A car travels at a constant speed of 15m/s. How many miles does it travel in 1hrs

The tiny spring displacement with a spring constant of 1.20N/m after strecting is C) 4.2 mm.Option C

To find the displacement of the spring when it is stretched by a 0.0050-N force, we can use Hooke's Law. Hooke's Law states that the force exerted on a spring (F) is equal to the spring constant (k) multiplied by the displacement (x):
F = k * x
We are given the spring constant (k) as 1.20 N/m and the force (F) as 0.0050 N. We need to solve for the displacement (x). Rearrange the equation to isolate x:
x = F / k
Now, plug in the given values:
x = (0.0050 N) / (1.20 N/m)
x = 0.00416667 m
To convert the displacement from meters to millimeters, multiply by 1,000:
x = 0.00416667 m * 1,000 mm/m
x = 4.16667 mm
Since the answer should be in two decimal places, we can round it to 4.17 mm. None of the given options match this value, but the closest option is C) 4.2 mm. Therefore, the answer is:C) 4.2 mm

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The moment of inertia of the 2.90 kg, 30.0-cm-diameter disk for rotation about an axis through the center is 0.1969 kg [tex]m^2[/tex].

The moment of inertia of a disk rotating about an axis through its center can be calculated using the formula:

I = (1/2) * M * [tex]R^2[/tex]

where I is the moment of inertia, M is the mass of the disk, and R is the radius of the disk.

In this case, the mass of the disk is M = 2.90 kg, and the radius of the disk is half of its diameter, R = 0.30 m / 2 = 0.15 m. Substituting these values into the formula, we get:

I = (1/2) * 2.90 kg * (0.15 [tex]m)^2[/tex]

I = 0.1969 kg [tex]m^2[/tex]

Therefore, the moment of inertia of the 2.90 kg, 30.0-cm-diameter disk for rotation about an axis through the center is 0.1969 kg [tex]m^2.[/tex]

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The kinetic energy of the photoelectron produced in the energy meter of the PAC device is 17.1 eV.

The energy of a photon is given by the equation:

E = hf

where E is the energy of the photon, h is Planck's constant (h = 4.1 × [tex]10^{-15[/tex] eV•s), and f is the frequency of the photon.

Since the photon is not absorbed in the solution, its energy is not used to overcome the work function of the metal, so the energy of the photon is equal to the kinetic energy of the photoelectron produced:

E = KE

The work function of the metal is given as 3.4 eV.

To calculate the kinetic energy of the photoelectron, we can use the following equation:

KE = E - work function

Substituting the given values, we get:

E = hf = (4.1 × [tex]10^{-15[/tex] eV•s)(5.0 × [tex]10^{15[/tex] Hz) = 20.5 eV

KE = E - work function = 20.5 eV - 3.4 eV = 17.1 eV

Therefore, the kinetic energy of the photoelectron produced in the energy meter of the PAC device is 17.1 eV.

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An electric motor is not similar to a radio receiver or an automobile battery, but it is similar to an electric generator.

Both electric motors and generators use the principles of electromagnetism to convert electrical energy into mechanical energy or vice versa.

However, while electric generators convert mechanical energy into electrical energy, electric motors convert electrical energy into mechanical energy.

Therefore, an electric motor is analogous to an electric generator rather than a radio receiver or car battery.

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The window loses approximately 38,080 W, or 38.08 kW, of heat each hour.

The rate of heat loss through a window can be calculated using the formula:

Q/t = kA(∆T/d)

where Q/t is the rate of heat transfer, k is the thermal conductivity of glass, A is the area of the window, ∆T is the temperature difference between the inner and outer surfaces of the glass, and d is the thickness of the glass.

Given that the glass is 9.0 mm thick, or 0.009 m, the area of the window is 2.7 m x 2.4 m =[tex]6.48 m^2[/tex], the temperature difference is 4°C, and the thermal conductivity of glass is approximately 1.05 W/mK, we can solve for Q/t:

Q/t = [tex](1.05 W/mK)(6.48 m^2)(4°C)/(0.009 m)[/tex]

Q/t = 38,080 W

Therefore, the window loses approximately 38,080 W, or 38.08 kW, of heat each hour.

