Why does the gravitational force between the Earth and moon predominate over electric forces? 1. Because the distance between the Earth and the moon is very large. 2. Because there is no electric charge on the moon. 3. Because both the Earth and the moon are electrically neutral. 4. Because the masses of the Earth and moon are very large.

a) The rate at which the sphere emits thermal radiation is 570 W.

b) The rate at which the sphere absorbs thermal radiation is 1310 W.

c) The sphere's net rate of energy exchange is -738 W.

(a) Rate at which the sphere emits thermal radiation:Stefan's law is given by,

Q = σAεT⁴

Where, σ = 5.67 x 10⁻⁸ W m⁻² K⁻⁴ (Stefan's constant)

A = 4πr² (Surface area of sphere)

r = 0.500 m (Radius of sphere)

ε = 0.921 (Emissivity of sphere)

T = 26.9 ∘ C = 300.9 K (Temperature of sphere)

Substitute all the given values in the above equation, we get

Q = σAεT⁴

Q = 5.67 x 10⁻⁸ x 4π(0.500)² x 0.921 x (300.9)⁴

Q = 5.70 x 10² W

Therefore, the rate at which the sphere emits thermal radiation is 570 W.

(b) Rate at which the sphere absorbs thermal radiation:We know that,Q = σAεT⁴

Where, T is the temperature of the environment, which is 77.0 ∘ C = 350.0 K

Substitute all the given values in the above equation, we get

Q = σAεT⁴

Q = 5.67 x 10⁻⁸ x 4π(0.500)² x 0.921 x (350.0)⁴

Q = 1.31 x 10³ W

Therefore, the rate at which the sphere absorbs thermal radiation is 1310 W.

(c) Sphere's net rate of energy exchange:As we know that,Q = σAε(T₁⁴ - T₂⁴)

Where, T₁ is the temperature of the environment, which is 77.0 ∘ C = 350.0 K, and T₂ is the temperature of the sphere, which is 26.9 ∘ C = 300.9 K.

Substitute all the given values in the above equation, we get

Q = σAε(T₁⁴ - T₂⁴)

Q = 5.67 x 10⁻⁸ x 4π(0.500)² x 0.921 x [(350.0)⁴ - (300.9)⁴]

Q = -7.38 x 10² W

Therefore, the sphere's net rate of energy exchange is -738 W.

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A **comparator** is a device commonly used in making a comparison between two objects.

A comparator is designed to measure and compare the properties or characteristics of two different objects or quantities. It can be a physical device, an instrument, or even a software-based tool. The purpose of a comparator is to determine the similarities or differences between the objects being compared.

Comparators are utilized in various fields and applications. For example, in metrology, comparators are used to measure and compare the dimensions, tolerances, or features of manufactured parts against established standards. In electronics, comparators are used to compare voltages or signals and determine their relationship (e.g., greater than, less than, equal to). In decision-making processes, comparators are employed to assess and evaluate different options or alternatives based on specific criteria.

Overall, a comparator serves as a valuable tool for conducting comparative analysis and aiding in decision-making processes across numerous disciplines.

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The unit of measurement of lightness or darkness of a color is called "value." Value is the degree of lightness or darkness of a color.

The concept of value is essential in art since it can be utilized to produce a strong sense of space. In art, artists employ a range of values to produce the illusion of light and shadow on a surface, resulting in the illusion of a three-dimensional shape.

The value scale is made up of a series of monochromatic grays that range from black to white. In the value scale, each step is an even change in luminosity. Dark colors have a low value, whereas light colors have a high value. In conclusion, value is the unit of measurement of the lightness or darkness of a color.

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When converging at an angle with another aircraft, it is essential to take appropriate actions to ensure safety. When you find yourself converging at an angle with another aircraft, it is crucial to prioritize safety.

The first step is to establish visual contact with the other aircraft, if possible. Then, follow the "see and avoid" principle, maneuvering to the right to avoid a potential collision. Maintain constant vigilance and communicate your intentions through radio transmissions if available.

When you find yourself converging at an angle with another aircraft, it is crucial to prioritize safety by taking immediate and appropriate actions. First, attempt to establish visual contact with the other aircraft. If visual contact is established, adhere to the "see and avoid" principle, which entails taking action to avoid a collision. In this scenario, it is recommended to maneuver to the right, as this is the standard practice. This ensures that both aircraft alter their paths in a predictable and consistent manner. Simultaneously, maintain a vigilant watch for any further changes in the situation and utilize radio communication, if available, to coordinate intentions and ensure mutual awareness. These proactive measures are critical for effective collision avoidance during converging flight paths.

