Answer:
0.1
Explanation:
1. A paper pinwheel is spinning in the wind. Which statement is correct about the forces responsible for the rotation?
The components of gravity and the force of wind that point through the pivot are responsible for the rotation.
Only the perpendicular component of wind is responsible for the rotation, because gravity points downward.
Only the perpendicular component of gravity is responsible for the rotation, because wind points toward the pivot. <- MY ANSWER
The perpendicular components of gravity and the force of wind are responsible for the rotation.
2. Which statement about kinetic and static friction is accurate?
Static friction is greater than kinetic friction, but they both act opposite the applied force.
Static friction is greater than kinetic friction, and they both act in conjunction with the applied force.
Kinetic friction is greater than static friction, but they both act opposite the applied force.
Kinetic friction is greater than static friction, and they both act in conjunction with the applied force.
3. A group of engineers is preparing a satellite to land by moving it 10% closer to Earth in each rotation. Which statement is correct about the rotational inertia of the satellite?
Rotational inertia decreases proportional to the decrease in the radius of rotation.
Rotational inertia increases proportional to the decrease in the radius of rotation. <- MY ANSWER
Rotational inertia first decreases and then increases as the satellite is ready to land.
Rotational inertia does not change because it is conserved.
4. In the experimental setup shown (see image attached), a car has one end of a string attached to it. The other end of the string has a hanger on which metal discs can be added or removed. The car moves along the table, and two photogate probes sense the motion of the car. The probes send information to a computer that displays the experimental results.
Which quantity is being held constant in this experiment?
mass
force
acceleration
velocity <- MY ANSWER
Answer:
1) Only the perpendicular component of wind is responsible for the rotation, because gravity points downward.
2) static friction is greater than kinetic friction and the applied force acts oppositely
3) Inertia decreases with decreasing radius of the orbit.
4) constant is the mass of the car
Explanation:
1) In order to know which statement is correct, let's analyze the movement of the mill.
The rotation of the blades due to a force perpendicular to each one, which creates a torque that results in its rotation.
The true force does not affect the movement because it is applied at the center of mass of the mill that coincides with its axis of rotation.
Of the statements, the correct one is:
The perpendicular to the wind is responsible for the movement
2) static friction is the result of microscopic bonds between two surfaces when there is no relative motion between them
kinetic friction is the result of the unions between the two surfaces when there is a relative movement between them.
From these statements it follows that friction is contrary to motion, static friction must be greater than kinetic friction.
Of the statements the correct one is:
static friction is greater than kinetic friction and the applied force acts oppositely
3) The inertia of a particle is
I = m R²
where m is the mass of the satellite and R is the distance from the center of the Earth, decreasing with each orbit
The correct answer is:
Inertia decreases with decreasing radius of the orbit.
4) From the figure it can be seen that the motorist subjected to a constant force, for which we can apply Newton's second law
F = m a
The force equals the applied weight.
The quantity that is held constant is the mass of the car
since the applied force and therefore the acceleration can be changed by changing the applied weight.
Answer:
1. The perpendicular components of gravity and the force of wind are responsible for the rotation.
2. Static friction is greater than kinetic friction, but they both act opposite the applied force.
3. Rotational inertia decreases proportional to the decrease in the radius of rotation.
4. mass
If an electron emits 3 eV of energy, what is the corresponding wavelength of the emitted photon?
Answer:
The wavelength of the emitted photon is 413.6 nm
Explanation:
Given;
energy of the emitted photon, E = 3 eV = 3 x 1.602 x 10⁻¹⁹ J
speed of light, c = 3 x 10⁸ m/s
The energy of the emitted photon is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f is the frequency of the light = c / λ
λ is the wavelength
[tex]E = \frac{hc}{\lambda}\\\\\lambda = \frac{hc}{E}\\\\\lambda = \frac{(6.626*10^{-34})(3*10^8)}{3*1.602*10^{-19}}\\\\\lambda = 4.136 *10^{-7} \ m\\\\\lambda = 413.6 *10^{-9} \ m\\\\\lambda = 413.6 \ nm[/tex]
Therefore, the wavelength of the emitted photon is 413.6 nm
If a planet was located approximately 24 thousand light-years from the center of a galaxy and orbits that center once every 194 million years, how fast is the planet traveling around the galaxy in km/hr? If needed, use 3.0 × 108 m/s for the speed of light.
