Which part of a wind-powered system ultimately produces the electricity? A. nacelle B. blade C. turbine D. generator

Answer:

the answer is c

Explanation:

i did it on usatestprep

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.

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.

https://brainly.com/question/10759891

#SPJ3

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]

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.

Answer:

the answer is c

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.

To know more about the Work, here

https://brainly.com/question/7250694

#SPJ3

Answer: The correct answer is transmission.

Explanation:

Sound is a form of energy. The energy that survives the transfer is called the sound transmission. If the energy gets absorb then the sound is not heard to the listener. In this case, the absorption will occur.

People are able to hear footsteps because the sound made by a foot hitting the floor travels through the air to reach their ears.  In this case, the sound transmission will occur as the footsteps can be heard. It means that the energy is transferred here.    

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.

https://brainly.com/question/758238

#SPJ6

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 distance between two parallel metal plates to create an electric field equivalent to that inside a 1.5V AA battery can be calculated using the formula E = V/d. Without a specific electric field strength value, the exact distance can't be provided. However, due to the low voltage of the battery, the necessary plate separation would be extremely small.

The question is asking how far apart two parallel metal plates must be to create an electric field equivalent to the electric field within a 1.5V AA battery, given that the electric field is uniform. To find out the distance between the plates, we need to use the formula for electric field strength, E = V/d, where E is the electric field strength, V is the potential difference, and d is the distance between the plates.

Since we know the voltage (V) from the AA battery is 1.5 V and we are looking for d, we can rearrange the formula to d = V/E. However, we don't have the value of the electric field strength E inside the battery. If we had a reference value, for example, electric field magnitude from another setup, we could use it to solve for d. For instance, if we knew that the electric field between two plates with a certain separation was, let's say, 4.50  imes 103 V/m for a 15.0 kV potential difference, we could find their distance and then extrapolate for the 1.5V scenario.

Without specific numbers, we cannot provide the exact distance for the plates. However, it's important to realize that the electric field strength would be very small compared to typical laboratory setups due to the low voltage of the AA battery. Hence, the plates separation would have to be extremely small to match the electric field inside the battery.

Answer:

Mutation is correct

Explanation:

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.

(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]

Answer:

D. Test a cell phone's reception in the exact same location under various atmospheric conditions

Explanation:

Components of an experiment:

Independent variable:

A manipulated variable, in an experiment or study, whose presence or degree incurs a change in the  dependent variable.

Dependent variable:

A variable which is being studied in the experiment and is supposed to change with respect to the independent variable.

Control variable:

Control variable includes the variables which are of no concern in the experiment and must be constant so that they do not interfere with the results of the experiment. Change in control variable might affect the results of an experiment.

In Anna's experiment, the independent variable is the atmospheric conditions. The dependent variable is cell phone reception that will change according to the atmospheric conditions. All other things related to the experiment are control variables like cellphone, location of the cell phone. Both of these must stay constant to collect the best data.

Hence, the best option is D.

Answer:

D. Test a cell phone's reception in the exact same location under various atmospheric conditions

Explanation:

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

Learn more about centripetal force and gravitational force here:

https://brainly.com/question/14021112?referrer=searchResults

https://brainly.com/question/16613634?referrer=searchResults

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

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.

potential is the first one kinetic is the second