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To accurately determine the tension force and the horizontal and vertical forces, we need more information about the specific scenario or system in question.

Could you please provide additional context or details about the situation? This will allow us to calculate the forces accurately.

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Option 3 is correct. High energy particles from the Sun do not hit the surface of the Earth because of the Earth's magnetic field.

The Earth's magnetic field is generated by the movement of molten iron in the Earth's core. The magnetic field is strongest at the poles and weakest at the equator. When high energy particles from the Sun enter the Earth's atmosphere, they are deflected by the magnetic field. The particles spiral around the Earth and eventually become lost in space.

The Earth's atmosphere also helps to protect us from high energy particles. The atmosphere is made up of a mixture of gases, including nitrogen, oxygen, and argon. These gases absorb high energy particles, preventing them from reaching the Earth's surface.

The Earth's magnetic field and atmosphere are two important factors that protect us from high energy particles from the Sun. These factors help to keep us safe from harmful radiation and allow us to live on the surface of the Earth.

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The rate of heat-loss from the steam pipe is 39.5 MW

To determine the rate of heat loss from the steam pipe, we can use the Stefan-Boltzmann law and the heat transfer equation. Here's how you can calculate it step by step:

Calculate the temperature difference between the surface of the pipe and the surrounding air:

ΔT = T_pipe - T_surrounding = 144°C - 10°C = 134°C

Convert the temperature difference to Kelvin:

ΔT_Kelvin = ΔT + 273.15 = 134°C + 273.15 = 407.15 K

Calculate the outer surface area of the pipe:

A = π * D * L

where D is the outer diameter and L is the length of the pipe.

A = π * 0.058 m * 57 m ≈ 10.395 m²

Calculate the rate of heat loss using the Stefan-Boltzmann law:

Q = ε * σ * A * ΔT^4

where ε is the emissivity of the outer surface, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m²·K^4)), and ΔT is the temperature difference in Kelvin.

Q = 0.7 * 5.67 x 10^-8 W/(m²·K^4) * 10.395 m² * (407.15 K)^4

Now let's calculate the result:

Q = 0.7 * 5.67 x 10^-8 W/(m²·K^4) * 10.395 m² * (407.15 K)^4

Q ≈ 0.7 * 5.67 x 10^-8 * 10.395 * 895008853763.12

Q ≈ 3.95 x 10^7 W

Therefore, the rate of heat loss from the steam pipe is approximately 39.5 MW (megawatts).

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The velocity of the object at t = 7 seconds is -196 units per time (depending on the units of the position function).

To find the velocity of the object at t = 7 seconds, we need to calculate the derivative of the position function with respect to time.

x(t) = 2t³ - 35t² + 10

To find the velocity, we differentiate the position function with respect to time (t):

v(t) = d/dt [x(t)]

Applying the power rule of differentiation, we differentiate each term separately:

v(t) = d/dt [2t³] - d/dt [35t²] + d/dt [10]

Differentiating each term:

v(t) = 6t² - 70t + 0

Simplifying, we have:

v(t) = 6t² - 70t

Now we can substitute t = 7 seconds into the velocity function to find the velocity at that time:

v(7) = 6(7)² - 70(7)

Evaluating the expression:

v(7) = 6(49) - 490

v(7) = 294 - 490

v(7) = -196

Therefore, the velocity of the object at t = 7 seconds is -196 units per time (depending on the units of the original position function).

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The voltage across the 31 μF capacitor in a circuit where a 15-VV battery is connected to three capacitors in series having capacitances of 4.7 μF, 13 μF, and 31 μF can be calculated using the formula;

[tex]$$V_C = \frac{C}{C_1+C_2+C_3}V_T$$[/tex]

where [tex]$C_1$, $C_2$ and $C_3$[/tex] represent the capacitances of the capacitors

[tex]$V_T$[/tex]is the total voltage across the capacitors.

The first step to obtain the answer is to find the total capacitance.$$
[tex]C_{total} = C_1 + C_2 + C_3$$$$[/tex]
[tex]C_{total} = 4.7\mu F + 13\mu F + 31\mu F$$$$[/tex]
[tex]C_{total} = 48.7\mu F$$[/tex]

Next, the total voltage across the capacitors can be found. In this case, the voltage is equal to the battery voltage;

[tex]$$V_T = 15 V[/tex]
[tex]$$[/tex]$$ Substituting these values in the formula above;

[tex]$$V_C = \frac{31 \mu F}{4.7\mu F + 13\mu F + 31\mu F} \times 15V$$$$[/tex]
[tex]V_C = \frac{31 \mu F}{48.7\mu F} \times 15V$$$$[/tex]
[tex]V_C = 9.59V$$[/tex]

The voltage across the [tex]31 μF[/tex] capacitor is 9.59 V.

