What does the law of conservation of energy state about energy in a closed system

In a solar-powered car, solar energy is converted to electrical energy by solar cells, which is then used to power an electric motor and generate mechanical energy for the car to move.

The correct order of energy transformations in a solar-powered car is:

For example, when sunlight hits the solar panel, the photovoltaic cells in the panel absorb the energy and generate an electric current. This electric current is used to power the motor, which converts electrical energy into mechanical energy that powers the car's movement.

To determine the longest wavelength of electromagnetic radiation that can eject an electron from the metal, one can use the equation E = hc / λ, where E equals the work function, h is Planck's constant, c is the speed of light, and λ is the wavelength. Rearranging it as λ = hc / φ and putting the given value of work function and constant values, one can find the required wavelength.

To calculate the longest wavelength of electromagnetic radiation that can eject an electron from the surface of the metal, we need to use the equation which describes the relationship between the energy of a photon (E) and its wavelength (λ). This equation is:

E = hc / λ

Where:

Given the work function φ (7.40×10-19J) and other constant values, we can rearrange this formula to calculate λ:

λ = hc / φ

The result will give you the longest wavelength of electromagnetic radiation that can eject an electron from the metal surface.

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The longest wavelength of electromagnetic radiation that can eject an electron from the surface of a piece of the metal is [tex]2.69 \times 10^{-7}\)[/tex] meters.

We use the photoelectric effect equation:

[tex]\[ E = h \nu \][/tex]

where [tex]\( E \)[/tex] is the energy of the photon, [tex]\( h \)[/tex] is Planck's constant, and [tex]\( \nu \)[/tex] is the frequency of the radiation.

The energy of the photon must be at least equal to the work function [tex](\( \phi \))[/tex] of the metal for the electron to be ejected. Therefore, we have:

[tex]\[ E = \phi \][/tex]

[tex]\[ h \nu = \phi \][/tex]

Since [tex]\( \nu = \frac{c}{\lambda} \)[/tex], where [tex]\( c \)[/tex] is the speed of light and [tex]\( \lambda \)[/tex] is the wavelength of the radiation, we can rewrite the equation as:

[tex]\[ h \frac{c}{\lambda} = \phi \][/tex]

Solving for [tex]\( \lambda \)[/tex], we get:

[tex]\[ \lambda = \frac{h c}{\phi} \][/tex]

Given that [tex]\( h = 6.626 \times 10^{-34}\)[/tex] Js (Planck's constant), [tex]\( c = 3.00 \times 10^8\)[/tex] m/s (speed of light), and [tex]\( \phi = 7.40 \times 10^{-19}\)[/tex] J (work function), we can plug in these values to find [tex]\( \lambda \)[/tex]:

[tex]\[ \lambda = \frac{6.626 \times 10^{-34} \text{ Js} \times 3.00 \times 10^8 \text{ m/s}}{7.40 \times 10^{-19} \text{ J}} \][/tex]

[tex]\[ \lambda = \frac{1.9878 \times 10^{-25} \text{ Jm/s}}{7.40 \times 10^{-19} \text{ J}} \][/tex]

[tex]\[ \lambda = 2.6862 \times 10^{-7} \text{ m} \][/tex]

The correct answer is B) Nitrogen and oxygen

Explanation:

The atmosphere refers to the gaseous layer that covers a body such as a planet. The composition of the atmosphere varies according to the planet or body and therefore the elements found on it are different. In the case of our planet, the atmosphere is mainly composed of nitrogen and oxygen as these are 99% of the atmosphere, while other elements such as carbon dioxide represent the 1% missing. Also, the atmosphere is divided into different layers according to heigh this includes the troposphere, stratosphere, among others. Thus, 99 percent of the Earth's atmosphere is made up only of nitrogen and oxygen.

The correct answer is D. none of the above. In all of the given options, work is done.

The correct answer is D. none of the above. In all of the given options, work is done.

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(a) What is its resistance? Express your answer to two significant figures and include the appropriate units."

(b) How much charge passes through it in 11 min ? Express your answer to two significant figures and include the appropriate units.

Solution

(a) We can calculate the resistance by using Ohm's law:

[tex]\Delta V= I R[/tex]

Since we know that [tex]\Delta V=120 V[/tex] and [tex]I=14.5 A[/tex], we can find R:

[tex]R= \frac{\Delta V}{I} = \frac{120 V}{14.5 A}=8.3 \Omega [/tex]

(b) The current is equal to the quantity of charge by unit of time:

[tex]I= \frac{Q}{t} [/tex]

In our problem, t=11min=660 s, so we can calculate the charge passed through the hair dryer:

[tex]Q=It=14.5 A \cdot 660 s = 9570 C[/tex]

Answer:

Explanation:

According to the conservation of energy, energy can neither be created nor be destroyed but can transform from one form to another.

