Hi there!
Begin by using Gauss' Law to find the electric field.
[tex]\oint {E \cdot} \, dA = \frac{Q_{encl}}{\epsilon_0}[/tex]
E = Electric field (N/C)
dA = differential area element
Q = enclosed charge (C)
ε₀ = Permittivity of free space (8.85 * 10⁻¹² C²/Nm²)
We can construct a large cylinder around the wire in order to determine the electric flux. The electric field lines will pass through the LATERAL surface area of the cylinder, so:
[tex]A = 2\pi rL[/tex]
Where 'L' is the length of the cylinder and 'r' is the distance from the capacitor.
The enclosed charge is equivalent to the charge per meter length (λ) multiplied by the length, so:
[tex]\oint {E \cdot} \, dA = \frac{\lambda L}{\epsilon_0}[/tex]
We can rewrite the dot product as EA (where cosθ = 1 since the normal vector points in the direction of the field).
A = the lateral surface area of a cylinder, so:
[tex]E * 2\pi rL = \frac{\lambda L}{\epsilon_0}[/tex]
Rearrange to solve for 'E'.
[tex]E = \frac{\lambda L }{2\pi r L \epsilon _0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}[/tex]
a)
Plug in the distance into 'r'.
[tex]E = \frac{\lambda }{2\pi r \epsilon_0} \\\\E = \frac{0.0000072}{2\pi * 5.5 * (8.85 * 10^{-12})} = \boxed{23542.18 \frac{N}{C}}[/tex]
b)
Repeat:
[tex]E = \frac{\lambda }{2\pi r \epsilon_0}\\\\E = \frac{0.0000072}{2\pi * 2.5 * (8.85 * 10^{-12})} = \boxed{51792.8 \frac{N}{C}}[/tex]
We can see that the distance from the wire is INVERSELY related to the electric field strength by a power of r⁻¹. The field strength DECREASES as the distance INCREASES.
There are two space ships traveling next to each other. The first one is 500
Kg and the second one is 498 Kg. Since they are 35 meters apart, what is
the force of gravity between the two space ships?
This question involves the concept of Newton's law of gravitation.
The force of gravity between the two spaceships is "1355.78 N".
Newton's Law Of GravitationAccording to Newton's Law of Gravitation:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
where,
F = force of gravity between ships = ?G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²m₁ = mass of first ship = 500 kgm₂ = mass of second ship = 498 kgr = distance between ships = 35 mTherefore,
[tex]F=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(500\ kg)(498\ kg)}{(35\ m)^2}\\\\[/tex]
F = 1355.78 N = 1.356 KN
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Two space ships traveling next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.
It is given that the First spaceship's weight ([tex]m_{1}[/tex]) is 500 kg,
The second spaceship's weight ([tex]\rm m_{2}[/tex]) is 498 kg.
The distance between spaceships (r) is 35 meters.
It is required to find the Force of gravity between these spaceships.
What is Gravitational force?It is defined as the force which attracts any two masses in the universe.
By Newton's law of Gravitation:
[tex]\rm F= \frac{Gm_1m_2}{r^2}[/tex] , Where
[tex]\rm F = The\ force \ of \ gravity \ between \ the \ spaceships\\\rm G= Universal\ Gravitational \ Constant = 6.67 \times 10^{-11} N.m^2/kg^2[/tex]
Putting values in the above formula:
[tex]\rm F = \frac{(6.67\times 10^{-11} N.m^2/kg^2)(500kg)(498kg)}{(35m)^2}[/tex]
F = 1355.78 N = 1.356 KN
Thus, Two spaceships travel next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.
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How can an athlete participating in a 40m sprint modify and improve their performance based on the kinematic variable of speed and acceleration?
The athlete can improve performance by building strength, coordination and balance.
Who is an Athlete?This is an individual who is proficient in sports and other forms of physical exercise.
Improvement of performance based on the kinematic variable of speed and acceleration can be achieved by building strength, coordination and balance by performing plyometric exercises etc.
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can somebody please help
The amplitude of the wave on the given sinusoidal wave graph is 10 cm.
What is amplitude of wave?
The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.
Amplitude of the wave is calculated as follows;
From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.
Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.
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If a 40 N block is resting on a rough horizontal table with a
coefficient of static friction p=0.60, then what is the force of
static friction acting on the block?
Based in the relationship between static frictional force and coefficient of static friction, the force of static friction is 24 N.
What is friction?Friction is a force that opposes the relative motion of an object moving over another at their surafce of contact.
Frictional force is constant for each type of material. This constant is known as coefficient of friction.
The coefficient of friction is given as follows:
Coefficient of friction = Frictional force/normal reactionFrom the data given:
coefficient of static friction p = 0.60Weight of block = 40 NForce of static friction = 0.60 × 40
Force of static friction = 24 N.
Therefore, the force of static friction is 24 N.
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1. What is the distance covered by a T-Rex that goes from 0 m/s to 9 m/s in 6.78 seconds? (10
points)
With the use of first and third equation of motion, the distance covered by a T-Rex is 30.51 m
Linear MotionWhen a body is in linear motion, the body is moving in a straight line. some of the parameters to consider are:
Distance coveredSpeedVelocityAccelerationE.T.CGiven that a T-Rex move from 0 m/s to 9 m/s in 6.78 seconds, the distance covered can be found by calculating the acceleration.