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The speed of the car is 24.4 m/s. The source frequency, denoted as fS, is the frequency of the sound wave emitted by the car as it moves.

The source frequency can be determined using the equation:

fS = f0(v + vo)/(v - vs) where f0 is the frequency of the sound wave as measured by a stationary observer, v is the speed of the sound wave in the medium, vo is the speed of the observer relative to the medium, and vs is the speed of the source relative to the medium.

Substituting the given values of f0 = 99.0 Hz, f = 58.0 Hz, v = 331 m/s, vo = 0, and solving for vs, we get:

vs = f0(v - vo)/(f0 - f)vs = 99.0 Hz(331 m/s - 0 m/s)/(99.0 Hz - 58.0 Hz)vs = 23.5 m/s.

This gives us the speed of the car relative to the medium.

To find the actual speed of the car, we need to add the speed of sound (331 m/s) to the speed of the car relative to the medium.

Thus, the speed of the car is:vc = vs + vvc = 23.5 m/s + 331 m/svc = 354.5 m/s ≈ 24.4 m/s.

Therefore, the speed of the car is 24.4 m/s.

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The length of the pendulum can be determined by analyzing its angular displacement and the time it takes to reach a certain position. Given an initial angular displacement of 7.1 degrees and a measured angular.

Displacement of 0.889866 degrees after 0.668966 seconds, the length of the pendulum can be calculated using the formula for the period of a simple pendulum.

The period of a simple pendulum is given by the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. In this case, we can determine the period based on the time it takes for the pendulum to move from an initial angular displacement of 7.1 degrees to a measured angular displacement of 0.889866 degrees.

First, we convert the angular displacements to radians by multiplying them by π/180:

Initial angular displacement: θ1 = 7.1 degrees × π/180 = 0.124 radians

Measured angular displacement: θ2 = 0.889866 degrees × π/180 = 0.0155 radians

Next, we calculate the period T using the time and the difference in angular displacements:

T = Δt / (θ2 - θ1)

Given that Δt = 0.668966 seconds, we substitute the values into the formula:

T = 0.668966 s / (0.0155 rad - 0.124 rad)

Simplifying the equation gives us:

T = 0.668966 s / (-0.1085 rad)

T ≈ -6.162 s/rad

Since the period is the time taken for one complete oscillation, we take the absolute value of T:

T ≈ 6.162 s/rad

Finally, we can rearrange the formula for the period of a pendulum to solve for the length L:

L = (T^2 * g) / (4π^2)

Given that g is approximately 9.8 m/s², we substitute the values:

L = (6.162 s/rad)^2 * 9.8 m/s² / (4π^2)

Simplifying the equation gives us:

L ≈ 1.592 m

Therefore, the length of the pendulum is approximately 1.592 meters.

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The four elements of the separation of powers are: Legislative, Executive, Judicial, and the Checks and balances.

The Separation of Powers is a constitutional doctrine that divides power among the three branches of government in order to avoid abuse of authority and protect liberty. These three branches are Legislative, Executive, and Judicial.

The legislative branch is a part of the government that is responsible for creating laws. It consists of two houses: the Senate and the House of Representatives.

The executive branch is responsible for enforcing laws and is headed by the President of the United States. The President is responsible for executing or carrying out the laws passed by Congress.

The judicial branch is responsible for interpreting the laws and making sure they are being applied correctly. It is composed of a system of federal courts and judges. The highest court in the United States is the Supreme Court.

The system of checks and balances is used to ensure that no single branch of government becomes too powerful. Each branch has the power to limit the powers of the other branches to prevent tyranny. For example, the president can veto a bill passed by Congress, but Congress can override the veto with a two-thirds majority vote.

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According to the question, the stream function (ψ) of the given flow field is: ψ = -3x^2y + f(x).

To find the stream function of a flow field, we can use the relationship between the stream function (ψ) and the velocity potential (φ). In two-dimensional flow, these two quantities are related by the following equation:

ψ = -∫(∂φ/∂y) dx + f(x)

Given that the velocity potential (φ) of the flow field is ø = 3xy^2 - xd, we need to find (∂φ/∂y) to calculate the stream function.Taking the partial derivative of φ with respect to y, we get:

(∂φ/∂y) = 6xy

Now, integrating (∂φ/∂y) with respect to x, we have:

-∫(∂φ/∂y) dx = -∫6xy dx = -3x^2y + g(y)

Here, g(y) is the integration constant with respect to x.