84km/hr
Explanation:
To find one light yr
= 3E8m/s x 24x 365/1000
= 996980E7kms
So
Radius of orbit for 24000light years
=996980E7kms x 24000
Total orbit distance is = 2πr
= 2 *π x 996980E7kms x 24000
Total time will be
= 194E6 x 24x365 hrs
Finally speed = distance/ time
= 2 *π x 996980E7kms x 24000/194E6 x 24x365 hrs
= 84km/hr
Select the correct answer. Physics is explicitly involved in studying which of these activities? A. the mixing of metals to form an alloy B. the metabolic functions of a living organism C. the motion of a spacecraft under gravitational influence D. the depletion of the atmospheric ozone layer due to pollutants E. the killing of cancerous cells by radiation therapy
Answer: C. the motion of a spacecraft under gravitational influence.
Explanation:
A is Metallurgy, B is Biology, C is astro-physics, I am not sure what D is, but it's safe to say it's not physics, E, micro-biology, and the study of radiation. C is the only one involving physics.
Asteroid X orbits moon XYZ with a speed of 500m/s and a centripetal force of 1500N. Asteroid X, Y and Z all have the same mass. If asteroid Y orbits moon XYZ with the same orbital radius as asteroid X but at a speed of 100m/s. What is the centripetal force acting acting on asteroid Y? If asteroid Z orbits moon XYZ with a speed of 1500m/s but on orbital radius that is half of asteroid X, what is the centripetal force acting asteroid Z?
Answer this question
Explanation:
A 3.0kg mass tied to a string
is swirled round with a uniform
speed of 20.0mls. If the length of
the string is 40cm Find the
tension in the string
Answer:
[tex] \boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN} [/tex]
Given:
Mass (m) = 3.0 kg
Uniform speed (v) = 20 m/s
Length of string (r) = 40 cm = 0.4 m
To Find:
Tension in the string (T)
Explanation:
Tension (T) is the string will be equal to centripetal force ([tex] \sf F_c [/tex]).
[tex] \boxed{ \bold{ T = F_c = \frac{m {v}^{2} }{r} }}[/tex]
Substituting value of m, v & r in the equation:
[tex] \sf \implies T = \frac{3 \times {20}^{2} }{0.4} \\ \\ \sf \implies T = \frac{3 \times 400}{0.4} \\ \\ \sf \implies T =3 \times 1000 \\ \\ \sf \implies T =3000 \: N \\ \\ \sf \implies T =3 \: kN[/tex]
[tex] \therefore[/tex]
Tension in the string (T) = 3 kN
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm. The explorer finds that the pendulum completes 105 full swing cycles in a time of 125 s.
What is the value of the acceleration of gravity on this planet?Express your answer in meters per second per second.
Answer:
The value is [tex]g = 16.104 \ m/s[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 53.0 \ cm = 0.53 \ m[/tex]
The number of cycle is [tex]n = 105[/tex]
The time taken is [tex]t = 125 \ s[/tex]
The period is mathematically represented as
[tex]T = 2 \pi \sqrt{\frac{ l}{g} }[/tex]
Also the period is mathematically represented as
[tex]T = \frac{ t}{n}[/tex]
[tex]T = \frac{125}{110}[/tex]
[tex]T = 1.14 \ s/cycle[/tex]
So
[tex]1.14 = 2 * 3.142 \sqrt{\frac{ 0.53}{g} }[/tex]
=> [tex]g = \frac{ 4 * 3.142^2 * 0.53}{1.14^2}[/tex]
=> [tex]g = 16.104 \ m/s[/tex]
Find the angles of the first three principal maxima above the central fringe when this grating is illuminated with 672 nm light.