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Kinetic energy, same speed, mass.

Answer:

We know that the kinetic energy of the rabbit and the Irish setter is the same. The formula for kinetic energy is:

Kinetic energy = (1/2) * mass * speed^2

Let's set the kinetic energy equal for both:

(1/2) * 5.0 kg * rabbit speed^2 = (1/2) * 12 kg * (2.8 m/s)^2

Simplifying:

2.5 * rabbit speed^2 = 47.04

Dividing both sides by 2.5:

rabbit speed^2 = 18.816

Taking the square root of both sides:

rabbit speed = 4.34 m/s

Therefore, the rabbit is running at a speed of 4.34 m/s.

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The rabbit is running at a speed of 3.21 m/s.

The kinetic energy of an object is given by the formula: KE = [tex]1/2 mv^2,[/tex]where m is the mass of the object and v is its velocity. Since the mass of the rabbit and the setter are the same, we can use the formula above to find the velocity of the rabbit.

To find the velocity of the rabbit, we can use the following equation:

KE rabbit = KE setter

1/2[tex]mrabbit^2[/tex] = 1/2 [tex]mvsetter^2[/tex]

where mrabbit is the mass of the rabbit and msetter is the mass of the setter.

Rearranging the equation, we get:

[tex]mrabbit^2[/tex] = [tex]mvsetter^2[/tex] + 2KERabbit

Substituting the values given in the problem, we get:

[tex]mrabbit^2[/tex] = (12 kg)[tex](3.6 m/s)^2[/tex] + 2*1/2 * (1/2)(1/2)(5 kg)[tex](3.6 m/s)^2[/tex]

[tex]mrabbit^2[/tex] = 153.6 [tex]kg m^2[/tex] + 3.36 m^2

[tex]mrabbit^2[/tex] = 157 [tex]m^2[/tex]

mrabbit = [tex](157 m^2)[/tex]/(5 kg)

mrabbit = 31.4 kg

The velocity of the rabbit can be found by dividing its mass by its acceleration due to gravity:

v = m/a

v =[tex]31.4 kg/9.8 m/s^2[/tex]

v = 3.21 m/s

Therefore, the rabbit is running at a speed of 3.21 m/s.

It's important to note that the units of velocity are m/s (meters per second), which is the same as the units used in the question.  

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For a multi-body system, the location of the center of mass (COOM) is not fixed and can change depending on the positions and masses of the individual bodies.

This is because the position of the COM depends on the distribution of mass within the system and the relative positions of the bodies in the system. As the bodies move or interact with each other, the distribution of mass and their relative positions can change, causing the COM to shift accordingly.

The center of mass is the average location of the mass in the system, and it can be calculated using the formula:

COOM = (m1r1 + m2r2 + ... + mn rn) / (m1 + m2 + ... + mn)

where m1, m2, ..., mn are the masses of the bodies and r1, r2, ..., rn are the positions of their centers of mass relative to some reference point. The center of mass is an important concept in physics, as it can help us understand how a system will move and behave under different conditions.

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To answer this question, we need to use the equation for entropy change:

ΔS = Q/T

Where ΔS is the change in entropy, Q is the energy transferred, and T is the temperature.

In this case, we know that ΔS = 7 J/K and Q = 5400 J. We can rearrange the equation to solve for T:

T = Q/ΔS

Plugging in the values, we get:

T = 5400 J / 7 J/K
T ≈ 771.4 K

Therefore, the approximate temperature of the system is 771.4 K.

How warm or chilly a body is has to do with temperature. In other words, when we think of the word temperature, we think of how hot or cold a body is.

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D. T2 = 8T1. Two stars orbit one another, but a planet normally travels farther from the system's center than either of the two stars.

The cube of a planet's semi-major axis is directly proportional to the square of its time period of revolution around the sun, according to Kepler's law of periods.

