The power generated by the engine was just 2.984 KW. How long would this engine have to run to produce 3.60 × 104 J of work?

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

\(k = \frac{1}{2} m {v}^{2} \\ \)

m is the mass

v is the velocity

From the question we have

\(k = \frac{1}{2} \times 74 \times {11}^{2} \\ = 37 \times 121\)

We have the final answer as

Hope this helps you

Answer:130

Explanation:just answered it

The orbital speed of the proton is 4.7 x 10⁶ m/s.

The orbital speed of the proton is calculated by applying the following equations as shown below.

Centripetal force of the proton = magnetic force of the proton

Fc = qVB

mv²/r = qvB

mv² = qvBr

v² = qvBr/m

v = qBr/m

where;

The orbital speed of the proton is calculated as follows;

v = (1.6 x 10⁻¹⁹ x 0.35 x 0.14) / (1.67 x 10⁻²⁷)

v = 4.7 x 10⁶ m/s

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Answer:

milligram 1000g

kilogram 1/1000 g

electrons negative particle outside nucleus

speed km/h

potential energy at rest

mixture sawdust + salt + iron filings

Explanation:

:D

1. 1000g -kilogram

2.1/1000 g - milligram

3. negative particle outside nucleus - electrons

4. km/h - speed

5.energy at rest - potential
6.  centimeter - cm

7. Sawdust + salt + iron filings - mixture

The energy by  virtue of its position is called the potential energy.

The km/h is the unit of speed.

Milligram is written as 1/1000 g.

1000 g is equal to 1 kilogram.

Electron are the negatively charged subatomic particles rotating outside the nucleus.

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The speed of the car traveling 58 meters in 20 seconds is 2.9 m/s.

Speed is the rate of chnage of distance. The S.I unit of speed is meter per seconds (m/s).

To calculate the speed of the car, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

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Answer:

Explanation:

according to the graph at B the potential energy of the particle is 2J

therefore we can use the kinetic energy equation to calculate the particle's velocity or speed.

\(E_{k} =1/2mv^{2}\)

2J= 1/2*1/2kg*v^2

8=v^2

v= 2√2 ms-1

Answer: Temperature

Explanation:

The average kinetic energy of the particles in a gas is proportional to the temperature of the gas. Because the mass of these particles is constant, the particles must move faster as the gas becomes warmer.

"The kinetic energy of an object is the energy that it possesses due to its motion."

"The product of the half of the mass of each gas molecule and the square of RMS speed."

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Answer:

using the formula -- V=u+at

u=12m/s, v=100m/s, t=30s, a=?

100=12+a(30)

100-12=a(30)

88=a(30)

88/30=a(30)/30

2.93=a

Answer:

10000

Explanation:

10 to the 5th power

He goes 400 m straight then goes 500m to the left side then goes backwards 400m so walking north is cancelled out he covers only 500 m to the west.

If u mean how much distance he travelled from starting point then this is the answer

Answer: He goes 400 m straight then goes 500m to the left side then goes backwards 400m so walking north is cancelled out he covers only 500 m to the west.

If you mean how much distance he travelled from starting point then this is the answer

Answer:

i need help with the same question

Explanation:

The internal energy released during the breakup is 5.26MeV.

The atomic number of uranium is 92. This means that the atomic structure has 92 protons and 92 electrons. The core of U-238 has 146 neutrons, but the number of neutrons varies between 141 and 146. Uranium is radioactive, so it constantly emits particles and transforms into other elements. A radioactive silvery metal.

Uranium is a very important element as it provides the nuclear fuel used to generate electricity in nuclear power plants. It is also the main material from which other synthetic transuranium elements are made. Neither you nor your customers are at risk of radiation emitted from the uranium in the Vaseline bottle whether it is held in hand or standing on a shelf or table. Accidentally ingesting or inhaling shavings does not put you or your customers at risk of uranium.

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Answer:

Therefore, the GPE of the fish from the given height is 26.19 J.

Explanation:

The gravitational potential energy (GPE) of an object at a height h is given by the formula:

GPE = mgh

where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

In this case, the mass of the fish is 0.3 kg, the height from which it was dropped is 9 meters, and the acceleration due to gravity on Earth is approximately 9.81 m/s². Thus, the GPE of the fish can be calculated as:

GPE = (0.3 kg) × (9.81 m/s²) × (9 m) = 26.19 J

Therefore, the GPE of the fish from the given height is 26.19 J.

I like puffins

Answer:

True

Explanation:

Energy decreases as it moves up trophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level.

(a) To find the horizontal distance d that the ball travels before landing, we can use the equation d = (Vx)t, where Vx is the horizontal velocity of the ball and t is the time it takes for the ball to fall from the ramp to the floor.

First, we need to find the horizontal velocity of the ball when it leaves the ramp. Since the ball is rolling without slipping, we can use the equation V = ωR, where V is the linear velocity, ω is the angular velocity, and R is the radius of the ball. The radius of the ball is 0.20 m / 2 = 0.10 m.