The form of energy is converted into another form is called the transformation of energy.

Here, the chemical energy of the gasoline is converted into kinetic energy of the car.

The force on an electron placed 10-9 m away from a nucleus with a charge of +40e can be calculated using Coulomb's law, considering the charge of the electron and the proton, the distance between them, and Coulomb's constant.

The question pertains to the force experienced by an electron in the vicinity of an atomic nucleus with a charge of +40e. To calculate this force, we will use Coulomb's law, which states that the electric force (F) between two point charges is directly proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. The formula is given by F = k * |q1 * q2| / r², where k is Coulomb's constant (8.9875 × 10⁹ N⋅m²/C²).

Given that the charge of a proton (and thus the atomic number Z) is +e and the charge of an electron is -e, the force will be attractive, and we can ignore the signs for magnitude calculation. The charge of a proton is e = 1.602 × 10⁻¹⁹ C. For a +40e charge, the total charge is 40 × e. Plug these values, along with the given distance of 10 × 10⁻¹ m into Coulomb's law to compute the force on the electron.

Therefore, the magnitude of the force on the electron by a nucleus with a charge of +40e located 10⁻¹ m away can be calculated using the steps above.

The mean distance between the center of the Jupiter and the center of the Sun is "7.85 x 10¹¹ m"

The force of gravitation between the Sun and Jupiter must be equal to the centripetal force between them, for the equilibrium revolution of Jupiter around the Sun.

[tex]Centripeta\ Force\ on\ Jupiter = Gravitational\ Force\ of Attraction\ \\\\\frac{M_{Jupiter}v^2}{r} = \frac{GM_{Jupiter}M_{Sun}}{r^2}\\\\v^2 = \frac{GM_{Sun}}{r}\ -------- eqn(1)\\\\[/tex]

where,

G = Gravitational  Constant = 6.67 x 10⁻¹¹ N.m²/kg²

[tex]M_{Sun}[/tex] = Mass of Sun = 1.99 x 10³⁰ kg

r = mean distance between the center of the Jupiter and the Sun = ?

v = linear speed of the Jupiter around the Sun = [tex]\frac{Circumference\ of Jupiter's\ Path}{Time\ Period\ of\ Revolution}[/tex]

[tex]v = \frac{2\pi r}{3.79\ x\ 10^8\ s}\\\\v^2 = \frac{4\pi^2 r^2}{14.36\ x\ 10^{16}\ s^2}[/tex]

Using the values in eqn (1), we get:

[tex]\frac{4\pi^2 r^2}{14.36\ x\ 10^{16}\ s^2} = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{r}\\\\r^3 = \frac{(14.36\ x\ 10^{16}\ s^2)(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{4.83\ x\ 10^{35}\ m^3}[/tex]

r = 7.85 x 10¹¹ m

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The attached picture shows the relationship between the centripetal force and the gravitational force acting on a planet (Jupiter) revolving around the sun.

Final answer:

A motorist traveling at 20 m/s and located 60 m from a yellow stoplight needs a steady deceleration of 4 m/s², after a reaction time of 0.50 s, to stop precisely at the light.

Explanation:

A motorist is traveling at 20 m/s and is 60 m from a stoplight when it turns yellow. The motorist's reaction time is 0.50 s before stepping on the brake. We need to calculate the steady acceleration (slowing down) required to stop the car right at the light.

First, calculate the distance covered during the reaction time. Since the car continues at its initial speed during the motorist's reaction time, the distance covered is:

D_{reaction} = v × t = 20 m/s × 0.50 s = 10 m

This means the remaining distance to be covered under deceleration is 60 m - 10 m = 50 m.

Next, we use the kinematic equation v^2 = u^2 + 2as, where

v = final velocity (0 m/s, since the car stops),

u = initial velocity (20 m/s),

a = acceleration,

s = distance covered under deceleration (50 m).

Rearranging the equation for a, we get:

a = (v^2 - u^2) / (2s) = (0^2 - 20^2) / (2 × 50) = -400 / 100 = -4 m/s².