We can use Kepler's Third Law,

[tex]T^{2} = (4\Pi^{2}/GM) * a^{3}[/tex]

G = gravitational constant

M = mass of the star

a = semi-major axis of the planet's orbit

While a is the distance from the planet to the focus of its elliptical orbit):

[tex]a1 = (2/3) * r1[/tex]

We can find the semi-major axis of planet 2:

[tex]r2 = 4r1\\a2 = (2/3) * r2 = (2/3) * 4r1 = (8/3) * r1\\T1^{2} = (4\Pi^2/GM) * a1^{3}\\T2^{2} = (4\Pi^{2}/GM) * a2^{3}[/tex]

We can solve for GM:

[tex]GM = (4\Pi^{2}/T1^{2}) * a1^{3}\\T2^{2} = (4\Pi^{2}/T1^{2}) * a2^{3} * (a1/a2)^{3}[/tex]

Simplifying the second equation using the known ratios of r2 to r1 and a2 to a1:

[tex]T2^{2} = (4\Pi^{2}/T1^{2}) * (8/3)^{3} * r1^{3} * (3/8)^{3}\\T2^{2} = (4\Pi^{2}/T1^{2}) * (27/64) * r1^{3}[/tex]

Solving for T2 in terms of T1:

[tex]T2 = (3/4) * T1[/tex]

Using the other given ratios, we can also find the remaining periods:

[tex]t2 = (1/2) * T1\\t1 = 2 * t2 = T1\\T2 = 8 * t1[/tex]

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The campers in Dr. Hewitt's story heard the sound of the campers across the lake more clearly at night than during the day due to the phenomenon known as sound refraction.

Refraction is the bending of sound waves as they pass through different mediums, such as air of different densities.

During the day, the air near the surface of the lake is warmer than the air at higher altitudes, creating a boundary between two layers of air with different densities.

This boundary acts as a barrier to sound waves, causing them to bend upward and away from the listener on the opposite shore.

However, at night, the air near the surface of the lake cools more quickly than the air at higher altitudes, creating a more uniform density throughout the air column.

This allows sound waves to travel straight across the lake without bending upward, making them easier to hear on the opposite shore.

Therefore, the campers in Dr. Hewitt's story were able to hear the sound of the campers across the lake more clearly at night due to the absence of the boundary layer that typically refracts sound waves during the day.

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Cockroaches are able to reach such fast speeds of 1.5 m/s because of their adaptation of using jet propulsion. Jet propulsion is when they eject air out of their abdomen to generate a force that propels them forward. This adaptation allows them to move quickly and efficiently without the need for wings or galloping like a horse.

Their body size also plays a role in their ability to move quickly. Cockroaches are small and streamlined, which reduces air resistance and allows them to move more easily. In comparison, larger animals may have a harder time reaching high speeds due to their size and the drag they create.

Cockroaches can run so fast because of their adaptation of using jet propulsion and their small body size, which reduces air resistance and allows for more efficient movement.

Cockroaches are able to reach speeds of 1.5 m/s, which is equivalent to 320 km/h when scaled to body size, due to the following adaptation: they gallop like a horse, alternating front and rear limbs striking the surface.

This galloping method allows them to effectively use their body size and limb coordination to achieve impressive speeds. Unlike the other options mentioned, cockroaches do not rely on wing flapping for lift, nor do they run on their hind limbs only or use jet propulsion by ejecting air out of their abdomen. Instead, their adaptation of galloping enables them to move rapidly and efficiently.

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The type of response seen in the rat is an example of classical conditioning.

Classical conditioning is a method of learning in which a neutral stimulus is repeatedly matched with an unconditioned stimulus, causing the neutral stimulus to elicit a conditioned response. The flash of light in this scenario was initially a neutral stimulus with no intrinsic significance or attachment for the rat.

When the light was consistently paired with the shock, an unconditioned stimulus that elicits an innate response (such as fear or pain), the rat began to associate the light with the shock and eventually came to respond to the light alone with the same fear response, which is the conditioned response.

The rat learnt to anticipate the shock when it saw the light by repeatedly matching it with the shock, and it responded by moving away to escape the shock. This is an example of adaptive behaviour in which the rat avoids potential risk or injury. The classical conditioning learning process is regarded to be a fundamental mechanism underpinning many types of behaviour, including emotional reactions and physiological reflexes.

