is the relationship between the intervals on the Celsius scale and the Kelvin scale

MAXImum [283]

An interval of 1 degree Celsius is equal to 1 Kelvin

Explanation:

There are different scales used for measuring temperatures. The most common ones are:

- Celsius scale: it is set by the following convention: the zero in this scale is placed at the freezing temperature of the water, while the 100 degrees is placed in corresponding to the boiling point of water. The the scale is divided into 100 equal intervals between these two points. Temperatures in this scale are indicated with the Celsius degree ( $^\circ{}C$ .

- Kelvin scale: it is set such that the absolute zero (the temperature at which particles in matter stop to have any motion) is placed at zero in this scale. A difference of 1 degree in this scale is equal to 1 degree Celsius. Temperatures in this scale are indicated with the Kelvin (K).

The conversion between the two temperatures is:

$T(^{\circ}) = T(K) - 273.15$

So for example, the absolute zero ( $0 K$ ) corresponds to $-273.15^{\circ}C$ in the Celsius scale. However, intervals between two temperatures are measured in the same way in the two scales: therefore, an interval of 1 degree in the Celsius scale is equal to an interval of 1 Kelvin in the Kelvin scale.

Learn more about temperatures here:

brainly.com/question/1603430

brainly.com/question/4889772

#LearnwithBrainly

4 0

3 years ago

A box is being pushed across the floor at a constant velocity with an

Oksana_A [137]

Answer:

Magnitude of frictional force = 30 N

Explanation:

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn=m.a

The net force is the vector addition of each individual force. If the sum of all the forces acting on an object is zero, then the acceleration is zero. That means the object moves at a constant speed or is at rest.

When an object is pushed across a horizontal rough surface, there are two forces acting in the direction of the motion: The applied force and the frictional force.

If the applied force is greater than the frictional force, then the object moves at a constant positive acceleration. If the frictional force is greater than the applied force, then the object won't move at all (if it was at rest) or will start a deaccelerated motion (braking).

Finally, if both forces are equal, the object will move at a constant speed or remains at rest. Since the box is moving at a constant speed, we can conclude the frictional force equals the applied force:

Magnitude of frictional force = 30 N

8 0

4 years ago

A sound from a source has an intensity of 270 dB when it is 1 m from the source.

Rufina [12.5K]

Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: $\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
where
$I_{2}$ is the intensity at distance 2
$I_{1}$ is the intensity at distance 1
$d_{2}$ is distance 2
$d_{1}$ is distance 1

We can infer for our problem that $I_{1}=270$ , $d_{1}=1$ , and $d_{2}=3$ . Lets replace those values in our formula to find $I_{2}$ :
$\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
$\frac{I_{2} }{270} =( \frac{1}{3} )^2$
$\frac{I_{2} }{270} = \frac{1}{9}$
$I_{2}= \frac{270}{9}$
$I_{2}=30$ dB

We can conclude that the intensity of the sound when is 3 m from the source is 30 dB.

8 0

3 years ago