What happens to the internal energy of a system when mechanical work is done on it? What happens to its temperature?

Answers

Answer 1

When the system performs mechanical work, internal energy decreases.

Explanation:

Likewise, U (internal energy) = Q (heat) + W (work completed). If the system performs work, then work is negative and internal energy declines.

If done quickly, the internal energy will drop, which will lead to a drop in temperature. Any internal energy loss, if gradual, is offset by heat entering the system to maintain the same temperature.

Answer 2

When mechanical work is done on a system, the internal energy of the system increases. The temperature of the system does not necessarily change.

What is temperature?
Temperature
is a physical property of matter that quantitatively expresses the common notions of hot and cold. It is measured by a thermometer and usually expressed in units of degrees Celsius (°C) or Fahrenheit (°F). Temperature is an important parameter in many natural processes and is an essential component of weather forecasting. Temperature is related to the amount of energy in a system; the higher the temperature, the greater the average energy of the system's constituent particles. The average kinetic energy of the particles in a system is proportional to the absolute temperature. Temperature also has an effect on the physical properties of matter, such as its phase and viscosity. Heat transfer from one system to another is a major component of thermodynamics, and is dependent on the temperature difference between the two systems.

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Related Questions

PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.



The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.


Which number label represents the cell membrane?


1

2

4

6

(this is middle school science)

Answers

Answer:

1. cell membrane

2. golgi body

3. mitochondrion

4. cytoplasm

5. nucleolus

6. nucleus

Explanation:

The correct answer to this question is Option A; 6.

Why?

In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.

~Thank you~

Fill in the question

Answers

4) 55m
5) 30 seconds
6) 1.83m/s

. [30%] We first showed that The electric field for a point charge radiating in 3-dimensions has a distance dependence of 1/r 2 (see Equation 1). In Problem 1 you showed that the electric field for a point charge radiating in 2-dimensions has a distance dependence of 1/r . Consider again the 2-dimensional case described in Problem 1. What distance dependence do you expect for the electric potential

Answers

Answer:

Answer is explained in the explanation section below.

Explanation:

Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.

Solution:

The relation between electric field and the electric potential is:

E = [tex]\frac{dV}{dr}[/tex]

So, making dV the subject, we have:

dV = E x dr

Integrating the above equation, we get.

V = [tex]\int\limits^_ {} \,[/tex]E x dr      Equation 1

Now, in 2-D

E is inversely proportional to the radius r.

E ∝ 1/r

So, we can write: replacing E ∝ 1/r in the equation 1

V ∝  [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr

Which implies that,

V ∝  log (r)

Hence, distance dependence expected for the electric potential =  ln (r)

Which example describes a nonrenewable resource?

Answers

Answer: Examples of nonrenewable resources include crude oil, natural gas, coal, and uranium. These are all resources that are processed into products that can be used commercially. For example, the fossil fuel industry extracts crude oil from the ground and converts it to gasoline.

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.

Answers

Answer:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Explanation:

Given

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]

Point: [tex](f(t0), g(t0), h(t0))[/tex]

[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information

Required

Determine the parametric equations

[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]

Differentiate with respect to t

[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]

Let t = 1 (i.e [tex]t0 = 1[/tex])

[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + k[/tex]

To solve for x, y and z, we make use of:

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]

This implies that:

[tex]r'(1)t = xi + yj + zk[/tex]

So, we have:

[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]

[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]

By comparison:

[tex]xi = it[/tex]

Divide by i

[tex]x = t[/tex]

[tex]yj = \frac{1}{3}jt[/tex]

Divide by j

[tex]y = \frac{1}{3}t[/tex]

[tex]zk = kt[/tex]

Divide by k

[tex]z = t[/tex]

Hence, the parametric equations are:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

what is the mathematical formula associated with newton's 2nd law of motion?​

Answers

Answer:

F= m x a

Explanation:

Force (f) = mass (m) x acceleration (a)

The voltage v(t) = 141.4 cos (ωt) is applied to a load consisting of a 10Ω resistor in parallel with an inductive reactance XL=ωL = 3.77Ω. Calculate the instantaneous power absorbed by the resistor and by the inductor. Also calculate the real and reactive power absorbed by the load, and the power factor. Draw all the voltage, current and power waveforms, also the draw the circuit and phasor diagrams.