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If the electron charge distribution or probability density of a particle is spherically symmetric, it means that the likelihood of finding the electron at given distance from the nucleus is same in all directions.

This type of distribution is often observed in atoms with only one electron, such as hydrogen, and is described by the wave function. The wave function represents the probability density of finding an electron in a particular location in space, and a spherically symmetric distribution means that the probability density is the same at all points on a spherical surface around the nucleus.
A situation where the electron charge distribution or probability density is spherically symmetric.
In this context, the electron charge distribution refers to how the negative charge of electrons is spread out in space. Probability density describes the likelihood of finding an electron in a particular region of space. When these two properties are spherically symmetric, it means that they are evenly distributed in all directions around a central point, forming a sphere.

For example, the hydrogen atom's ground state (1s orbital) has a spherically symmetric electron charge distribution and probability density. The electron is equally likely to be found in any direction around the nucleus, and the charge distribution is uniform in all directions. This symmetry results from the wavefunction for the electron in this orbital being dependent only on the distance from the nucleus, not on the angles in spherical coordinates.

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The formula also holds for k+1. By mathematical induction, the formula f(n) = -(n-1) holds for all n ≥ 1.

The proposed definition is a valid recursive definition of a function f from the set of nonnegative integers to the set of integers.

We can use the recursive definition to calculate the values of f(n) for different values of n:

[tex]f(0) = 1 \\f(1) = f(0) - 1 = 1 - 1 = 0 \\f(2) = f(1) - 1 = 0 - 1 = -1 \\f(3) = f(2) - 1 = -1 - 1 = -2 \\f(4) = f(3) - 1 = -2 - 1 = -3[/tex]

and so on.

We can prove this formula by mathematical induction:

Base case: When n = 1, we have f(1) = -(1-1) = 0,

Inductive step: Assume that the formula holds for some arbitrary value k, i.e., f(k) = -(k-1). Then we need to show that the formula also holds for k+1, i.e., f(k+1) = -(k).

Using the recursive definition:

[tex]= f(k+1) = f(k) - 1\\= -(k-1) - 1\\= -k[/tex]

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The power output of the light bulb is option (c)60 W.

Power is the rate at which work is done or energy is transferred. It is usually measured in watts (W) and is calculated by dividing the amount of work or energy by the time taken to do the work or transfer the energy. To calculate the power output of the light bulb, we'll use the following formula:
Power = Emissivity × Area × σ × [tex]Temperature^4[/tex]
Given:
Emissivity (ε) = 0.68
Area (A) = 1.0 cm² = 1.0 x [tex]10^{-4}[/tex] m² (converting to square meters)
σ (Stefan-Boltzmann constant) = 5.67 x [tex]10^{-8}[/tex] W/m²⋅K⁴
Temperature (T) = 2100 K
Power = 0.68 × 1.0 x [tex]10^{-4}[/tex] × 5.67 x [tex]10^{8}[/tex] × [tex](2100)^4[/tex]
Power ≈ 59.97 W
The closest option to this value is:
c. 60 W

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For materials A and B, the total hemispherical emissivity varies with temperature due to their spectral hemispherical emissivities' dependence on wavelength.

As temperature increases, the peak wavelength of emitted radiation shifts to shorter wavelengths, according to Wien's displacement law. Material A and B may have different responses to this shift, leading to variations in their total hemispherical emissivities. In summary, the temperature affects the total hemispherical emissivity of both materials because it influences the distribution of emitted radiation across different wavelengths.

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The ball's velocity just after the force is applied is 19.0 m/s in the positive direction.

We can use the impulse-momentum theorem to solve this problem. The impulse on the ball is given as 45.9 N s, and the mass of the ball is 0.61 kg. The initial velocity of the ball is 27 m/s in the positive direction, and the force is applied in the negative direction of the x-axis.

The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. We can use this theorem to find the final velocity of the ball:

Solving for vf, we get:

Therefore, the ball's velocity just after the force is applied is 19.0 m/s in the positive direction.

