200 J of heat is added to two gases, each in a sealed container. Gas 1 is in a rigid container that does not change volume. Gas 2 expands as it is heated, pushing out a piston that lifts a small weight. Which gas has the greater increase in its thermal energy?Which gas has the greater increase in its thermal energy?Gas 1Gas 2Both gases have the same increase in thermal energy.

Answers

Answer 1

Answer:

Gas 1

Explanation:

The reason for this is that for gases attached to both gases or containers, with a heat of 200 J, the change in volume is only observed in gas 2, whereas the volume of gas 1 is the same as that of gas. Therefore, the internal energy (heat) or thermal energy of the system is not utilized for Gas 1 and hence the absorption and transfer of energy is the same, whereas Gas 2 is propagated by the use of additional heat of heat. Thus there is a large increase in the thermal energy of Gas1.

Related Questions

A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion, after which a horizontal force of 60 N is required to keep the block in moving with constant speed. Find the coefficient of static and kinetic friction between the block and the surface.

Answers

Answer:

μs = 0.30

μk = 0.24

Explanation:

In order to calculate the kinetic friction and static friction between the block and the surface, you take into account that the kinetic friction is important when the block is moving and the static friction when the block is at rest.

You use the following formula to find the coefficient of static friction:

[tex]F_1=\mu_s Mg[/tex]       (1)

F1 = 75N

μs: coefficient of static friction = ?

M: mass of the block = 25.0kg

g: gravitational acceleration = 9.8m/s^2

You solve for μs in the equation (1):

[tex]\mu_s=\frac{F_1}{Mg}=\frac{75N}{(25.0kg)(9.8m/s^2)}=0.30[/tex]

For the coefficient of kinetic friction you have:

[tex]F_2=\mu_k Mg[/tex]       (2)

F2 = 60N

μk: coefficient of kinetic friction = ?

You solve for μk in the equation (2):

[tex]\mu_k=\frac{F_2}{Mg}=\frac{60N}{(25.0kg)(9.8m/s^2)}=0.24[/tex]

Then, you have:

coefficient of static friction = 0.30

coefficient of kinetic friction = 0.24


3. Which is a general chemical equation for an endothermic, double-replacement reaction?

O AB+ CD + energy-> AD+ CB
O AB+ CD AD + CB + energy
O AB+ C + energy - A+ CB
O AB+C- A+ CB + energy

Answers

The correct answer is A.  AB+ CD + energy-> AD+ CB

Explanation:

In chemistry, a reaction is endothermic if the reaction involves absorption of heat or energy and this is necessary for the reaction to start. In terms of the chemical equation, this implies energy is part of the reactants or initial substances. Besides this, if the reaction is a double-replacement reaction this means two ions of the original substances are swapped or replaced, which means new substances in the products.

According to this, option A is the correct chemical equation because energy is part of the reactants, which shows the reaction is endothermic and the reactants AB + CD lead to the products AD + CB which shows two ions of the compounds were replaced (double replacement).

Answer:

The proper Answer is A) AB+CD + Energy --> AD + CB

Explanation:

The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?

Answers

Answer:

[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]

Explanation:

Hello,

In this case, since pressure is defined as the force applied over a surface:

[tex]P=\frac{F}{A}[/tex]

We can associate the force with the weight of the needle computed by using the acceleration of the gravity:

[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]

And the area of the the tip (circle) in meters:

[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]

Thus, the pressure exerted on the record turns out:

[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]

Which is truly a large value due to the tiny area on which the pressure is exerted.

Best regards.

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 3.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible

Answers

Answer:

1089.74 Hz

Explanation:

Using Pythagoras theorem, we can find the distance from the point to the second speaker.

Thus;

d2 = √(3² + 2.3²)

d2 = √(9 + 5.29)

d2 = √14.29

d2 = 3.78 m

Then, the path distance which is the extra distance travelled would be;

Δd = d2 - d1

Δd = 3.78 - 3

Δd = 0.78 m

Now, the destructive interference condition is given by the formula;

Δd = (m + ½)λ

λ is the wavelength

m is a non - negative integer.

