The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its

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

2 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

2 years ago

If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy.

Alina [70]

If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy. It is a true statement .

Kinetic energy is the energy that an object possesses due to its motion. It is basically the energy of mass in motion. Kinetic energy can never be negative and it is a scalar quantity i.e. it provides only the magnitude and not the direction.

According to law of conservation of mechanical energy change in potential energy is equal and opposite to the change in the kinetic energy.

According to the principle of conservation of mechanical energy, The total mechanical energy of a system is conserved i.e., the energy can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature.

since, potential energy is stored in the form of work done

Work done = Fs cos (theta)

If force always acts perpendicular to an object's direction of motion

theta = 90 °

cos (90 ) = 0

Work done = 0

since , there is no work done , hence kinetic energy will not change

To learn more about kinetic energy here

brainly.com/question/12669551

#SPJ4

7 0

1 year ago

What's the simplest approach to solve a physics dimensional question

Darina [25.2K]

Explanation:

If Q is the unit of a derived quantity represented by Q = MaLbTc, then MaLbTc is called dimensional formula and the exponents a, b and, c are called the dimensions.

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7 0

3 years ago

A 2-kg box is pushed to the right by a force of 4 N for a distance of 32 m. It has an initial velocity of 4 m/s to the right. NO

rewona [7]

Answer: a) 8 Kg m/s b) 16 Kg m/s c) 24 Kg m/s d) 16 J e) 128 J f) 144 J

g) 4 s

Explanation:

a) As momentum by definition is the product of mass times the velocity (is a vector quantity), we can write in this case the following:

pi = m. v₀ = 2 Kg . 4 m/s = 8 Kg. m/s

b) In order to get the change in momentum, we need to get first the final speed of the object.

As we have the total distance travelled, and we could find the acceleration, we could use a kinematic equation to solve the question, but later we will need the kinetic energy, it would be better to apply the work-energy theorem, and calculate ΔK as the work done by external force F, as follows:

ΔK = F . d = 1/2 m (vf² - v₀²)

As we know F, d, m, and v₀, we can solve the equation above for vf:

vf = 12 m/s

So, we can compute the final momentum as follows:

pf = m. vf = 2 Kg. 12 m/s = 24 Kg. m/s

Finally, we can find the change in momentum, as the difference between the final momentum and the initial one, calculated in a):

Δp = pf - pi = 24 Kg. m/s - 8 Kg. m/s = 16 Kg. m/s

c) As we have already found, final momentum is as follows:

pf = m . vf = 2 Kg. 12 m/s = 24 Kg. m/s

d) By definition the initial kinetic energy of the box is as follows:

Ki = 1/2 m v₀² = 1/2. 2 Kg .4² m²/s² = 16 J

e) We can find the change in the kinetic energy taking directly the difference between the final and initial ones, as follows:

ΔK = Kf - Ki = 1/2. 2 Kg (12² - 4²) m²/s² = 128 J

f) From above, we have Kf = 1/2 m. vf² = 1/2 . 2 Kg. 12² m²/s² = 144 J

g) As we know the magnitude of F, and the value of m, we can find the acceleration (assumed constant) , applying Newton's Second Law, as follows:

Fext = m .a ⇒ a = F/m = 4 N / 2 Kg = 2 m/s²

Appying the definition of acceleration, we can solve for t, as follows:

t = (vf-v₀) / a = (12 m/s - 4 m/s) / 2 m/s² = 4 s

6 0

2 years ago