How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.00×10−15 m (a typical nuclear distance)?

Answers

Answer 1

Answer:

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field. The work would it take to push two protons will be 7.7×10⁻¹⁴.

What is electric potential?

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electricfield.

The given data in the problem is;

q is the charge= 1.6 ×10⁻¹⁹ C

V is the electric potential

r₁ is the first separation distance= 2.00×10−10 m

r₂ is the second separation distance= 3.00×10−15 m

The electric potential generated by the proton at rest at the two points, using the formula:

Firstly the electric potential at loction 1

$\rm V=(Kq)/(r) \n\n v_i= 9* 10^9 * (1.6*10^(-19))/(2.0*10^(-10))$

The electric potential at loction 2

$V_f = 9 * 10^9 (1.6 * 10^(-19))/(3.0*10^(-15)) \n\n \rm v_f= 4.8 *10^5 \ V$

The product of difference of electric potential and charge is defined as the workdone.

$\rm W= q \triangle V \n\n \rm W= 1.6 * 10^-19 *( 4.8*10^5 -7.2) \n\n \rm W= 7.7 * 10^(-14)$

Hence the work would it take to push two protons will be 7.7×10⁻¹⁴.

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Answer 2

Answer:

We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

$W=q \Delta V$

where $q=1.6 \cdot 10^(-19)C$ is the charge of the proton and $\Delta V$ is the potential difference between the final position and the initial position of the proton. To calculate this $\Delta V$ , we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

$V=k(Q)/(r)$

where $k=9.0 \cdot 10^9 N m^2 C^(-2)$ is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

$V_i = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(2.0 \cdot 10^(-10))=7.2 V$

$V_f = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(3.0 \cdot 10^(-15))=4.8 \cdot 10^5 V$

Therefore, the work done is

$W=q \Delta V=(1.6 \cdot 10^(-19)C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^(-14) J$

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The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
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4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
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Answers

Final answer:

The physics of wind instruments is based on standing waves. The lowest frequency of a sound wave produced in an open-open pipe can be calculated. When a hole is drilled through the pipe, the fundamental frequency is lower than before. Only odd multiples of the fundamental frequency will be present in a pipe with a hole halfway down its length. An open-closed pipe needs to be twice the length of an open-open pipe to achieve the same fundamental frequency. The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

Explanation:

The lowest frequency f of the sound wave produced when blowing into an open-open pipe can be calculated using the formula f = v/2L, where v is the speed of sound and L is the length of the pipe. Plugging in the values, we get f = 343/(2*0.8), which equals 214.375 Hz.

When a hole is drilled through the side of the pipe, the fundamental frequency of the sound wave generated in the pipe is lower than before. This is because the effective length of the pipe has been changed, resulting in a lower frequency.

The fundamental frequency of the sound that can be produced in the original pipe with a hole drilled halfway down its length can be calculated as f = v/L, where L is the new effective length of the pipe. Since the hole is halfway down, the effective length becomes half of the original length, resulting in a frequency equal to the original fundamental frequency.

When blowing air into the pipe with a hole halfway down its length, only the odd multiples of the fundamental frequency will be present. Therefore, the frequencies that can be created are only the odd multiples of the fundamental frequency.

The length of an open-closed pipe needed to achieve the same fundamental frequency as an open-open pipe is twice the length of the open-open pipe. This is because an open-closed pipe has only odd harmonics, which are spaced twice as far apart as the harmonics in an open-open pipe.

The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

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The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.

Answers

Answer:

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

Final answer:

The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.

Explanation:

To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.

Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.

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Maggots feed on dead and decaying organisms for energy. What are maggots?autotrophs
producers
decomposers
heterotrophs

Answers

Answer:

Explanation:

Decomposers is the correct answer

Answer:

Decomposers is the right answer

Explanation:

Maggots are decomposers because they eat the dead bodys for energy

I don't know if the thing I wrote it truse so ya

An advantage of J.J. Thomson's Plum Pudding Model was that it _____. A. was a much less expensive way to study atoms
B. simplified the calculations necessary to describe an atom
C. clearly explained where electrons were located in an atom
D. is much less expensive to bake a plum pudding than to look at an atom

Answers

the answer is d i think

Answer: plz mark brainliest

the answer is C

Explanation:

bc he used the plums or whatever's inside of the pudding to identify were electrons could be located and since it was a well known deserte many people where able to understand his analogy

An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Answers

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

$\alpha=(L_(1)-L_(0))/(L_(0)(T_(1)-T_(0)))$

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is $25*10^(-6)C^(-1)$

Substitute, $L_(0)=17.400cm,T_(1)=100,T_(0)=20,\alpha=25*10^(-6)C^(-1)$

$25*10^(-6)C^(-1) =(L_(1)-17.400)/(17.400(100-20))\n\n25*10^(-6)C^(-1) = (L_(1)-17.400)/(1392) \n\nL_(1)=[25*10^(-6)C^(-1) *1392}]+17.400\n\nL_(1)=17.435cm$

Hence, The new length of aluminum rod is 17.435 cm.

Learn more:

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Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

pls give brainliest

A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?

Answers

the mass would be 75kg

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