Since the integration constant g(y) depends only on y, we can write it as f(x) to match the notation used in the stream function equation. Therefore, the stream function (ψ) of the given flow field is:

ψ = -3x^2y + f(x)

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Given the data, we have the mass of the first car, m1, as 2000 kg, and the mass of the second car, m2, as 3000 kg. The velocities before the collision are u1 = 10.0 m/s for the first car and u2 = 0 m/s for the second car. The distance moved by both cars after the collision is d = 2.00 m.

Using the conservation of momentum principle, we can set up the equation m1u1 + m2u2 = (m1 + m2)v, where v is the common final velocity of both cars after the collision. Substituting the given values, we have 2000 × 10.0 + 3000 × 0 = (2000 + 3000)v, which simplifies to 20000 = 5000v. Solving for v, we find v = 4.0 m/s.

The total distance moved by both cars after the collision is d = 2.00 m. Therefore, the average velocity of both cars after the collision, vavg, is calculated as (final velocity)/2, which in this case is 4.0/2 = 2.0 m/s.

The time taken for both cars to stop, t, can be determined using the equation 2.00 = (final velocity)/2 × t. Solving for t, we find t = 1 s.

The negative acceleration of both cars after the collision, a, is given by (final velocity)/(time taken), which in this case is 4.0/1 = 4.0 m/s².

The normal force, Fn, acting on both cars is given by Fn = (m1 + m2)g, where g = 9.81 m/s² is the acceleration due to gravity. Substituting the given values, we have Fn = (2000 + 3000) × 9.81 = 49050 N.

The force of friction acting on both cars, f, can be calculated as f = μkFn, where μk is the coefficient of kinetic friction. However, since the coefficient of static friction, μs, is not provided, we cannot determine μk. Therefore, the answer cannot be provided with the given information.

In summary, the given data allows us to calculate the final velocity, average velocity, time taken to stop, negative acceleration, and normal force. However, without the coefficient of static friction, we cannot determine the force of friction or provide a complete answer.

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The potential energy is highest on a marble roller coaster at the highest point of the track.

The potential energy of an object is directly related to its height and its position relative to the reference point. In the case of a marble roller coaster, as the marble climbs up the track, it gains potential energy due to its increased height.

At the highest point of the roller coaster track, the marble reaches its maximum elevation, and thus, its potential energy is at its highest point.

As the marble moves downhill from the highest point, its potential energy decreases and is converted into kinetic energy, which is the energy of motion.

At the bottom of the track, where the marble reaches its lowest point, the potential energy is at its minimum because the height is at its lowest and the marble has converted most of its potential energy into kinetic energy.

The potential energy is highest on a marble roller coaster at the highest point of the track. This is where the marble reaches its maximum elevation and has the greatest amount of potential energy due to its height. As the marble moves downhill, its potential energy decreases and is converted into kinetic energy.

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we can use the magnitude and components of the vector B to find its y-component. Let's consider B vector in standard position (starting at the origin). Its coordinates are (6, 4, 0).

The given vectors are:
a = (-3, -6, 2)
b = (6, 4, 0)
c = (-1, 2, -2)
d = (-2, 3, 4)
Here, the magnitude of vector B is 61. So, ||B|| = 61
Therefore, we have:
[tex]||B||² = (6)² + (4)² + (0)²[/tex]
[tex]=> ||B||² = 36 + 16 + 0[/tex]
[tex]=> ||B||² = 52[/tex]
The formula to find the y-component of a vector is given by:
[tex]$y$-component $= ||\vec{v}||\cdot\sin\theta$[/tex]
where,[tex]$||\vec{v}||$[/tex] is the magnitude of vector [tex]$\vec{v}$[/tex] and [tex]$\theta$[/tex] is the angle that vector [tex]$\vec{v}$[/tex] makes with the positive[tex]$x$-axis[/tex].

Here, we can use the following equation to calculate the angle that vector B makes with the positive x-axis:
[tex]$\tan\theta = \frac{y}{x}$[/tex]
[tex]=> $\theta = \tan^{-1}\left(\frac{y}{x}\right)$[/tex]
Thus, the angle made by the vector B with the positive x-axis is:
[tex]$\theta = \tan^{-1}\left(\frac{4}{6}\right)$[/tex]
[tex]$\theta = \tan^{-1}\left(\frac{2}{3}\right)$[/tex]
Hence, the y-component of vector B is given by:
[tex]$y$-component $= ||\vec{B}||\cdot\sin\theta$[/tex]
[tex]$= 61 \cdot \sin(\tan^{-1}(2/3))$[/tex]
[tex]$= 61 \cdot \frac{2}{\sqrt{13^2+2^2}}$[/tex]
[tex]$= 61 \cdot \frac{2}{\sqrt{173}}$[/tex]

Therefore, the +Y component of vector B is $\frac{122}{\sqrt{173}}$, which is approximately equal to 9.265 units (rounded to three decimal places).

