"A proton is placed in a uniform electric field of 2750 N/C. You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Electron in a uniform field. Calculate the magnitude of the electric force felt by the proton. Express your answer in newtons.(F = ? )Calculate the proton's acceleration.
( a= ? m/s2 )

Calculate the proton's speed after 1.40 {\rm \mu s} in the field, assuming it starts from rest.
( V= ? m/s )"

Answers

Answer 1

Answer:

To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.

PART A ) For the electrostatic force we have that is equal to

$F=qE$

Here

q= Charge

E = Electric Force

$F=(1.6*10^(-19)C)(2750N/C)$

$F = 4.4*10^(-16)N$

PART B) Rearrange the expression F=ma for the acceleration

$a = (F)/(m)$

Here,

a = Acceleration

F = Force

m = Mass

Replacing,

$a = (4.4*10^(-16)N)/(1.67*10^(-27)kg)$

$a = 2.635*10^(11)m/s^2$

PART C) Acceleration can be described as the speed change in an instant of time,

$a = (v_f-v_i)/(t)$

There is not $v_i$ then

$a = (v_f)/(t)$

Rearranging to find the velocity,

$v_f = at$

$v_f = (2.635*10^(11))(1.4*10^(-6))$

$v_f = 3.689*10^(5)m/s$

Answer 2

Answer:

Final answer:

The magnitude of the electric force felt by the proton is 4.4 x 10^-16 N. The proton's acceleration is 2.64 x 10^11 m/s^2. The proton's speed after 1.40 μs in the field is 3.70 x 10^5 m/s.

Explanation:

The charge of a proton is 1.6 x 10-19 coulombs and the electric field strength is 2750 N/C. Therefore, the magnitude of the electric force felt by the proton is (1.6 x 10-19 C)(2750 N/C) = 4.4 x 10-16 N. The mass of a proton is approximately 1.67 x 10-27 kilograms. Therefore, the proton's acceleration is (4.4 x 10-16 N)/(1.67 x 10-27 kg) = 2.64 x 1011 m/s2. Since the proton starts from rest, its initial velocity (u) is 0. Therefore, the proton's speed after 1.40 μs is v = (2.64 x 1011 m/s2)(1.40 x 10-6 s) = 3.70 x 105 m/s.

Learn more about Proton in Electric Field here:

brainly.com/question/32825186

#SPJ3

Related Questions

Calculate the linear momentum per photon,energy per photon, and the energy per mole of photons for radiation of wavelength; (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm (violet), (d) 200 nm (ultraviolet), (e) 150 pm (X-ray), and (f ) 1.0 cm (microwave).
A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?
A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate. (f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?
A train travels 64 kilometers in 5hours and then 93 kilometers in 2hours. What is it’s average speed?
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

John, who has a mass of 65kg stands at rest on the ice. He catches a 10kg ball that is thrown to him at 5m/s.

Answers

The momentum of John after catching the ball is 50 kg.m/s.

"Your question is not complete, it seems to be missing the following information";

find John's momentum

The given parameters;

mass of John, m = 65 kg
mass of the ball caught by John, m' = 10 kg
initial velocity of John, u = 0
initial velocity of the ball, v = 5 m/s

Apply the principles of conservation of linear momentum to determine the momentum of John.

The momentum of John is calculated as follows;

P = mu + mv

P = (65 x 0) + (10 x 5)

P = 0 + 50

P = 50 kg.m/s

Thus, the momentum of John after catching the ball is 50 kg.m/s.

Learn more here:brainly.com/question/24159955

Why a switch is connected in phase wire and never is neutral wire?

Answers

If you had the wire connected to a neutral wire, you would never get a charge. You could receive a charge pulse from the phase wire.

Answer:

ifyou have a wre connect you should not have to connected

i think that is the answer

A curve of radius 35 m isbanked; therefore, no friction
is required at a speed of 7
m/s of a car. What is the?
banking angle

Answers

Answer:

The banking angle is 23.98 degrees.

Explanation:

We have,

Radius of a curve is 35 m

Speed of a car is 7 m/s

It is required to find the banking angle. At equilibrium, net force is equal to the centripetal force between vehicle and the road such that the banking angle is given by :

$\tan\theta=(v^2)/(rg)$

g is acceleration due to gravity

$\tan\theta=(7^2)/(35* (22)/(7))\n\n\theta=\tan^(-1)\left(0.445\right)\n\n\theta=23.98^(\circ)$

So, the banking angle is 23.98 degrees.

When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?

Answers

The extension of the spring in the elevator is 60 mm.

For the extension of the spring to be zero, the elevator must be moving downwards under free fall.

The given parameters;

mass of the box, m = 5 kg
extension of the spring, x = 50 mm = 0.05 m

The spring constant is calculated as follows;

F = kx

mg = kx

$k = (mg)/(x) \n\nk = (5 * 9.8)/(0.05) \n\nk = 980 \ N/m$

The tension on the spring in an elevator accelerating upwards is calculated as follows;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 2)

T = 59 N

The extension of the spring is calculated as follows;

$T = kx\n\nx = (T)/(k) \n\nx = (59)/(980) \n\nx = 0.06 \ m\n\nx = 60 \ mm$

For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.

For free fall, a = g

T = m(g - a) = 0

Learn more here:brainly.com/question/4404276

Answer:

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

we know that;

mg = kX

5 × 9.81 = k(0.05)

k = 981 N/m

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:i. The current density ii. The drift velocity

Answers

Answer:

The current density is $J = 2.04 * 10^(6) A /m^2$

The drift velocity is $v_d = 1.5 * 10^(-4) m/s$

Explanation:

From the question we are told that

The nominal diameter of the wire is $d = 1.02 mm= (1.02)/(1000) = 0.00102 \ m$

The current carried by the wire is $I = 1.67 A$

The power rating of the lamp is $P = 200 W$

The density of electron is $n = 8.5 * 10^(28) \ e/m^3$

The current density is mathematically represented as

$J = (I)/(A)$

Where A is the area which is mathematically evaluated as

$A = \pi (d^2)/(4)$

Substituting values

$A = 3.142 * ((1.02 * 10^(-3))^2 )/(4)$

$A = 8.0*10^(-4)m^2$

$J = (1.67)/(8.0*10^(-4))$

$J = 2.04 * 10^(6) A /m^2$

The drift velocity is mathematically represented as

$v_d = (J)/(ne)$

Where e is the charge on one electron which has a value $e = 1.602 *10^(-19) C$

$v_d =(2.04 * 10^6 )/(8.5 *10^(28) * 1.6 * 10^(-19))$

$v_d = 1.5 * 10^(-4) m/s$

A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)