Answers

Answer:

A) P(t) = 2651.25 [ 1 - cos2wt ] W

B)  Real power = 999.79 watts

    Reactive power = 2652.86 VA

c) power factor = 0.3526

Explanation:

Given data:

V(t) = 141.4 cos (ωt)

R(t) = 10 Ω

Inductive reactance XL = ωL = 3.77 Ω

Ir(t) = V(t) / R(t) = 14.14

A) Calculate the instantaneous power absorbed by the resistor and by inductor

By resistor :

Pr(t) = V(t) * Ir(t) = 141.4 * 14.14 [tex]cos^{2} wt[/tex] = 1999.396 [tex]cos^{2} wt[/tex]

      hence Pr  = 999.698 (cos2ωt + 1) w

By Inductor :

Pl(t) = V(t) I'L(t) = 141.4 cosωt * 37.5 cos(ωt - 90)  

                        =  5302.5 [tex]sin^2 wt[/tex]

Hence Pl(t) = 5302.5 [tex]sin^2 wt[/tex]   w  =  2651.25 [ 1 - cos2wt ] W

B) calculate the real and reactive power

First we have to determine the power factor

Given that : V(t) = 141.4 cosωt  v ,   Ir(t) = 14.14 cosωt A

IL(t) = 37.5 cos (ωt - 90° )

The phasor representation of the above is :

V = [tex]\frac{141.4}{\sqrt{2} } <0^{0} v[/tex] = 141.4 ∠0° ,  Ir = 10 ∠ 0° , IL = 37. 5 ∠ -90°

Total load current = Ir + IL = 28.35 ∠ -69.35°

power factor = cos -69.35° = 0.3526

Next we will determine the Real and reactive power using the relation below

S = VI = 100 ∠ 0°  * 28.35 ∠ -69.35°

         = 2835 ∠ 69.35°

S = P + jQ = 999.79 + 2652.85 j

Real power  = 999.79 watts

Reactive power = 2652.85 VA

Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift depending on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].

One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.

a. Stretch
b. Vibrate
c. Contract
d. Expand

Answers

Answer:

The correct answer is option D.

Explanation:

It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.

I hope this helps.

A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then it is cooled isochorically until the pressure is 20 kPa at point 3. The gas is now compressed isothermally until its volume is back to 5 L (point 4). Finally, the gas is heated isochorically to return to point 1.
a. Draw the four processes and label the points in the pV plane.
b. Calculate the work done going from 1 to 2.
c. Calculate the pressure and temperature at point 2.
d. Calculate the temperature at point 3.
e. Calculate the temperature and pressure and point 4.
f. Calculate the work done going from from 3 to 4.
g. Calculate the heat flow into the gas going from 3 to 4. g

Answers

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its surface. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these

Answers

Answer:

a) E = 0

b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

From which we have;

[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]

E = 0/A = 0

E = 0

b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]

[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]

By Gauss theorem, we have;

[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]

Therefore, we get;

[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]

The electrical field outside the spherical shell

[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]

[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]

Therefore, we have;

[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on a tall ring that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration and the normal force on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride

Answers

Answer:

Answer is explained in the explanation section below.

Explanation:

In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.

a) You pass through the highest point:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.

And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.

b) You pass through the lowest point of the ride:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.

And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.

ALOT OF POINTS PLZ HURRYQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQWhat does Newton's third law say about why momentum is conserved in collisions?
A: Equal Forces act in equal times, so the change in momentum for both objects must be equal.
B: Unequal forces act for unequal times, so the change in momentum for both objects must be unequal.
C: Equal forces act for unequal times, so the change in momentum for both objects must be equal.
D: Unequal forces act for equal times, so the change in momentum for both objects must be equal.

Answers

Answer:

A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.

(Hope this helps! Btw, I am the first to answer.)

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 60-m-diameter blades at that location. Take the air density to be 1.25 kg/m3. Cengel, Yunus; Cengel, Yunus. Thermodynamics: An Engineering Approach (p. 98). McGraw-Hill Higher Education. Kindle Edition.

Answers

Answer:

1767Kw

Explanation:

Velocity of wind = 10 m/s

diameter of the blades= 60m

ρ= air density = 1.25 kg/m3

Acceleration due to gravity= 9.81 m/s^2

Mechanical energy of the wind can be calculated using the expression below

Energy= (e*m)

= ρ V A e............eqn(1)

Where A= area

ρ= air density

e= wind energy per unit mass of air

e= (v^2)/2..........eqn(2)

If we substitute the values into eqn (2) we have

e= [(10)^2]/2

=50J/Kg

But Area=A= (πd^2)/4

Area= ( π× 60^2)/4

Area=2827.8m^2

If we input substitute the values into eqn (1) we have

Energy= 1.25 ×10 × 50×2827.8

=1767145.7W

We can convert to kilo watt

=1767145.7W/ 1000

= 1767Kw

Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw

At what tempreture will the of and oC
be the same​

Answers

Answer:

-40 degrees

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

Explanation:

Hope it is helpful.....

Answer:

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1 m3 of N2 gas at 500 kPa and 120oC while the other side contains 1 m3 of He gas at 500 kPa and 40oC. Assume the piston is made of 8 kg of copper initially at the average temperature of the two gases on both sides. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move

Answers

Answer:

The answer is "[tex]\bold{83.8^{\circ} \ C}[/tex]".