Complete Question
A grating has 767 lines per centimetre.Find the angles of the first three principal maxima above the central fringe when this grating is illuminated with 672 nm light.
Answer:
The first is
[tex]\theta_1 = 2.92^o[/tex]
The second is
[tex]\theta _2 = 5.93^o[/tex]
The third is
[tex]\theta _3 = 8.92^o[/tex]
Explanation:
From the question we are told that
The number of lines per cm is [tex]I = 767 \ lines/cm = 76700 \ lines / m[/tex]
The wavelength is [tex]\lambda = 672nm = 672 *10^{-9} \ m[/tex]
Generally the condition for constructive interference is
[tex]dsin (\theta ) = n\lambda[/tex]
Here d is the distance of separation of the slit which is mathematically represented as
[tex]d = \frac{1}{l}[/tex]
[tex]d = \frac{1}{76700}[/tex]
[tex]d = 1.30*10^{-5} \ m[/tex]
So
[tex]\theta_1 = sin^{-1} [\frac{ n \lambda }{ d} ][/tex]
For the first angle n = 1
So
[tex]\theta_1 = sin^{-1} [\frac{ 1 * 672 *10^{-9} }{ 1.30 *10^{-5}} ][/tex]
[tex]\theta_1 = 2.92^o[/tex]
For the second angle n = 2
So
[tex]\theta_2 = sin^{-1} [\frac{ 2 * 672 *10^{-9} }{ 1.30 *10^{-5}} ][/tex]
[tex]\theta _2 = 5.93^o[/tex]
For the second angle n = 3
So
[tex]\theta_3 = sin^{-1} [\frac{ 3 * 672 *10^{-9} }{ 1.30 *10^{-5}} ][/tex]
[tex]\theta _3 = 8.92^o[/tex]
What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is [tex]E =248.2 \ N/C[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 12 \ cm = 0.12 \ m[/tex]
The charge is [tex]Q = 4.4 nC = 4.4 *10^{-9} \ C[/tex]
The distance from the center is [tex]k = 0.1 \ mm = 1*10^{-4} \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.12}{2}[/tex]
=> [tex]r = 0.06 \ m[/tex]
Generally electric field is mathematically represented as
[tex]E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ][/tex]
substituting values
[tex]E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ][/tex]
[tex]E =248.2 \ N/C[/tex]
4. A car increases velocity or speed from 25 m/s to 50 m/s in 5 seconds. What is the acceleration of the car ?
Answer:
5m/s²
Explanation:
v²-v¹ =50 - 25 = 25m/s = 5m/s²
t 5 5
it is better to sell used paper then to burn them
Answer:
Using sold paper is better than to burn it because burning paper may cause pollution.
Explanation:
Hope it will help ^_^
When light travels from air into water, Group of answer choices its velocity and wavelength change, but its frequency does not change.
Explanation:
When light travels from air into water, due to the change in refractive index of the two medium, refraction of light occurs.
Velocity of light in a medium is given by :
[tex]v=\dfrac{c}{n}[/tex]
c is speed of light
n is refractive index
The frequency remains the same in every medium. It means that the its velocity and wavelength change, but its frequency does not change.
A stationary police car emits a sound of frequency 1205 Hz that bounces off of a car on the highway and returns with a frequency of 1255Hz. The police car is right next to the highway, so the moving car is traveling directly toward or away from it.
1. How fast was the moving car going?
2. Was the car moving away or towards the police?
3. What frequency would the police car have received if it had been traveling toward the other car at 25.0m/s?
Answer:
Let's take the frequency receive by police officer to be
f¹/f²= u-v/u+v
So 1255/1205= 340-v/340+v
V= 6.9m/s positive value implies that the car is moving toward the police
2 to find frequency we use
f/1205= 340+25/340-25
f= 1396Hz
When 10 N force applied at 30 degrees to the end of a 20 cm handle of a wrench, it was just able to loosen the nut. What magnitude of the force would require to just loosen the nut, of the force apply perpendicularly at the end of the handle
Answer:
5 N
Explanation:
From the question,
The magnitude of the force that would be required to just loosen the nut when the force is applied perpendicularly at the end of the handle is
Fy = Fsinθ................. Equation 1
Where Fy = force acting perpendicular at the end of the handle, F = Force applied to the handle, θ = angle of inclination of the force to the end of the handle.