In this case, m = 2

Thus;

0.78 = (2 + ½)λ

λ = 0.78/(2½)

λ = 0.312 m

Now the formula for frequency of a wave is given by;

f = v/λ

Where v is speed of sound.

Thus;

f = 340/0.312

f = 1089.74 Hz

An accelerating voltage of 2.25 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 36.4 cm away. (a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field

Answers

Answer:

s= 8.28×10⁻¹⁶m

Explanation:

given

V= 2.25×10³V

from conservation of energy

mv²/2=qΔV

v=√(2qΔV/m)

v= √(2×1.6×10⁻¹⁹×2.25×10³/9.1×10⁻³¹)

=√7.9×10¹⁴m/s

=2.8×10⁷m/s

the deflection of electron beam is

S= gt²/2

recall t= d/v

s=g([tex]\frac{d}{v}[/tex])²/2

s= [tex]\frac{1}{2}[/tex]×9.8×(0.364/2.8×10⁷)²

s= 8.28×10⁻¹⁶m

what tools use cut wood

Answers

Answer:

hand saws

power saws

Circular Saw

Explanation:

that is all that i know

Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)

Answers

Answer:

*If the particles are deflected in opposite directions, it implies that their charges must be opposite

*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Explanation:

When a charged particle enters a magnetic field, it is subjected to a force given by

        F = q v x B

where bold letters indicate vectors

   

this expression can be written in the form of a module

        F = qv B sin θ

and the direction of the force is given by the right-hand rule.

In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1

If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.

Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N . Rolling friction can be neglected. You may want to review (Pages 165 - 168) . Part A What is the magnitude of the force of the car on the truck

Answers

Answer:

a) 3344 N

b) 3344 N

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = 3344 N

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is 3344 N

To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch

Answers

Answer:

By a factor of about 0.23

Explanation:

Pressure is force over an area: P=F/A

Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that we can say:

P₁=(4/0.9025)P₂=4.4P₂   or

P₂=(0.9025/4)P₁=0.23P₁

What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.

By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.

Pressure

Friction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.

Pressure exists as force over an area: P=F/A

Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

let's divide P₁ by P₂

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that, we can say:

P₁=(4/0.9025)P₂=4.4P₂ or

P₂=(0.9025/4)P₁=0.23P₁

Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,

To learn more about Pressure refer to:

https://brainly.com/question/912155

#SPJ2

Passengers in a carnival ride move at constant speed in a circle of radius 5.0 m, making a complete revolution in 4.0 s. As they spin, they feel their backs pressing against the wall holding them in the ride. A. What is the direction of the passengers' acceleration? a. No direction (zero acceleration) b. Directed towards center c. Directed away from center d. Directed tangentially B. What is the passengers' linear speed in m/s? C. What is the magnitude of their acceleration in m/s^2? D. What is their angular speed in rad/s?

Answers

Answer:

A. b) Directed towards center

B. [tex]v = 7.854\ m/s[/tex]

C. [tex]a_c = 12.337\ m/s^2[/tex]

D. [tex]w = 1.57\ rad/s[/tex]

Explanation:

The "force" that they feel pressing their backs against the wall is because the reaction to the  centripetal acceleration .

A.

This acceleration has its direction towards the center of the circle. (option b)

B.

Their linear speed can be calculated with the equation:

[tex]v = (\theta/t)*r[/tex]

Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:

[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]

C.

The centripetal acceleration is given by the equation:

[tex]a_c = v^2/r[/tex]

[tex]a_c = 7.854^2/5[/tex]

[tex]a_c = 12.337\ m/s^2[/tex]

D.

Their angular speed is given by the equation:

[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]

In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0215 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.15 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.

Answers

Answer:

h = 8.48*10^-3m

Explanation:

In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.

The total momentum of the system before the collision is equal to the total momentum after:

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]        (1)

Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.

m1: mass of the marble = 0.0215kg

m2: mass of the pendulum concentrated at its end = 0.250kg

v1: horizontal speed of the arble before the collision = 5.15m/s

v2: horizontal speed of the pendulum before the collision = 0m/s

v: horizontal speed of both marble and pendulum after the collision = ?

