At which position is the northern hemisphere experiencing winter?

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:

<em>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)</em>

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

2 years ago

How does Lenz's Law illustrate the concept that "you can't get

Crazy boy [7]

Answer:

When there is a change in magnetic flux linkage through a loop of wire, an electromotive force is induced in the loop, according to the Faraday-Newmann-Lenz Law:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

The negative sign in the formula represents Lenz's Law, and tells us about the direction of the electromotive force.

In fact, the negative sign means that the direction of the induced emf is such that to oppose to the change in the magnetic flux that originated the induced emf.

This is a consequence of the law of conservation of energy: no energy can be created out of nowhere. In fact, when the emf is induced in the loop, electrical energy appears in the circuit; however, this electric energy cannot come out of nowhere. Instead, it is just "created" from the transformation of some other form of energy (for instance, the mechanical energy that is used to move the loop in the magnetic field, and changing its magnetic flux).

The negative sign in Lenz's Law tells exactly this: the direction of the induced emf is such that it opposes the initial change in magnetic flux that generated the induced emf, so that overall the total energy is conserved.

5 0

2 years ago

A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s

lisabon 2012 [21]

Answer:

Magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Given

Contact Time = t = 0.05 seconds

Mass (of ball) = 0.80kg

Initial Velocity = u = 25m/s

Final Velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is given by;

F = ma

Where m = 0.8kg

a = Average Acceleration

a = (u + v)/t

a = (25 + 25)/0.05

a = 50/0.05

a = 1000m/s²

Average Force = Mass * Average Acceleration

Average Force = 0.8kg * 1000m/s²

Average Force = 800kgm/s²

Average Force = 800N

Hence, the magnitude of the average force exerted on the wall by the ball is 800N

3 0

3 years ago

A scientist plans to find out the percent of teenagers who like science. She interviews 500 teenagers leaving a science museum a

Georgia [21]

The source of information was biased. It was like walking along a river bank in the country and asking everybody you meet whether they like fishing. Or asking 500 people sitting in the bleachers whether they like baseball.
I'm sure the scientist would have gotten different data if she interviewed 500 teenagers at neighborhood basketball courts, or 500 teenagers at a rock concert.

5 0

2 years ago

Read 2 more answers

Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m

const2013 [10]

Answer:

The answer to your question is vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

hmax = vo²sinФ/ 2g

Solve for vo²

vo² = 2ghmax / sinФ

Substitution

vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

vo² = 19.62(1.3) / 0.866

vo² = 25.51 / 0.866

vo² = 29.45

Result

vo = 5.43 m/s

5 0

2 years ago