Explanation:

Formula for calculating the mass in He:

[tex]\to m = \frac{PV}{RT}\\[/tex]

        [tex]= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg[/tex]

Formula for calculating the mass in [tex]N_2[/tex]:

[tex]\to m = \frac{PV}{RT}\\[/tex]

        [tex]= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg[/tex]

by using the temperature balancing the equation:

[tex]T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}[/tex]

    [tex]= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C[/tex]


what is a vector quantity?

Answers

Answer:

A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude

Vector Quantity

A physical Quantity, which has magnitude, direction and units But must follow the traingle law of vector addition

Example:- Force, velocity acceleration displacement etc.

If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest to the Earth. While moving towards the Earth (along the path from D to A) there is a component of force in the __________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow down, speed up). While moving away from the Earth (along the path from A to D) there is a component of force in the _________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow

Answers

Answer:fastest,same,slow down,opposite,slow

Explanation:

A satellite move fastest when its closest to the Earth. The other correct options are same direction, speed up, opposite direction and slow.

Velocity of a satellite around the planet.

If a satellite is orbiting the Earth in elliptical motion, then it will move fastest when its closest to the Earth (based on Kepler's, law).

While moving towards the Earth (along the path from D to A) there is a component of force in the  same direction as the motion; this causes the satellite to speed up.

While moving away from the Earth (along the path from A to D) there is a component of force in the opposite direction as the motion; this causes the satellite to slow.

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In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position

Answers

Answer:

Explanation:

The relation between electric field and potential difference is as follows

E = - dV / dr

That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .

Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means

dV / dr = constant

E = constant

Electric field is constant .

So the option which is correct is

" points east and does not vary with position " .

A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.

Answers

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

[tex]V_{A}[/tex] = 1 m³

[tex]T_{A}[/tex] = 10°C = 283 K

[tex]P_{A}[/tex] = 350 kPa

[tex]m_{B}[/tex] = 3 kg

[tex]T_{B}[/tex] = 35°C = 308 K

[tex]P_{B}[/tex] = 150 kPa

Now, lets apply the ideal gas equation;

[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]

[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150

[tex]V_{B}[/tex] = 265.188 / 150  

[tex]V_{B}[/tex] = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, [tex]m_{A}[/tex] =  [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221

[tex]m_{A}[/tex]  = 4.309 kg

Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg

Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex]  = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

[tex]P_{f}[/tex] =  [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]

given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K

we substitute

[tex]P_{f}[/tex] =  ( 7.309 × 0.287 × 293)  / 2.77

[tex]P_{f}[/tex] =  614.6211119 / 2.77

[tex]P_{f}[/tex] =  221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

please help me I'm begging you Define and give examples of elements and compounds the structure of atoms (electrons, neutrons, and protons)

Answers

An element is a substance whose atoms all have the same number of protons: another way of saying this is that all of a particular element's atoms have the same atomic number. ... When atoms of the same element have different numbers of neutrons, they are called isotopes. Examples of Compounds:
Water - Formula: H2O = Hydrogen2 + Oxygen. ...
Hydrogen Peroxide - Formula: H2O2 = Hydrogen2 + Oxygen2 ...
Salt - Formula: NaCl = Sodium + Chlorine. ...
Baking Soda - Formula: NaHCO3 = Sodium + Hydrogen + Carbon + Oxygen3 ...
Octane - Formula: C8H18 = Carbon8 + Hydrogen18 Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged). The electron is a subatomic particle, symbol e⁻ or β⁻ , whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure.

is a step in the scientific method. The step that follows this step involves forming

Answers

Answer: read this hope this helped

Explanation: A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. ... The next step in the scientific method is to test the hypothesis by designing an experiment. This includes creating a list of materials and a procedure— a step-by-step explanation of how to conduct the experiment.

A high-voltage direct-current generating station delivers 10 MW of power at 250 kV to a city, as depicted in Fig. P2.12. The city is represented by resistance RL and each of the two wires of the transmission line between the generating station and the city is represented by resistance RTL. The distance between the two locations is 2000 km and the transmission lines are made of 10 cm diameter copper wire. Determine (a) how much power is consumed by the transmission line and (b) 12 V I0 _

Answers

Answer:

The answer is below

Explanation:

The resistivity of copper is ρ = 1.72 * 10⁻⁸ Ωm, diameter d = 10 cm = 0.1 m

The resistance (R) of transmission line is given as:

Rtl = ρL / A; where ρ = resistivity of copper = 1.72 * 10⁻⁸ Ωm, L = length of transmission line = 2000 km = 2000000 m, A is the area of the wire = πd²/4 = π(0.1)²/4