You solve the equation (1) for v, and replace the values of the other parameters:

[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]

Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:

[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]

v = 0.40m/s

g: gravitational acceleration = 9,8m/s^2

[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]

Then, the height reached by marble and pendulum is 8.48*10^-3m

You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your normal weight is w= 635 N. If the elevator has an initial acceleration of magnitude 2.45 m/s2, what does the scale read? Express your answer in newtons.

Answers

Answer: 479. 425 N

Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.

It is given by N= m(g+a)

When it is accelerating downward, the scale reading is less than the true weight.

It so given by N = m(g-a)

The answer to the above questions is in the attached photo

Answer:

the scale will read 476.414 N

Explanation:

Weight = 635 N

mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)

mass m = 635 ÷ 9.81 = 64.729 kg

initial acceleration of the elevator a = 2.45 m/s^2

the force produced by the acceleration of the elevator downwards = ma

your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss

apparent weight = weight - ma

apparent weight = 635 - (64.729 x 2.45)

apparent weight =  635 - 158.586  = 476.414 N

A student has made the statement that the electric flux through one half of a Gaussian surface is always equal and opposite to the flux through the other half of the Gaussian surface. This is:_______.

a. never true.

b. never false.

c. true whenever enclosed charge is symmetrically located at a center point, or on a center line or centrally placed plane

d. true whenever no charge is enclosed within the Gaussian surface.

e. true only when no charge is enclosed within the Gaussian surface.

Answers

Answer:

E.true only when no charge is enclosed within the Gaussian surface.

Explanation:

Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.

Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop

Answers

Answer:

v = 1.7 m/s

Explanation:

By applying conservation of energy principle in this situation, we know that:

Loss in Potential Energy of Car = Gain in Kinetic Energy of Car

mgΔh = (1/2)mv²

2gΔh = v²

v = √(2gΔh)

where,

v = velocity of car at top of the loop = ?

g = 9.8 m/s²

Δh = change in height = 45 cm - Diameter of Loop

Δh = 45 cm - 30 cm = 15 cm = 0.15 m

Therefore,

v = √(2)(9.8 m/s²)(0.15 m)

v = 1.7 m/s

What is meant civilized?

Answers

Answer:

at an advanced stage of social and cultural development. "a civilized society"

Explanation:

polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)

Stress is a factor that contributes to heart disease risk.true or false

Answers

The answer for this question would be True

A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that is 0.60 m from the center is closest to

Answers

Complete Question

The complete question is  shown on the first uploaded image  

 

Answer:

The electric field at that point is  [tex]E = 7500 \ N/C[/tex]

Explanation:

From the question we are told that  

       The  radius of the inner circle is [tex]r_i = 0.80 \ m[/tex]

        The  radius of the outer circle is  [tex]r_o = 1.20 \ m[/tex]

       The  charge on the spherical shell [tex]q_n = -500nC = -500*10^{-9} \ C[/tex]

      The magnitude of the point charge at the center is  [tex]q_c = + 300 nC = + 300 * 10^{-9} \ C[/tex]

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 [tex]E = \frac{k * q_c }{x^2}[/tex]

substituting values  

                  [tex]E = \frac{k * q_c }{x^2}[/tex]

where  k is  the coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

     substituting values

                  [tex]E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}[/tex]

                 [tex]E = 7500 \ N/C[/tex]

On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?

Answers

Answer:

T = 2.06h

Explanation:

In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:

[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex]         (1)

T: time for a complete orbit = ?

r: radius of the orbit

G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2

Mm: mass of the moon = 7.34*10^22 kg

The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:

[tex]r=R_m+160km\\\\[/tex]

Rm: radius of the moon = 1737.1 km

[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]

Then, you replace all values of the parameters in the equation (1):

[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]

In hours you obtain:

[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]

The time that the Apollo takes to complete an orbit around the moon is 2.06h

Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals

Answers

Answer:

producers make its own energy frominorganic substances.