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The speed of the electron is approximately -3.67 x 10^6 m/s

The force on a charged particle moving through both electric and magnetic fields is given by the Lorentz force equation: F = q(E + v x B), where F is the force, q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.

In this case, we are told that the electron experiences no net force, which means that the Lorentz force equation simplifies to q(E + v x B) = 0. Since the charge of the electron is negative, we can write this as q(-E - v x B) = 0.

This tells us that the electric field and the magnetic field must be balanced by the velocity of the electron. Specifically, the velocity must be perpendicular to both the electric and magnetic fields, and its magnitude must be such that v x B = -E.

We can solve for the velocity by taking the cross product of both sides of the equation with B: v = -E/B. Plugging in the given values, we get:

v = -(1.41 kV/m)/(0.384 T) = -3.67 x 10^6 m/s

Note that the negative sign indicates that the electron is moving in the opposite direction to the magnetic field. The magnitude of the velocity is about 3.67 million meters per second, which is a significant fraction of the speed of light.

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The maximum static friction force on the sled is 46.15 N for a 53.0 kg sled on snow with us = 0.0888.

The maximum static friction force (f_s) on the sled can be calculated using the following formula:

f_s = μ_s × N

where μ_s is the coefficient of static friction and N is the normal force acting on the sled.

First, we need to calculate the normal force N acting on the sled. The normal force is equal to the weight of the sled, which can be calculated using the formula:

W = m × g

where m is the mass of the sled and g is the acceleration due to gravity, which is approximately 9.81 m/s².

Substituting the values, we get:

W = 53.0 kg × 9.81 m/s² = 519.93 N

Therefore, the normal force acting on the sled is 519.93 N.

Now, we can calculate the maximum static friction force using the formula:

f_s = μ_s × N

Substituting the given coefficient of static friction (μ_s = 0.0888) and the calculated value of N, we get:

f_s = 0.0888 × 519.93 N

Simplifying, we get:

f_s = 46.15 N

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When creating inclined surfaces in an isometric drawing, use isometric lines that follow the established axes to maintain accurate proportions. Non-isometric lines may be used to represent other details or surfaces that don't align with the main axes.

To create inclined surfaces in an isometric pictorial drawing or sketch, you would first establish the isometric axes. Isometric axes are the three main lines that represent the object's width, height, and depth, with each axis at a 120-degree angle from the others. These axes create an isometric grid that helps maintain proper proportions when drawing the object.

An isometric line is a line that follows one of the isometric axes. These lines maintain a consistent 30-degree angle to the horizontal plane in isometric drawings. To create inclined surfaces, you would draw isometric lines along the appropriate axes to represent the edges of the inclined surface. It's essential to maintain the correct angles and proportions to accurately represent the object's dimensions.

A non-isometric line is a line that does not follow the isometric axes and is not parallel to any of the three main axes. These lines may be used to show hidden details, irregular shapes, or curved surfaces in an object. To include a non-isometric line, draw the line at the necessary angle, while still maintaining the object's proportions in relation to the isometric grid.

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To solve this problem, we need to use the concept of rotational motion and the relationship between linear speed and angular speed.

First, we need to find the linear speed of any point on the tire. We know that the angular speed of the tire is 8.0 rad/s, and the radius of the tire is 0.55 m. Therefore, the linear speed of any point on the tire can be found using the formula:
v = r * ω

where v is the linear speed, r is the radius of the tire, and ω is the angular speed.

Plugging in the values, we get:
v = 0.55 m * 8.0 rad/s
v = 4.4 m/s

This means that any point on the tire is moving with a linear speed of 4.4 m/s.

Now, we need to find the speed of the pebble relative to the road when it is at the top of the tire. Since the pebble is stuck in the treads of the tire, it is also moving with the same linear speed as the tire, which is 4.4 m/s.

However, we also need to consider the fact that the pebble is moving in a circular path around the center of the tire, which means that it has a certain velocity vector that is tangent to the circle at any given point. At the top of the tire, the pebble's velocity vector is pointing horizontally, perpendicular to the direction of motion of the tire.