Given: F = 10 N, θ = 30°
Substitute these values into equation 1
Fy = 10(sn30°)
Fy = 10(0.5)
Fy = 5 N.
Two scientist did the same experiment but arrived at different results.the results would most likely
Answer:
provide new testble ideas
On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per hour and drive for an additional two hours. What is your average x component of velocity
Answer:
The average velocity is 180 km/hr
Explanation:
Given;
initial velocity, u = 60 km per hour
final velocity, v = 120 km per hour
initial time = 1 hour
final time = 2 hour
Initial position = 60 km/h x 1 hour = 60 km
final position = 120 km/h x 2 hour = 240 km
The average velocity is given by;
[tex]V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr[/tex]
Therefore, the average velocity is 180 km/hr
How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m
Answer:
F=2496 N
Explanation:
Given that,
Mass of SUV, m = 1600 kg
Initial speed, u = 0
Final speed, v = 25 m/s
Distance, d = 200 m
We need to find the net force. Firstly, let's find acceleration using equation of motion.
[tex]v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2[/tex]
Net force, F = ma
[tex]F=1600\times 1.56\\\\F=2496\ N[/tex]
So, the net force is 2496 N.
The amount of net force that will be required to accelerate a 1600kg SUV from rest to a speed of 25 m/s in a distance of 200m is 2500N
HOW TO CALCULATE NET FORCE:
The net force of a body can be calculated by multiplying its mass by acceleration. However, the acceleration of this SUV needs to be calculated using the following equation of motion:v² - u² = 2asa = v² - u²/2sWhere:
a = acceleration (m/s²)v = final velocity (m/s)u = initial velocity (m/s)s = distance (m)a = 25² - 0²/2(200)a = 625/400a = 1.563m/s²Since a = 1.563m/s²
F = 1600 × 1.563F = 2500NTherefore, the amount of net force that will be required to accelerate a 1600kg SUV from rest to a speed of 25 m/s in a distance of 200m is 2500N.Learn more at: https://brainly.com/question/18031807?referrer=searchResults
A rock is thrown horizontally from a cliff with a speed of 15 m/s. It falls half the height of the cliff in the last three seconds of its fall.
a. What is the total fall time?
b. How high is the cliff?
c. What is the horizontal distance from the cliff when it hits the ground?
Answer:
A) 10.243 s
B) 514.64 m
C) 153.645 m
Explanation:
From projectile motion;
y(t) = h - ½gt²
Where;
h is the height of cliff
t is the time taken to fall from top of cliff down to half the height of the cliff
y(t) is height of the rock as function of time
We are told the rock falls half way of the cliff height.
Thus;
y = h/2
So;
h/2 = h - ½gt²
This gives;
h - h/2 = ½gt²
h/2 = ½gt²
h = gt² - - - - (1)
Also,the total free fall time from top of the cliff to ground would be;
T = t + 3
Thus, distance at this point is zero.
So;
0 = h - ½gT²
h = ½gT²
Putting t + 3 for T, we have;
h = (1/2)g(t + 3)² - - - 2
Inspecting eq 1 and eq 2,we have;
gt² = (1/2)g(t + 3)²
g will cancel out to give;
2t² = t² + 6t + 9
t² - 6t - 9 = 0
The roots are -1.243 or 7.243.
We will pick the positive value as time can't be negative.