A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice​

Answers

Answer:

nvbnncbmkghbbbvvvvvvbvbhgggghhhhb

What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)

Answers

Answer:

1.99*10-4sec

Explanation:

Signal propagation speed=0.82∗2.46∗108m/s

d=2000 m

Tp=20000/0.82∗2.46∗108 sec

ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8

= 1.99* 10^-4seconds

A block is attached to a horizontal spring and it slides back and forth in simple harmonic motion on a frictionless horizontal surface. At one extreme end of the oscillation cycle, where the block comes to a momentary halt before reversing the direction of its motion, another block is placed on top of the first block without changing its zero velocity. The simple harmonic motion then continues. What happens to the amplitude and the angular frequency of the ensuing motion of the two-block system

Answers

Answer:

A = A₀ ,   w = w₀/√2

Explanation:

This is a problem that we must solve with Newton's second law. They indicate that at the end of the initial movement where the speed is zero, add a mass to the block, we assume that it has the same mass, therefore the total mass is m_total = 2 m. Let's write Newton's second law at this point

                   [tex]F_{e}[/tex] = m_total a

the elastic force is

                   F_{e} = - k x

acceleration is

                   a = d²x / dt²

we substitute

                   - k x = m_total   d²x / dt²

                     d²x / dt² + (k / m_total) x = 0

we substitute

                     d²x / dt² + (k /2m) x = 0

the solution to this differential equation is

                    x = A cos (wt + Ф)

where

                  w = √ (k / 2m)

to find the constant Ф we use the velocity

                    v = dx / dt = - Aw sin (wt + Ф)

                   

At the most extreme point and when the new movement begins (t = 0) they indicate that v = 0

                   0 = - A w sin Ф

for this expression to be zero the sine must be zero therefore Ф = 0

when replacing

                  x = A cos (wt)

                  w = 1 /√2  √ (k / m)

if we want to relate to the initial movement (before placing the block)

                 w₀ = √ (k / m)

                 w = w₀ /√ 2

The amplitude of the movement is the distance from the equilibrium point to where the movement begins, in this case it is the same as in the initial movement

                  A = A₀

the subscript is used to refer to the oscillations before placing the second block

we substitute to have the final equation

                 x = A₀ cos (w₀ t /√2)

                     

                 A = A₀

                 w = w₀/√2

The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20-m wide and averages 20-m deep.
(a) What is the average speed of the river in the gorge?
(b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?

Answers

Answer:

(a) V = 0.75 m/s

(b) V = 0.125 m/s

Explanation:

The speed of the flow of the river can be given by following formula:

V = Q/A

V = Q/w d

where,

V = Speed of Flow of River

Q = Volume Flow Rate of River

w = width of river

d = depth of river

A = Area of Cross-Section of River = w d

(a)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 20 m

d = 20 m

Therefore,

V = (300 m³/s)/(20 m)(20 m)

V = 0.75 m/s

(b)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 60 m

d = 40 m

Therefore,

V = (300 m³/s)/(60 m)(40 m)

V = 0.125 m/s

If the person is shaking her hand up-and-down 12 times per second, what is the wave speed?

Answers

Welllll, first of all, it would take incredible muscular development and control to be able to do that, and I don't believe it's actually humanly possible.

But for Math and Physics problems, that's OK.  We don't mind suspending our disbelief, accepting a temporary alternate reality, and working with the hand that is dealt.

The speed of a wave doesn't depend on how the wave is created.  A puppy wagging its tail, a fly batting its wings, or a person shaking her hand up and down, are moving the air.  The wave that travels away from the vibration is a sound wave in air.  Its speed depends only on the characteristics of the air it travels through.

For some typical combination of temperature, pressure, and humidity, this speed (of sound) is taken to be 343 meters per second.

Notice that the 'sound' of shaking her hand up and down will not be 'heard' by anyone, no matter how close she stands to them.  12 Hz (12 per second) is not a fast-enough wiggle to be sensed as sound by human ears.  If the person senses the wave at all, it will only be as some kind of pulsating breeze.

Bromine, a liquid at room temperature, has a boiling point

Answers

Yes it does !  The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided.  The boiling point is higher than room temperature.