[tex]R_{tl}=\frac{\rho L}{A}=\frac{1.72*10^{-8}*2000000}{\pi*0.1^2/4}=4.4 \ ohm[/tex]

Power = [tex]\frac{V_L^2}{R_L}[/tex]

Power = 10 MW = 10 * 10⁶ W

[tex]10*10^6=\frac{(250*10^3)^2}{R_L} \\\\R_L=\frac{(250*10^3)^2}{10*10^6} \\\\R_L=6250\ ohm[/tex]

[tex]I_L=\frac{V_L}{R_L} \\\\I_L=\frac{250*10^3}{6250} =40\ A[/tex]

a) Since there are two tranmission lines, the power consumed by the lines is:

[tex]P_{TL}=2*I_L^2*R_{TL}=2*40^2*4.4=14080\ W[/tex]

b) The energy generated by the source = 10 * 10⁶ W + 14080 W = 10014080 W

Fraction used = 10 * 10⁶ / 10014080 * 100% = 99.86%


The string will break if the tension in
it exceeds 0.180 N. What is the
smallest possible value of d (in cm)
before the string breaks?

Answers

Answer:

define d first?

you need to list more variables

Answer:

list more valuable unit

please answer correctly

Answers

Answer:

616000 J.

Explanation:

Bi. Determination of the work done.

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:

Workdone = Force × distance

Wd = F × s

With the above formula, we can obtain the Workdone as follow:

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Wd = F × s

Wd = 220 × 2800

Wd = 616000 J.

A battery has an emf of ε = 15.0 V. THe terminal voltage of the battery is Vt = 11.6 V when it is delivering P = 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance r of the battery?

Answers

AnswerHM???

Explanation:

I dONT KNOW

What is the acceleration of a .3 kg mass when there is a net force of 25.9 N on it?

Answers

Answer:

86.33m/s^2

Explanation:

Acceleration = Force/Mass

= 25.9/0.3

= 86.33

A transformer has 150 turns in the primary coil and 350 turns in its secondary coil. If the primary coil has a voltage of 200 volts, how many volts will the secondary coil have?
242 volts
288
353
467

Answers

Answer:

467 volts

Explanation:

Vs/Vp = Ns/Np

Vs = Ns/Np × Vp

Vs = 350/150 × 200 = 7/3 × 200

Vs = 467 volts

a car acceleration from rest to 90km/h in 10 seconds. what is its acceleration in meter per second square?​

Answers

Answer:

2.5 m/s^2

Explanation:

First, convert 90 km/hr into m/s:

90/3.6 = 25 m/s

vf = final velocity = 25 m/s

vi = initial velocity = 0 m/s

t = time = 10 seconds

a = acceleration, unknown

Then, find a using the following equation:

(vf - vi)/t = a

(25 m/s)/10 s = 2.5 m/s^2

a = 2.5 m/s^2

Hope this helps!! :)

A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 3.0 m/s. What is the magnitude of the potential difference V? *
1 point
900 V
400 V
200 V
400 kV

Answers

Answer:

The magnitude of the potential difference is 900 V.

Explanation:

Given;

mass of the bead, m = 4.0 g = 0.004 kg

charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C

final velocity of the bead, v = 3 m/s

What is the magnitude of the potential difference V?

Apply the principle of conservation of energy;

The electric potential energy at the beginning is equal to kinetic energy of the bead  at the end of the journey.

qV = ¹/₂mv²

[tex]V = \frac{mv^2}{2q} \\\\V = \frac{0.004 \ \times \ (3)^2}{2(20 \times 10^{-6})}\\\\V = 900 \ V[/tex]

Therefore, the magnitude of the potential difference is 900 V.

The magnitude of the potential difference (V) is equal to 900 Volts.

Given the following data:

Mass of bead = 4.0 g to kg = 0.004 kgCharge of bead = 20 μC = [tex]20 \times 10^{-6} \;C[/tex]Final velocity of bead = 3 m/s

To determine the magnitude of the potential difference (V):

How to calculate the potential difference (V).

We would apply the law of conservation of energy, which states that the electric potential energy possessed by the bead at the beginning is equal to the kinetic energy possessed by the bead at the end of the journey:

[tex]qV = \frac{1}{2} mv^2\\\\V = \frac{\frac{1}{2} mv^2}{q} \\\\V = \frac{\frac{1}{2} \times 0.004 \times 3.0^2}{20 \times 10^{-6}} \\\\V = \frac{ 0.002 \times 9}{20 \times 10^{-6}}[/tex]

V = 900 V.

Read more on kinetic energy here: https://brainly.com/question/1242059

For a given substance, the molecules
move fastest when the substance is

Answers

Heat makes the molecules move faster

Answer:GAS

Explanation:

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