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The statement "heat rises" is not accurate in explaining the temperature decrease with altitude.

The main reason why it is colder at higher altitudes is because of the decrease in air pressure with increasing altitude. As air rises in the atmosphere, the pressure decreases, and this decrease in pressure is accompanied by a decrease in temperature. It is known as adiabatic cooling.

When air molecules rise to higher altitudes, they expand due to the lower atmospheric pressure. As the air expands, it does work against the surrounding air molecules, leading to a decrease in its internal energy and, consequently, a drop in temperature. This adiabatic cooling causes the temperature to decrease with increasing altitude.

In summary, the decrease in temperature with higher altitudes is primarily due to adiabatic cooling resulting from the expansion of air as it rises and experiences lower atmospheric pressure.

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The correct statement is: The effect of viscosity in the fluid close to the plate causes the drag.

When a flat plate is pulled through a stationary fluid, it experiences drag. Drag is caused by the effect of viscosity in the fluid close to the plate. Viscosity is a property of fluids that determines their resistance to flow. As the fluid flows over the surface of the plate, the viscous forces between the fluid layers create shear stress, which opposes the motion of the plate.

The fluid in direct contact with the plate moves slowly due to the no-slip condition, where the fluid velocity is zero at the surface. As the fluid moves away from the surface, its velocity increases gradually. This variation in fluid velocity creates a velocity gradient, causing viscous shear stresses that result in drag on the plate.

Therefore, the effect of viscosity in the fluid close to the plate is the main cause of the drag experienced by the flat plate.

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The simulation of using one battery (ε=17 V) and two resistors with the same resistance (R=7Ω) to construct a circuit where the resistors are in series with the battery is as follows:

A circuit can be constructed with a resistor, a battery, and wires connecting them, which will conduct current when the circuit is closed. The current in a circuit is proportional to the voltage across the circuit and inversely proportional to the resistance. Thus, the current can be calculated using Ohm's Law, which states that

I = V/R where I is the current, V is the voltage, and R is the resistance.

In this circuit, the voltage across one of the resistors can be calculated by using the formula

V = IR,

where V is the voltage, I is the current, and R is the resistance. Since the two resistors are in series, the current through both of them is the same, and the voltage across each resistor is proportional to its resistance .According to Ohm's law, the current through the circuit is

I = V/R = 17/14 = 1.214 A

The voltage across one of the resistors is

V = IR = 1.214 x 7 = 8.5 V

The voltage across one of the resistors is 8.5 V when using one battery (ε=17 V) and two resistors with the same resistance (R=7Ω) to construct a circuit where the resistors are in series with the battery.

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We know that the units of acceleration are m/s², and the units of time are seconds (s).

[tex]a(t) = pt² - qt³So, m/s² = p (m/s)² - q (m/s)³, m/s² = m²/s² - m/s³.[/tex]S

ince these two expressions have the same units, we can set them equal to each other:

[tex]m/s² = m²/s² - m/s³⇒ m/s³ = m²/s² - m/s²⇒ m/s³ = (m/s²)(m - 1)⇒ 1/m² = m/s³⇒ m⁵/s⁶ = 1[/tex]

So, p has units of m/s and q has units of m²/s.

Acceleration is the rate of change of velocity with respect to time: a(t) = v'(t)dv/dt = pt² - qt³ Integrating both sides:[tex]∫dv = ∫pt² - qt³ dtv = pt³/3 - qt⁴/4 + C[/tex]Given that the initial velocity is 0, v = pt³/3 - qt⁴/4(c) We can obtain the position as a function of time by integrating the velocity function over time.∫ds = ∫v(t) dt

The initial position is 0, so:[tex]s = ∫v(t) dt = ∫pt³/3 - qt⁴/4 dt= p/12 t⁴ - q/20 t⁵ + C[/tex]We obtain the position of the particle as a function of time by adding a constant of integration C.

The position function is given as [tex]s = p/12 t⁴ - q/20 t⁵.[/tex]

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In order to find the position of the object when it changes its direction, we need to find the point where its velocity is zero.

Velocity is given by the derivative of position with respect to time, that is, v = dy/dt. Thus, we can find the velocity function by taking the derivative of the given position function:[tex]y = 8t² - 6t - 5v = dy/dt = 16t - 6.[/tex]

At the point where the velocity is zero, we have:[tex]16t - 6 = 0t = 0.375[/tex] sSubstituting this value of t into the position function gives us the position vector when the object changes direction:

[tex]y = 8(0.375)² - 6(0.375) - 5 = -5.72ˆj,[/tex] the position vector when the object changes direction is -5.72ˆj.