Thus, T = 7.243 + 3 = 10.243 s
h = gt² = 9.81 × 7.243²
h = 514.64 m
Formula for horizontal distance is given by;
x = uT
Where u is initial speed given as;
u = 15 m/s
Thus;
x = 15 × 10.243
x = 153.645 m
if you walk at 0.7 m/s, how long would it take to walk a mile which is 1609 m
Answer:
[tex]\Huge \boxed{\mathrm{2298.57 \ seconds}}[/tex]
[tex]\rule[225]{225}{2}[/tex]
Explanation:
[tex]\displaystyle \sf Speed = \frac{Distance \ covered }{Time \ taken}[/tex]
[tex]\displaystyle \sf s = \frac{d }{t}[/tex]
The speed is 0.7 m/s.
The distance covered is 1609 m.
[tex]\displaystyle \sf 0.7 = \frac{1609 }{t}[/tex]
Multiplying both sides by t.
Then dividing both sides by 0.7.
[tex]\displaystyle \sf t = \frac{1609 }{0.7}[/tex]
[tex]\sf t= 2298.57[/tex]
It would take 2298.57 seconds.
[tex]\rule[225]{225}{2}[/tex]
Answer:
[tex]\huge\boxed{\sf t = 2298.57\ secs}[/tex]
Explanation:
Given:
Speed = v = 0.7 m/s
Distance = S = 1 mile = 1609 metre
Required:
Time = t = ?
Formula :
v = S/t
Solution:
v = S/t
t = S/v
t = 1609 / 0.7
t = 2298.57 secs
Help!!! PLEASEEEEEEE
Answer:
I think the answer is D.
How many time constants must elapse if an initially charged capacitor is to discharge 90% of its stored energy through a resistor
Answer:
t = log9/10 x RC
Explanation:
Let's say initial energy is
E= 1/2Q²/C
But if 90% is discharged what will be remaining will be
E= E"/100= 1/2Q²/C
So if we divide the first by the second equation we have
Q'= Q/10
Then using
q=q0(1-e^-t /RC)
So that
Q= Q/10(1-e^-t /RC)
t/RC= log9/10
So t = log9/10 x RC
What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged particle as it travels through water (which has an index of refraction of n
Answer:
2.3*10^8m/s
Explanation:
Using
V = c/n
Where c= speed of light
n = refractive index of water
By substituting
We have
V= 3*10^8m/s/1.33
= 2.3*10^8m/s
A weightlifter holds a 1,700 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 800 N. How much work (in J) is required to lift the barbell to a height of 2 m
Answer:
W = 5000 J
Explanation:
Weight hold by a weightlifter is 1700 N, 1 meter above the ground.
One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 800 N.
We need to find the work is required to lift the barbell to a height of 2 m.
Total force is : 1700 N + 800 N = 2500 N
Work done in Physics is given by :
W = F×d
So,
W = 2500 N × 2 m
W = 5000 J
Hence, 5000 J of work is required to lift the barbell to a height of 2 m.
What is the average velocity of a car starting from 0 and travels 7 meters in 9 seconds?
Your question has been heard loud and clear.
Averge velocity formula= Total distance travelled / total time taken.
Total distance=7meters
Total time taken=9 seconds.
Average velocity = 7/9= 0.77 metres/second
Average velocity= 0.77m/s
Thank you.
what was the average speed in km/h of a car that travels 490.0 km in 4.2 h?
Answer:
BelowExplanation:
[tex]Distance = 490 \:km\\Time = 4.2 \:hours\\\\Average\:speed = \frac{Distance}{Time} \\\\A.V = \frac{490}{4.2} \\\\A.V = 116.67km/h[/tex]
Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the volume flow rate of the fluid in m^3/
Answer:
Volume flow rate = [tex]1.41 \times 10^{-3}m^3/s[/tex]
Explanation:
The volume flow rate through a channel can be gotten by multiplying its area and its velocity.