Therefore, the speed of the pebble relative to the road at the top of the tire is simply the horizontal component of its velocity vector, which is equal to the linear speed of the tire, or 4.4 m/s.

In summary, the speed of the pebble relative to the road when it is at the top of the tire is 4.4 m/s.

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The acceleration of the 2.61 kg block sliding down a 15.0 degree incline with a coefficient of kinetic friction of 0.220 is 2.13 m/s², and the force of friction on the block is 21.3 N.

The formula to find the acceleration of the block on the incline is a = gsinθ - ukcosθ, where acceleration due to gravity is g, θ is the angle of the incline, and uk is the coefficient of kinetic friction. Putting all the values we get

a = (9.81 m/s²)*sin(15.0°) - (0.220)*cos(15.0°)

= 2.13 m/s².

The force of friction will be found using the formula f = ukN. N is the normal force of the block. Here, the normal force of the block will be given as N = mgcosθ. Plugging in the given values, we get N = (2.61 kg)(9.81 m/s²)cos(15.0°) = 24.8 N, and

f = (0.220)(24.8 N)

= 21.3 N. Therefore, the force of friction acting on the block is 21.3 N.

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The magnitude of the average braking force is 6.0x10^3 N.

To find the magnitude of the average braking force for an automobile with a linear momentum of 3.0x10^4 kg m/s that is brought to a stop in 5.0 s, you can use the formula:

Force = Change in momentum / Time

Since the final momentum is 0 (as the automobile comes to a stop), the change in momentum is equal to the initial momentum. So,

Force = (3.0x10^4 kg m/s) / 5.0 s

Force = 6.0x10^3 N

The magnitude of the average braking force is 6.0x10^3 N.

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The latent heat of vaporization refers to the energy required to vaporize one kilogram of liquid at its boiling point.

The heat energy transfer required to melt one kilogram of ice when its temperature is already at its melting point is called the latent heat of fusion. This is because during the phase change from solid to liquid, the temperature of the substance does not change, but energy is still required to overcome the intermolecular forces holding the solid together and allow the molecules to move freely in the liquid phase. The amount of energy required to melt one kilogram of ice at its melting point is 334 kJ/kg. Similarly, the latent heat of vaporization refers to the energy required to vaporize one kilogram of liquid at its boiling point.

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The string produces a sound with a frequency of 168.1 Hz in its fundamental mode of vibration.

The frequency of the fundamental mode of vibration of a string is given by:

f = (1/2L) * sqrt(T/μ)

where:

L = length of the string

T = tension in the string

μ = mass per unit length of the string

We can first calculate the mass per unit length of the string using its mass and length:

μ = m/L

where:

m = mass of the string = 8.75 grams = 0.00875 kg

L = length of the string = 75.0 cm = 0.75 m

μ = 0.00875 kg / 0.75 m = 0.0117 kg/m

Substituting the values into the frequency equation, we get:

f = (1/2L) * sqrt(T/μ)

f = (1/2 * 0.75 m) * sqrt(590 N / 0.0117 kg/m)

f = 168.1 Hz

Therefore, the string produces a sound with a frequency of approximately 168.1 Hz in its fundamental mode of vibration.

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You can hear sound through an open window from sources that are not in your line of sight because of a physical phenomenon called diffraction.

Diffraction occurs when sound waves encounter an obstacle, such as a wall or a window, and bend around it, allowing the sound to reach the other side. This means that even if you cannot see the source of the sound, you can still hear it because the sound waves are able to bend and travel around the obstacle.

In the case of an open window, sound waves from outside can easily travel through the window and into the room. These sound waves can then diffract around objects in the room, such as furniture or walls, and reach your ears. This is why even if the source of the sound is not directly in your line of sight, you can still hear it through the open window.

It is important to note that the extent of diffraction depends on factors such as the size of the obstacle and the frequency of the sound waves. Higher frequency sounds have shorter wavelengths and are more easily diffracted, while lower frequency sounds have longer wavelengths and are less likely to diffract.

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The total flux through the cube for each of these fields is 1875, -1875. The correct option is D.