To find the object's velocity when it returns to its original position at t = 0, we need to substitute t = 0 into the velocity function that we found in part (a):v = 16t - 6v = 16(0) - 6 = -6, the velocity vector when the object returns to its original position at t = 0 is 6.00ˆj (since velocity is a vector, it has a magnitude of 6 m/s and points in the positive y direction).

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If a runner accelerates steadily at [tex]0.04m/s^2[/tex] to a maximum speed of 18km/hr in the last 20m of a half marathon. The velocity before acceleration began was 4.84 m/s.The acceleration will take 4 seconds.

To find the initial velocity of the runner before the acceleration began, we can use the equation for uniformly accelerated motion:

[tex]v^2 = u^2 + 2as[/tex]

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Given:

Final velocity (v) = 18 km/hr = 5 m/s

Acceleration (a) = [tex]0.04 m/s^2[/tex]

Displacement (s) = 20 m

Substituting the given values into the equation, we can solve for the initial velocity (u):

[tex]v^2 = u^2 + 2as[/tex]

[tex](5)^2 = u^2 + 2(0.04)(20)[/tex]

[tex]25 = u^2 + 1.6[/tex]

[tex]u^2 = 25 - 1.6[/tex]

[tex]u^2 = 23.4[/tex]

[tex]u ≈ 4.84 m/s[/tex]

Therefore, the runner's velocity before the acceleration began was approximately 4.84 m/s.

To calculate the time it takes for the acceleration to occur, we can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given:

Final velocity (v) = 18 km/hr = 5 m/s

Initial velocity (u) = 4.84 m/s

Acceleration (a) = [tex]0.04 m/s^2[/tex]

Substituting the given values into the equation, we can solve for the time (t):

v = u + at

5 = 4.84 + 0.04t

0.04t = 5 - 4.84

0.04t = 0.16

t = 0.16 / 0.04

t = 4 seconds

Therefore, the acceleration will take 4 seconds.

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To find the distance you should walk towards or away from an isotropic source for the sound level to change by a specific value, we can use the formula provided:

ΔL = B2 - B1 = 20Log(r1/r2)

Where ΔL represents the change in sound level, B1 and B2 represent the initial and final sound levels respectively, and r1 and r2 represent the initial and final distances from the source.

a) If you are standing at a distance R = 105 m from the isotropic source and want the sound level to increase by 2.0 dB, we can rearrange the formula:

2.0 = 20Log(r1/105)

Dividing both sides by 20 gives:

0.1 = Log(r1/105)

By taking the antilog of both sides, we get:

r1/105 = 10^0.1

r1/105 = 1.2589

Multiplying both sides by 105 gives:

r1 ≈ 132.37 m

Therefore, you should walk approximately 132.37 m towards the source for the sound level to increase by 2.0 dB.

b) If you are standing at a distance R = 105 m from the isotropic source and want the sound level to decrease by 2.0 dB, we can use the same formula:

-2.0 = 20Log(r2/105)

Dividing both sides by 20 gives:

-0.1 = Log(r2/105)

By taking the antilog of both sides, we get:

r2/105 = 10^(-0.1)

r2/105 ≈ 0.7943

Multiplying both sides by 105 gives:

r2 ≈ 83.38 m

Therefore, you should walk approximately 83.38 m away from the source for the sound level to decrease by 2.0 dB.

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The velocity of the rock 4.9 seconds after launch is 15.22 m/s downward. The speed of the CD just before it lands is 6.68 m/s.The problem states that the astronaut on Jupiter drops a CD straight downward from a height of 0.900 m.

To find the velocity of the CD just before it lands, we need to use the equation of motion given byv^2 = u^2 + 2as where, v is the final velocity u is the initial velocity a is the acceleration of the object and s is the displacement of the object.

The acceleration of the object is the acceleration due to gravity, which is 24.8 m/s².

The initial velocity of the object is 0 since it is dropped from rest.

The displacement is the height from which the object is dropped, which is 0.9 m.

Therefore, we havev² = 0 + 2 x 24.8 x 0.9v² = 44.64v = sqrt(44.64)v = 6.68 m/s.

Therefore, the speed of the CD just before it lands is 6.68 m/s.

b) The initial velocity of the rock can be calculated using the formula,v = u + at where, v is the final velocity u is the initial velocity a is the acceleration of the object t is the time taken.