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20×10−6s3.20×10 −6 s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Answers

Answer:

E = 326.17 N/C

Explanation:

(a) In order to calculate the magnitude of the electric field between the parallel plates you first calculate the acceleration of the proton. You use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]         (1)

vo: initial speed of the proton = 0m/s

t: time that the proton takes to cross the space between the plates = 3.20*10^-6 s

a: acceleration of the proton = ?

x: distance traveled by the proton = 1.60cm = 0.016m

You solve the equation (1) for a, and replace the values of all parameters:

[tex]a=\frac{2x}{t^2}=\frac{2(0.016m)}{(3.20*10^{-6}s)^2}=3.125*10^{10}\frac{m}{s^2}[/tex]

Next, you use the Newton second law for the electric force, to find the magnitude of the electric field:

[tex]F_e=qE=ma[/tex]           (2)

q: charge of the proton = 1.6*10^-19C

m: mass of the proton = 1.77*10^-27kg

You solve the equation (2) for E:

[tex]E=\frac{ma}{q}=\frac{(1.67*10^{-27}kg)(3.125*10^{10}m/s^2)}{1.6*10^{-19}C}\\\\E=326.17\frac{N}{C}[/tex]

The magnitude of the electric field in between the parallel plates is 326.17N/C

A d'Arsonal meter with an internal resistance of 1 kohm requires 10 mA to produce full-scale deflection. Calculate thew value of a series

Answers

Question:

A d’Arsonval meter with an internal resistance of 1 kΩ requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.

Answer:

4kΩ

Explanation:

Given;

internal resistance, r = 1kΩ

current, I = 10mA = 0.01A

Voltage of full scale, V = 50V

Since there is full scale voltage of 50V, then the combined or total resistance (R) of the circuit is given as follows;

From Ohm's law

V = IR

R = [tex]\frac{V}{I}[/tex]                 [substitute the values of V and I]

R = [tex]\frac{50}{0.01}[/tex]

R = 5000Ω = 5kΩ

The combined resistance (R) is actually the total resistance of the series arrangement of the series resistance([tex]R_{S}[/tex]) and the internal resistance (r) in the circuit. i.e

R = [tex]R_{S}[/tex] + r

[tex]R_{S}[/tex] = R - r                 [Substitute the values of R and r]

[tex]R_{S}[/tex] = 5kΩ - 1kΩ

[tex]R_{S}[/tex] = 4kΩ

Therefore the series resistance is 4kΩ

Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is

N

100

m

What is the elastic potential energy stored from the spring's compression?

Choose 1 answer:

-3.0J

-0.045 J

0.090 J

0.045 J

Answers

Answer:

0.045 J

Explanation:

From the question,

The elastic potential energy stored in a spring is given as,

E = 1/2ke²...................... Equation 1

Where E = elastic potential energy, k = spring constant, e = compression.

Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m

Substitute these values into equation 1

E = 1/2(100)(0.03²)

E = 50(9×10⁻⁴)

E = 0.045 J

Hence the right option is 0.045 J

Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m.The spring constant k is 100 N/m.

What is the elastic potential energy stored from the spring’s compression?

Answer: 0.045 J

When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?

Answers

Answer: If we have equilibrium, the magnitude must be zero.

Explanation:

If the charges are in equilibrium, this means that the total charge is equal to zero.

And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero

Based on what you know about electricity, hypothesize about how series resistors would affect current flow. What would you expect the effective resistance of two equal resistors in series to be, compared to the resistance of a single resistor?

Answers

Answer:

Effective resistance of two equal resistors in series is twice that of a single resistor and in essence will reduce the amount of current flowing in the circuit.

Explanation:

When two resistors are connected in series, their effective resistance is the sum of their individual resistances. For example, given two resistors of resistance values R₁ and R₂, their effective resistance, Rₓ is given by;

Rₓ = R₁ + R₂            --------------(1)

If these resistors have equal resistance values, say R, then equation 1 becomes;

Rₓ = R + R

Rₓ = 2R

This means that their effective resistance is twice of their individual resistances. In other words, when two equal resistors are in series, their effective resistance is twice the resistance of each single one of those resistors.

Now, according to Ohm's law, voltage(V) is the product of current (I) and resistance (R). i.e

V = IR

I = [tex]\frac{V}{R}[/tex]

We can deduce that current increases as resistance decreases and vice-versa.

So, if the two equal resistors described above are connected in series, the amount of current flowing will be reduced compared to having just a single resistor.

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