Next, we need to find the angular velocity of the ball. We can use the equation ω = √(2gh/R), where g is the acceleration due to gravity (9.8 m/s^2), h is the vertical height that the ball drops (0.49 m), and R is the radius of the ball (0.10 m). Plugging in the values gives us:

ω = √(2(9.8)(0.49)/0.10) = 31.3 rad/s

Now we can find the horizontal velocity of the ball:

V = ωR = (31.3)(0.10) = 3.13 m/s

Finally, we can find the time it takes for the ball to fall from the ramp to the floor using the equation h = 0.5gt^2, where h is the vertical height (1.31 m) and g is the acceleration due to gravity (9.8 m/s^2). Rearranging the equation and plugging in the values gives us:

t = √(2h/g) = √(2(1.31)/9.8) = 0.52 s

Now we can find the horizontal distance d that the ball travels before landing:

d = (Vx)t = (3.13)(0.52) = 1.63 m

Answer: The ball travels 1.63 m before landing.

(b) To find the number of revolutions that the ball makes during its fall, we can use the equation θ = ωt, where θ is the angular displacement, ω is the angular velocity, and t is the time. The angular displacement is equal to the number of revolutions times 2π, so we can rearrange the equation to solve for the number of revolutions:

θ / 2π = ωt / 2π = (31.3)(0.52) / 2π = 2.59 rev

Answer: The ball makes 2.59 revolutions during its fall.

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The angular momentum of the particle in the x-component, y-component, and z-component are 3.12, 2.84, and 1.19 kg.m/s² respectively.

Angular momentum can be defined as the rotational form of linear momentum. Angular momentum is a physical conserved quantity.

The angular momentum in the terms of linear momentum can be represented as:

\(\displaystyle \vec L= \vec r\times \vec p = m\vec r \times \vec v\)

Given, the mass of the particle, m = 70 Kg

The displacement of the vector of the particle, \(\vec r = (5.0, -5.5)\)

The velocity of the vector of the particle, \(\vec u = (3.2, 0, -8.1)\)

The angular momentum of the particle, L can be calculated as:

\(\displaystyle \vec L= m\vec r \times \vec u\)

\(\displaystyle \vec L= m(5.0,-5.5, 0)\times (3.1,0,-8.1)\)

\(\displaystyle \vec L= 0.070 \times (44.55, 40.5, 17.05) .Kg.m/s^2\)

\(\displaystyle \vec L= (3.12, 2.84, 1.19) \; kg.m/s^2\)

Therefore, the x-component of angular momentum = 3.12 kg.m/s²

The y-component of angular momentum = 2.84 kg.m/s²

The z-component of angular momentum = 1.19 kg.m/s²

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Answer:

answer here

Explanation:

the unit of work is fundamental unit because it doesn't depend on other units.

__________________

Thx

Answer:

The SI unit of work is joule (J)

Explanation:

Joule is a derived unit. ∴ unit of work is a derived unit

Answer:

hellooo

Explanation:

i think ummm

. . .

The amount of work done by the gas is proportional to the pressure and the change in volume, as well as the efficiency of the process. If the pressure and volume are known, the work done by the gas can be calculated by multiplying these values by the efficiency of the process.

The amount of work done by a gas when it expands is proportional to the change in volume, pressure, and temperature. According to the first law of thermodynamics, the energy of a closed system is conserved, so the work done by the expanding gas is equal to the energy transferred from the gas to the environment in the form of work. Therefore, the work done by the gas is equal to the change in energy of the system. Assume that the process is 10% efficient. Then, only 10% of the energy available to the system is converted into work. This means that the remaining 90% of the energy is lost to the environment in the form of heat. As a result, the amount of work done by the gas expanding into the atmosphere is given by the formula

W = E x η, where W is the work done by the gas, E is the energy available to the system, and η is the efficiency of the process. The energy available to the system is determined by the difference between the internal energy of the gas before and after the expansion. The internal energy of a gas is determined by its temperature, pressure, and volume.

Assuming that the temperature and pressure are constant, the change in internal energy is proportional to the change in volume. Therefore, the energy available to the system is equal to the product of the pressure and the change in volume: E = P x ΔV, where P is the pressure of the gas and ΔV is the change in volume during the expansion. Substituting this equation into the formula for work, we get W = P x ΔV x η.

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Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}\)            (you has an mistake in the formula)

         \(\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}\)

         \(\frac{1}{C_{eq1}}\) = 0.1   10⁶

         \(C_{eq1}\) = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          \(\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}\)

          \(\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6\)

          \(\frac{1}{C_{eq2} }\) = 0.2 10⁶

          \(C_{eq2}\) = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = \(\frac{Q^2}{2 C_3}\)

          U₃ =\(\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}\)

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = \(\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}\)

          U₄ = 0.5 J

The weight of the object at distance 3R from the Earth is 10 N if the weight at center of the Earth is 90N.

According to Newton's law of gravitation,

Weight W = m × g; where m is the mass of the object and

g is the acceleration due to gravity.