The final velocity is 19 m/s, the acceleration is -9.8 m/s² (since the object is moving upward and the acceleration due to gravity is in the opposite direction), and the time taken is 1.5 seconds.

Therefore,v = u + at19 = u - 9.8 x 1.5u = 19 + 14.7u = 33.7 m/s

(a) At launch, the velocity of the rock is equal to the initial velocity u, which is 33.7 m/s.

(b) To find the velocity of the rock after 4.9 seconds, we can again use the formula,v = u + at where, v is the final velocity u is the initial velocity a is the acceleration of the object t is the time taken.

The initial velocity is 33.7 m/s, the acceleration is -9.8 m/s², and the time taken is 4.9 seconds.

Therefore,v = u + atv = 33.7 - 9.8 x 4.9v = -15.22 m/s (Note that the velocity is negative since the rock is now moving downward).

Therefore, the velocity of the rock 4.9 seconds after launch is 15.22 m/s downward.

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The ball lands approximately 51.96 meters beyond the ground level.

To determine how far beyond the ground level the ball lands, we need to analyze the ball's motion. It is thrown at an angle of 30° with the horizontal from a point 60 meters away from the edge of a building that is 49 meters high above the ground.

First, we can break down the ball's motion into horizontal and vertical components. The horizontal component of the ball's velocity remains constant throughout its trajectory. The vertical component is affected by the acceleration due to gravity.

Using the given information, we can calculate the time it takes for the ball to reach its highest point. At the highest point, the vertical velocity becomes zero. By using the equation for vertical motion, we can determine the time taken.

Next, we can calculate the horizontal displacement of the ball using the horizontal component of the initial velocity and the time of flight. Since the horizontal component remains constant, the horizontal displacement is equal to the product of the horizontal velocity and the time of flight.

Finally, by subtracting the initial horizontal distance of 60 meters from the calculated horizontal displacement, we can determine how far beyond the ground level the ball lands.

It's important to note that this calculation assumes ideal conditions and neglects air resistance. Additionally, more precise calculations would require additional information about the initial velocity or launch angle of the ball.

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Suppose that each component of a certain vector is doubled. In such a scenario, the magnitude and the direction of the vector changes as discussed below.

Part ABy what multiplicative factor does the magnitude of the vector change? Express your answer using one significant figure.SolutionThe magnitude of a vector is given by the formula below:

|A| = sqrt(A_x² + A_y² + A_z²)

where

A_x, A_y, A_z

are the components of the vector.Now suppose each component of the vector is doubled. Therefore the new components of the vector are

2A_x, 2A_y, 2A_z.

Then the new magnitude of the vector is given by:

|A'| = sqrt((2A_x)² + (2A_y)² + (2A_z)²) = 2sqrt(A_x² + A_y² + A_z²)

Therefore the magnitude of the vector is doubled. The multiplicative factor is

2.Part BBy what multiplicative factor does the direction angle of the vector change?

Express your answer using one significant figure.SolutionThe direction of a vector can be obtained from the angle it makes with one of the coordinate axes.

The direction angle of a vector in 2D space is given by:

θ = tan⁻¹(A_y/A_x)In 3D

space, the direction angle can be expressed in terms of θ and ϕ where θ is the angle made with the positive x-axis and ϕ is the angle made with the positive z-axis.

θ = tan⁻¹(A_y/A_x)ϕ = tan⁻¹((A_y² + A_x²)/A_z)

Therefore if each component of the vector is doubled, the direction angles of the vector will remain the same. The multiplicative factor is 1.

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True, wind turbines don't emit air pollution.

Wind turbines generate electricity by harnessing the power of wind, and in the process, they do not emit air pollution. Unlike fossil fuel-based power plants, wind turbines do not burn any fuel, which means they don't release harmful pollutants such as carbon dioxide (CO_2), sulfur dioxide (SO2), nitrogen oxides (NOx), or particulate matter into the atmosphere. The operation of wind turbines produces clean, renewable energy without contributing to air pollution or greenhouse gas emissions.

However, it's important to note that the manufacturing, transportation, installation, and maintenance of wind turbines can have environmental impacts. The production of wind turbine components and the construction of wind farms may involve the use of energy and resources, which can result in some emissions and environmental footprint. Additionally, wind turbines can pose certain challenges related to noise pollution for nearby residents and potential impacts on bird and bat populations. However, when considering overall air pollution, wind turbines themselves do not contribute to it.

In summary, wind turbines do not emit air pollution during their operation, making them a clean and environmentally friendly source of electricity generation.

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The net flux on the surface of the charged body is zero due to the cancellation of positive and negative charges.