The channel under consideration is a circular channel. Hence, the cross-sectional area can be calculated by using the relation for calculating the area of a circle
Cross-sectional area = [tex]\pi \times d^{2}/4[/tex]
[tex]A = \pi \times (3\times 10^-2)^2 /4 = 7.07 \times 10 ^-4 m^2[/tex]
volume flow rate= [tex]A = 7.07 \times 10 ^-4 m^2 X 2.0 = 1.41 \times 10^{-3}m^3/s[/tex]
Volume flow rate = [tex]1.41 \times 10^{-3}m^3/s[/tex]
Would hurt if u have a baby?
Answer:
yes it would.
Explanation:
How can Newton’s laws explain why the brain is in danger of injury when there is impact to the head? A. The brain is fixed in one place inside the skull and cannot move away from the impact. B. The brain moves inside the skull with a force equal and opposite to that of the impact. C. The blood vessels of the brain expand, so more blood flows to the brain. D. The blood vessels of the brain shrink, so less blood flows to the brain.
Answer:
B. The brain moves inside the skull with a force equal and opposite to that of the impact.
Explanation:
Just took the test and got it correct.
Applying Newton's laws the brain is in danger of injury when an impact is made to our head because ; ( B ) The brain moves inside the skull with a force equal and opposite to that of the impact
According to Newton's law an object will continue to be at rest except a force ( impact ) acts on it. also the third law of Newton states that when two objects interact ( The brain and the impact to the head ) they apply forces of equal magnitude to each other but in opposite directions from one another.
The Impact to the head will have a direct effect on the brain because the brain will move in opposite direction to the direction of the impact and with the same magnitude which will lead to an injury to the brain because the brain is suppose to be at rest at all times and not in motion.
Hence we can conclude that Applying Newton's laws the brain is in danger of injury when an impact is made to our head because The brain moves inside the skull with a force equal and opposite to that of the impact.
Learn more : https://brainly.com/question/13532462
A large tanker truck carrying Blazblue extract has a capacity of 29053 liters. The density of Blazblue extract is 0.770 grams per milliliter. What is the weight, in pounds, of a full load of Blyzblah extract.
Answer:
1319.14 lb
Explanation:
From the question above,
Density of the Blazblue extract = mass of Blazblue extract/volume of Blazblue extract.
D = m/v.............. Equation 1
make m the subject of the equation
m = D×v.................... Equation 2
Given: D = 0.770 grams per millilitre, v = 29053 liters = 29053000 millilitres
Substitute these values into equation 2
m = 0.770×29053000
m = 22370810 grams
If,
1 gram = 0.00220462 pounds,
Then,
22370810 grams = (22370810×0.00220462) = 41319.14 lb
A proton at location A makes an electricfield E-->1 atlocation B. A different proton, placed atlocation B, experiences a force F--->1.
If |E-->|= 300 N/C, what is |F-->1|?
|F-->1|= N
Now the proton at B is removed and replaced by alithium nucleus, containing three protons and four neutrons. Theproton at location A remains in place.
What is the magnitude of the electric force on the lithium nucleus?
|F-->Li| = N
Now the Lithium nucleus is removed, and an electron is placed atlocation B. The proton atlocation A remains in place. What is the magnitude of the electric force on the electron?
|F-->1| = N
Answer:
Explanation:
|F-->1| = E-->1 x charge on proton
= 300 x 1.6 x 10⁻¹⁹
Magnitude of electric force on proton
|F-->1| = 480 x 10⁻¹⁹ N .
Inside lithium nucleus , there are three protons so charge on it
= 3 x 1.6 x 10⁻¹⁹ = 4.8 x 10⁻¹⁹ C .
Magnitude of electric force on lithium nucleus
|F-->Li| = E-->1 x charge on lithium nucleus
= 300 x 4.8 x 10⁻¹⁹
= 1440 x 10⁻¹⁹ N .
Electron has same charge as that possessed by proton so charge on electron = 1.6 x 10⁻¹⁹ C
Magnitude of force on electron
|F-->1| = E-->1 x charge on electron
= 300 x 1.6 x 10⁻¹⁹
= 480 x 10⁻¹⁹ N .