A proton is at the origin and an electron is at the point x = 0.36 nm , y = 0.39 nm . Find the electric force on the proton.Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Answers

Answer 1
Answer:

Answer:

The electric force on the proton is 8.2x10^-10 N

Explanation:

We use the formula to calculate the distance between two points, as follows:

r = ((x2-x1)^2 + (y2-y1)^2)^1/2, where x1 and x2 are the x coordinate, y2, y1 are the y coordinate. replacing values:

r = ((0.36-0)^2 + (0.39-0)^2)^1/2 = 0.53 nm = 5.3x10^-10 m

We will use the following expression to calculate the electrostatic force:

F = (q1*q2)/(4*pi*eo*r^2)

Here we have:

q1 = q2 = 1.6x10^-19 C, 1/4*pi*eo = 9x10^9 Nm^2C^-2

Replacing values:

F = (1.6x10^-19*1.6x10^-9*9x10^9)/((5.3x10^-10)^2) = 8.2x10^-10 N


Related Questions

A 200​-lb object is released from rest 600 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is minus10​v, where v is the velocity of the object in​ ft/sec, determine the equation of motion of the object. When will the object hit the​ ground? Assume that the acceleration due to gravity is 32 ft divided by sec squared and let​ x(t) represent the distance the object has fallen in t seconds.
A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).
The elasticity of demand for gasoline has been estimated to be 2.0, and the standard error is 1.0. The upper and lower bounds on the 95 percent confidence interval for the elasticity of demand for gasoline are:
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?4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?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.A. Only the odd multiples of the fundamental frequency.B. Only the even multiples of the fundamental frequency.C. All integer multiples of the fundamental frequency.E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?A. Half the length of the open-open pipe.B. Twice the length of the open-open pipe.C. One-fourth the length of the open-open pipe.D. Four times the length of the open-open pipe.E. The same as the length of the open-open pipe.F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the thirdharmonic.
A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.9 m. The total mass of the empty balloon and basket is mb = 121 kg and the total volume is Vb = 0.073 m3. Assume the average person that gets into the balloon has a mass mp = 73 kg and volume Vp = 0.077 m3. 1)What is the volume of helium in the balloon when fully inflated? m3 2)What is the magnitude of the force of gravity on the entire system (but with no people)? Include the mass of the balloon, basket, and helium. N

PLEASE HELP IT'S DUE IN LIKE 2 MINUTES

Answers

Answer:

1kg

Explanation:

this box is the smallest and weighs the least. Hope this helps :]

Question 7 of 10A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s

Answers

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

E=(\lambda)/(2\pi \epsilon_o r)

It is clear that the electric field is inversely proportional to the distance. So,

(E)/(E')=(r')/(r)

E'=(Er)/(r')

E'=(125* 3.5)/(1.5)  

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

"Which gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string?"

Answers

Answer:

the second derivative of y with respect to time gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string

Explanation:

This is because the transverse wave movement of particles take place in direction 90° to direction of movement of the wave (x) itself, so second derivative of y with respect to time (t)is what will be required

A local ice hockey pond is at 4°C when the air temperature falls, causing the water temperature to drop to 0°C with 10.9 cm of ice forming at the surface at a temperature of 0°C. How much heat was lost if the unfrozen pond is 1 m deep and 50 m on each side?

Answers

Answer: 114.4 GJ

Explanation:

Heat loss Q=U×A×ΔT

Heat loss of size A is determined by the U value of materials and the difference in temperature.

From 10.9cm from the ice

50m= 5000cm

A= 5000×5000

Q== (10.9) (5000) (5000)(4.184)(1×4 + 80)

Q = 95,771,760,000J

Q≈ 95.8 GJ

Linear gradient from the bottom of the pond to the ice:

Q = (89.1)(5000)(5000)(4.184)(1*2)

Q = 18,639,720,000J

Q ≈ 18.6 GJ

Total heat loss:

Q= 95.8GJ + 18.6GJ

Q= 114.4 GJ

Convert 7 (gcm^2)/(min^2) into a value in standard S.I. units. Be sure to use scientific notation if necessary. You do not need to answer units.

Answers

The required value is required in SI units.

The required answer is 1.94*10^(-10)\ \text{kg m}^2/\text{s}^2

SI units

The SI unit of mass, length and time is kg, m and s respectively.

In order to convert one unit into another it has to be multiplied or divided by the conversion factors.

A definite magnitude which has some quantity which is defined by convention or law is called a unit.

The conversion factors are

1\ \text{g}=10^(-3)\ \text{kg}

1\ \text{cm}=10^(-2)\ \text{m}

1\ \text{cm}^2=10^(-4)\ \text{m}^2

1 min = 60 s

1\ \text{min}^2=60*60\ \text{s}^2

So,

7\ \text{g cm}^2/\text{min}^2=7* (10^(-3)* 10^(-4))/(60* 60)\n =1.94*10^(-10)\ \text{kg m}^2/\text{s}^2

Learn more about SI units:

brainly.com/question/16393390

Hi hi hi hi hi hi hi