The flux on the surface of a charged body is determined by the net electric field passing through it. In this case, the body contains charges of +5nC, -8nC, +12nC, and -2nC. Each charge creates an electric field, and the net electric field at any point is the vector sum of the electric fields due to individual charges.

When calculating the flux, we consider Gauss's law, which states that the total electric flux through a closed surface is proportional to the net charge enclosed by that surface. In this case, since the body is not enclosed within any specific surface, we consider the entire body as the surface.

Given that the charges have different magnitudes and signs, the electric fields they create will have different directions and cancel each other out. The positive charges will create electric fields pointing outward, while the negative charges will create electric fields pointing inward. The magnitudes of these fields will depend on the distances from the charges.

Considering the net effect of all the charges, the positive and negative charges will cancel each other out, resulting in a total flux of zero on the surface of the body.

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The cart is released from rest, its initial velocity is zero in both the x and y directions. The x-component of the acceleration ( [tex]a_x[/tex] ) is approximately 4.24 m/[tex]s^2[/tex] , and the y-component of the acceleration ( [tex]a_y[/tex] ) is 2.45 m/[tex]s^2[/tex] . The cart's position 1.50 seconds after it is released is (9.54 m, 2.94 m).

To find the x- and y-components of the cart's initial velocity, we can use the given information. Since the cart is released from rest, its initial velocity is zero in both the x and y directions.

Therefore, the x-component of the initial velocity ([tex]v_{0x}[/tex]) is 0 m/s, and the y-component of the initial velocity ([tex]v_{0y}[/tex]) is also 0 m/s.

b. The acceleration of the cart points down the ramp, parallel to the ramp. Since the ramp is tilted at an angle of 30 degrees from the horizontal, we can decompose the acceleration into its x- and y-components.

The magnitude of the acceleration (a) is given as 4.90 m/[tex]s^2[/tex] . The x-component of the acceleration ([tex]a_x[/tex]) is given by [tex]a_x[/tex] = a * cos(30°), and the y-component of the acceleration ([tex]a_y[/tex]) is given by [tex]a_y[/tex] = a * sin(30°).

Using these formulas, we can calculate the x- and y-components of the acceleration as follows:

[tex]a_x[/tex] = 4.90 m/[tex]s^2[/tex] * cos(30°) = 4.90 m/[tex]s^2[/tex] * √3/2 ≈ 4.24 m/[tex]s^2[/tex]  (approximately)

[tex]a_y[/tex] = 4.90 m/[tex]s^2[/tex]  * sin(30°) = 4.90 m/[tex]s^2[/tex]  * 1/2 = 2.45 m/[tex]s^2[/tex]

Therefore, the x-component of the acceleration ( [tex]a_x[/tex] ) is approximately 4.24 m/[tex]s^2[/tex] , and the y-component of the acceleration ( [tex]a_y[/tex] ) is 2.45 m/[tex]s^2[/tex] .

c. To find the cart's position and velocity 1.50 seconds after it is released, we can use the kinematic equations.

The x-position of the cart can be calculated using the formula:

x = [tex]x_0[/tex] + [tex]v_{0x}[/tex]* t + (1/2) *  [tex]a_x[/tex]  * [tex]t^2[/tex]

Since the initial position ( [tex]x_0[/tex] ) is given as 0 m and the initial x-velocity ([tex]v_{0x}[/tex]) is 0 m/s, the equation simplifies to:

x = (1/2) *  [tex]a_x[/tex]  * [tex]t^2[/tex]

Plugging in the values:

x = (1/2) * 4.24 m/[tex]s^2[/tex]  * [tex](1.50 s)^2[/tex] = 9.54 m

The y-position of the cart can be calculated using the formula:

y =[tex]y_0[/tex] + [tex]v_{0y}[/tex] * t + (1/2) * [tex]a_y[/tex] * t^2

Since the initial position ([tex]y_0[/tex]) is given as 0.500 m and the initial y-velocity ([tex]v_{0y}[/tex]) is 0 m/s, the equation simplifies to:

y = [tex]y_0[/tex] + (1/2) *  [tex]a_y[/tex]  * [tex]t^2[/tex]

Plugging in the values:

y = 0.500 m + (1/2) * 2.45 m/[tex]s^2[/tex]  * [tex](1.50 s)^2[/tex] = 2.94 m

Therefore, the cart's position 1.50 seconds after it is released is (9.54 m, 2.94 m).

Since the initial velocity in both the x and y directions is 0 m/s, the velocity of the cart after 1.50 seconds is the same as its acceleration. So the velocity vector is (4.24 m/s, 2.45 m/s).

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