Therefore, a steady deceleration of 4 m/s² would be necessary for the motorist to stop right at the stoplight.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of [tex]v_f = 16.5 m/s[/tex], so [tex]\Delta v=16.5 m/s[/tex], in a time interval of [tex]\Delta t = 181 ms=0.181 s[/tex]. The acceleration of the ball in the horizontal direction (x-axis) is therefore

[tex]a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2 [/tex]

And the distance covered by the ball during this time interval, before it is released, is:

[tex]S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m [/tex]

(b) For this part we need to consider also the weight of the ball, which is [tex]W=mg=2.28 N[/tex]

From this, we find its mass: [tex]m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg [/tex]

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

[tex]F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N[/tex]

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

[tex]F_y=W=mg=2.28 N[/tex] (upward)

So, the magnitude of the force is

[tex]F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N [/tex]

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

[tex]\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11 [/tex]

From which we find [tex]\alpha=6.2^{\circ}[/tex]

In a parallel circuit, the brightness of the remaining bulbs isn't affected when one bulb goes bad or is disconnected. This is because each bulb has its own path to the voltage source, and neither the voltage nor the redistributed current through each bulb is affected by the removal of one bulb. Contrarily, in a series circuit, all bulbs would go dark.

In a parallel circuit with several identical light bulbs of equal resistance, when one bulb goes bad or is disconnected, it doesn't affect the brightness of the other bulbs. This is because in a parallel circuit, each bulb has its own separate path to the voltage source. Therefore, the removal of one path (the bad or disconnected bulb) doesn't affect the potential difference or voltage across the other bulbs.

Let's take the two primary variables determining the brightness of a bulb - the voltage and current. In a parallel circuit, the voltage across each bulb remains the same, regardless of whether a bulb is removed or not. On the other hand, current, though different for each bulb depending on its resistance, isn't affected by the removal of one bulb because the total current just gets redistributed through the remaining paths, maintaining the original brightness of the other bulbs.

In contrast, if this were a series circuit, the scenario would be different. In a series circuit, the removal or failure of one bulb would disrupt the entire circuit (since there's only one path for the current to flow), causing all other bulbs to go dark.

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Final answer:

The minimum constant acceleration required for an aircraft to become airborne after a takeoff run of 277 meters is [tex]2.25 m/s^2[/tex] when converting the liftoff speed from km/h to m/s and applying the kinematic equation.

Explanation:

To determine the minimum constant acceleration required for an aircraft to become airborne after a takeoff run of 277 meters, we can use the kinematic equation:

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

Where:

v is the final velocity (liftoff speed), which needs to be converted from 127 km/h to m/s.

u is the initial velocity, which is 0 m/s since the aircraft starts from rest.

a is the acceleration, which we are trying to find.

s is the displacement, which is the distance of the takeoff run (277 m).

First, we convert the liftoff speed to m/s:

[tex]$$ 127 km/hr \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 35.28 m/s $$[/tex]

Now we can use the kinematic equation to solve for a:

[tex]$$ 35.28^2 = 0^2 + 2a \times 277 $$ $$ a = \frac{35.28^2}{2 \times 277} = \frac{1244.6784}{554} = 2.2465 \frac{m}{s^2} $$[/tex]

The minimum constant acceleration required for the aircraft to be airborne is [tex]2.25 m/s^2[/tex] (rounded to two decimal places).

Sam does 1260 J of work to stop the boat.

The work done by Sam to stop the boat is equal to the change in kinetic energy of the boat.

Work = ΔKE

[tex]Work = KE_f - KE_i\\\\\\Work = 0 - \dfrac{1}{2}mv_i^2\\\\Work = -\dfrac{1}{2}(900 kg)(1.4 m/s)^2\\\\\\Work = -1260 J[/tex]

The negative sign indicates that the work is done by the boat on Sam.

We know that the mass of the boat and its riders is 900 kg. We also know that the initial velocity of the boat is 1.4 m/s. The final velocity of the boat is 0 m/s, since the boat stops. We can use the equation for kinetic energy to calculate the change in kinetic energy of the boat. The change in kinetic energy is equal to the work done by Sam to stop the boat. The work done by Sam is equal to -1260 J.

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Answer:

Decreases

Explanation:

Coulomb found that the electric force between two point charges is inversely proportional to the square of the distance that separates them. Therefore, in this case, when the distance increases the repulsion force decreases and when the distance decreases the repulsion force increases.

Answer:

Mutation is correct

Explanation:

Answer:

A. Common

Explanation:

A personal fall arrest system is the most common type of fall arrest in construction. Personal fall arrest systems are used as protection for OSHA workers who work on construction sites and are exposed to vertical drops of six feet or more. These systems consist of a body harness, anchorage and connector.