Electric flux is the measure of the electric field lines passing through a surface. It is a scalar quantity and is proportional to the number of electric field lines crossing the surface per unit area. Electric flux plays a crucial role in determining the electric field and charge enclosed by a closed surface.

a) The electric flux through the right face of the cube can be calculated using the formula:

flux = E * A * cos(theta)

where E is the electric field, A is the area of the face, and theta is the angle between the electric field and the normal to the face.

In this case, the electric field is given by E = 3.00 i N/C. The area of the right face is A = (2.50 m)^2 = 6.25 m^2. Since the electric field is parallel to the face, the angle between the electric field and the normal to the face is zero, so cos(theta) = 1.

Now, the electric flux through the right face is:

flux = E * A * cos(theta) = (3.00 N/C) * (6.25 m^2) * (1) = 18.75 Nm^2/C = 1.875 × 10^4 Nm^2/C = 1.875 × 10^7 pNm^2/C

b) The total flux through the cube for each of these fields can be calculated using the formula:

total flux = E * A * 6

where E is the electric field, A is the area of each face, and 6 is the number of faces on the cube.

For part (a), the total flux is:

total flux = E * A * 6 = (3.00 N/C) * (2.50 m)^2 * 6 = 1875 Nm^2/C = 1.875 × 10^6 pNm^2/C

For part (b), the total flux is:

total flux = E * A * 6 = (-2.00 N/C) * (2.50 m)^2 * 6 = -750 Nm^2/C = -7.50 × 10^5 pNm^2/C

Therefore, the answer is (D) 1875, -1875.

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In July, the Earth's rotation axis points in the direction of a star that is 23.5 degrees away from Polaris.

In January, Earth's rotation axis points in the direction of the star Polaris, also known as the North Star. However, in July, the rotation axis of the Earth points in a different direction. This is because the Earth's rotation axis is tilted at an angle of approximately 23.5 degrees relative to its orbital plane. As the Earth orbits around the Sun, this tilt causes the direction of the rotation axis to change relative to the stars in the sky.

In July, the Earth's rotation axis points in the direction of a star that is 23.5 degrees away from Polaris. This star is known as Vega and is located in the constellation Lyra. While Vega is not as famous as Polaris, it is still a bright star that can be easily seen in the summer sky in the northern hemisphere.

It is important to note that the direction of the Earth's rotation axis is constantly changing due to the Earth's orbit around the Sun. This means that over time, the direction of the rotation axis will point towards different stars at different times of the year. The changing direction of the rotation axis is what causes the seasons and is an important factor in the Earth's climate and weather patterns.

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The magnification is approximately 0.1047 or 0.105, given an object 1.60 cm high is held 3.00 cm from a person's cornea, and its reflected image is measured to be 0.167 cm high.

The magnification is the ratio of the height of the image to the height of the object. In this case, the height of the object is 1.60 cm and the height of the image is 0.167 cm. Therefore, the magnification can be calculated as:

Magnification = height of image / height of object
Magnification = 0.167 cm / 1.60 cm

Simplifying this fraction gives a magnification of approximately 0.1047 or 0.105, to three significant figures.

This means that the image appears smaller than the actual object, as the magnification is less than 1. In other words, the image is reduced in size when it is reflected off the cornea.

It is worth noting that this question relates to the field of optics, which involves the study of light and how it interacts with different materials. The measurement of the reflected image in this question is an example of how light can be reflected off surfaces and used to create images, which is a key concept in optics.

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If the plates of a charged parallel plate capacitor are moved further apart, while the charge on the plates is kept constant, the potential energy of each of the charges on the capacitor would increase. This is because potential energy is directly proportional to the distance between the charges and the electric field.

The electric field between the plates of the capacitor is inversely proportional to the distance between them. When the distance between the plates is increased, the electric field decreases.

This reduction in the electric field causes the charges to have more potential energy as they are farther away from each other. The increase in potential energy would also result in a decrease in capacitance, as capacitance is inversely proportional to the distance between the plates.

Therefore, moving the plates further apart while keeping the charge constant would result in an increase in the potential energy of each of the charges on the capacitor.

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