Also,

\(g = \dfrac{GM}{R^{2}}\)

where G=gravitational constant, M=mass of earth, R=radius of earth.

Hence,

\(W = m \times \dfrac{GM}{R^{2}}\)

Weight is inversely proportional to the square of radius R.

Weight of objects at the surface of the earth is 90 N.

Let W be the weight at a distance of 3 R from the center.

\(\dfrac{W}{90} = \dfrac{R^{2}}{(3R)^{2}}\\\dfrac{W}{90} = \dfrac{1}{9}\\W = 10\)

Weight at a distance 3 R is 10 N.

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Answer:

B

Explanation:

because when the air rises the density increases

We want to predict what will happen to the density of the air on the surface of the Earth when it warms up.

We will see that the correct option is D: "It decreases, then increases"

We know that the temperature of a given object (in this case a mass of air) is related to the kinetic energy of the particles that conform it.

As the temperature increases, the kinetic energy also increases, thus, the amount of motion of each particle increases, thus, the volume of the object increases.

Now remember that:

density = Mass/Volume.

So if the volume of something increases, we will see that the density decreases (as the volume is in the denominator).

Then if the temperature of the air increases, we will see that the density of the air in the surface decreases.

But it does not end there, as only the air near the surface suffers this change of density, we will have a denser mass of air (colder air) above it. And because it is denser (has more mass in the same volume) we can say that it is heavier.

Then eventually the hot air will rise, and the cold air will fall down, thus the density of the air in the surface increases again, as the colder and denser air comes near the surface.

This is one way of how wind currents are born.

Concluding we can see that the correct option is D: "It decreases, then increases"

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Answer:

Δλ = 3*10⁻³ m.

Explanation:

       \(v = \lambda* f (1)\)

       where v is the speed, λ is the wavelength and f is the frequency.

       \(\lambda = \frac{v}{f} (2)\)

        \(\lambda_{1} =\frac{v}{f_{1}} = \frac{343m/s}{440(1/s)} = 0.779 m (3)\)

        \(\lambda_{2} =\frac{v}{f_{2}} = \frac{343m/s}{442(1/s)} = 0.776 m (4)\)

       \(\Delta \lambda = \lambda_{1} - \lambda_{2} = 0.779 m - 0.776m = 3e-3 m (5)\)

       ⇒ Δλ = 3*10⁻³ m.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

d) electrons removed.

Explanation:

Two insulators are electrically neutral, that is, the number of electrons and protons are equal. When one insulator is rubbed against another, electrons are removed from one of the insulators due their high mobility. The insulator which lost the electrons becomes positively charged while the one that gained the electron becomes negatively charged.

Therefore, when rubbing two insulators together, the one that gets a positive charge has had electrons removed.

The process of grouping master cartons into one physical unit for materials handling or trnasport is referred to as an unitization.

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The absolute pressure at an ocean depth of 1.0 x 10^3 m is 1.002 x 10^8 Pa.

Hydrostatic pressure is the pressure that a fluid exerts on a surface due to the weight of the fluid above it. It is the result of the force of gravity acting on a column of fluid, and it is directly proportional to the height of the column of fluid and the density of the fluid.

The absolute pressure at an ocean depth of 1.0 x 10^3 m can be calculated using the hydrostatic pressure equation:

P = ρgh + Po

where:

P is the absolute pressure at the given depth

ρ is the density of the water

g is the acceleration due to gravity (assumed to be 9.81 m/s²)

h is the depth of the ocean

Po is the atmospheric pressure at the surface (assumed to be 1.01 x 10^5 Pa)

Substituting the given values, we get:

P = (1.025 x 10^3 kg/m³) x (9.81 m/s²) x (1.0 x 10^3 m) + 1.01 x 10^5 Pa

P = 1.025 x 9.81 x 10^6 Pa + 1.01 x 10^5 Pa

P = 1.002 x 10^8 Pa.

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The average frequency will be 465.5 Hz.

The frequency of light can be described as the number of waves passing through a point in a medium (or vacuum) per second.

Frequency generated by generator 1 = f1 = 460 Hz

Frequency produced by generator 2 = f2 = 471 Hz

the average frequency will be (f1 + f2) / 2

f1 + f2 = 460 + 471

f1 + f2 = 931 Hz

Average frequency = A = 931/2

A = 465.5 Hz

The average frequency we hear will be 465.5 Hz.

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The energy transferred in Joules by the radio when it has been switched on for 2 minutes would be 6000 Joules.

Power is defined as the rate of energy transfer or the rate at which work is done, and is given by the equation:

Power = Energy transferred / Time

Rearranging the equation to solve for energy transferred, we get:

Energy transferred = Power x Time

We are given:

Power = 50 W

Time = 2 minutes = 120 seconds

Therefore, the energy transferred by the radio when it has been switched on for 2 minutes is:

Energy transferred = Power x Time = 50 W x 120 s = 6000 J

In other words, the energy transferred by the radio is 6000 Joules.

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