Let us use equation 1
V = U + at
9 = 0 + 6.78a
a = 9 / 6.78
a = 1.33 m/[tex]s^{2}[/tex]
Now let us use equation 3
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
[tex]9^{2}[/tex] = 2 x 1.33 x S
81 = 2.655S
S = 81/2.655
S = 30.51 m
Therefore, the distance covered by a T-Rex is 30.51 m.
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The weight of an object on the moon is less because the _______ on the moon is less.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight
Answer:
acceleration of the gravity
Explanation:
The weight of an object on the moon is less because the acceleration of the gravity on the moon is less.
The amount of energy released when 45 g of -175° C steam is cooled to 90° C is
A. 101,700 J
B. 317,781 J
C. 419,481 J
D. 417,600 J
Answer:
The answer should be choice B.
A curve is banked at an angle of 29.1 degrees above the horizontal and the road surface has a coefficient of static friction of 0.4. What must the radius of curvature be for the safe minimum speed of 27.1 m/s?
Hi there!
We can begin by summing the forces acting on the car.
Along the axis of the incline, the forces of friction and gravity are working. The force of friction points towards the top of the ramp, while the force of gravity works towards the bottom.
We can use trigonometry to determine the force due to gravity along the ramp.
[tex]F_g = Mg sin\theta[/tex]
The force due to friction is equal to:
[tex]F_f = \mu N[/tex]
The normal force is the vertical component of the force due to gravity, so:
[tex]F_f = \mu Mgcos\theta[/tex]
Now, the combination of these two forces produces a component of the centripetal force. Drawing a diagram, we see that the true centripetal force is the HYPOTENUSE, while these forces sum up to its horizontal component along the ramp.
Therefore:
[tex]F_c sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]
The centripetal force is equivalent to:
[tex]F_c = \frac{Mv^2}{r}[/tex]
m = mass (kg)
v = velocity (m/s)
r = radius (m)
Rewrite:
[tex]\frac{Mv^2}{r}sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]
Cancel out 'M'.
[tex]\frac{v^2}{r}sin\theta = gsin\theta - \mu gcos\theta[/tex]
Rearrange to solve for 'r'.
[tex]r = \frac{v^2sin\theta}{gsin\theta - \mu gcos\theta}[/tex]
Plug in values and solve.
[tex]r = \frac{(27.1^2)sin(29.1)}{(9.8)sin(29.1) - 0.4(9.8)cos(29.1)} = \boxed{266.365 m}[/tex]
Need help with the following - question 2
Answer:
Find the change in momentum of the upper stage, that is:
∆p = m(vf - vi)
m being the mass of the upper stage
vf being the final velocity which was given
vi being the initial which was also given
find ∆p
then use ∆p in the same equation
∆p being the answer you got above
m being the mass of the lower stage (given)
vi being the initial velocity (given)
vf being the final velocity of the lower stage which you were asked to find
Explanation:
During a collision the change in momentum (∆p) for both objects is equal regardless of their speeds or masses before or after the equation
What is the x component of a vector that is defined as
45m at -35°?
the x- component of the vector is 36.86 m.
What is a vector?Vectors are quantities that have both magnitude and direcion
To calculate the x-component of the vector, we use the formula below.
Formula:
dx = dcosθ.......... Equation 1Where;
dx = x-component of the vectord = vector between the x-y componentθ = Angle of the vector to the horizontal.From the question,
Given:
d = 45 mθ = -35°Substitute these values into equation 1
dx = 45cos(-35°)dx = 45×0.918dx = 36.86 m.Hence, the x- component of the vector is 36.86 m.
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The resultant of two vectors is of magnitude 3 units and 4 units is 1 units, what is the value of their dot product?
Answer:
A dot B = C is the vector equation for this expression
A · B = A B cos θ
3 * 4 cos θ = 1 the value 1 is their dot product
cos θ = 1 / 12 = .083 θ = 85.2 deg
A grasshopper jumps at a 63.0 angle with an initial velocity of 4.22 m/s. how far away does it land
The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.
To calculate the distance of the landing point of the grasshopper to the point where its jumps, we use the formula of range.
What is horizontal range?Range can be defined as the horizontal distance between the point of projection to the point where the projectile hit the plain again.
R = u²sin2∅/g........... Equation 1Where:
R = Distance between the point of jump and the point at which it landsu = initial velocity∅ = angleg = acceleration due to gravityFrom the question,
Given:
u = 4.22 m/s∅ = 63°g = 9.8 m/s²Substitute these values into equation 1
R = (4.22)²sin(2×63)/9.8R = 1.47 mHence, The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.
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A train moves from rest to a speed of 22 m/s in 34.0 seconds. What is its acceleration?
Answer: a = 0.647 m/s^2
Explanation:
Acceleration = change in speed / time → a = 22 / 34 → a = .647 m/s^2
cornvet 500000grams in short form of using suitable prefix.
Answer:
0.5 mega grams
Explanation:
A source of light emits photons with a wavelength of 8.1 x 10-8 meters. What is the frequency of this light
Answer:
Explanation:
Speed of light v = 3 x 10⁸ m/s
wavelength λ = 8.1 x 10⁻⁸ m
frquency f = v/λ = 3.7 x 10¹⁵ Hz
If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
C = λν
As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,
The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸
=3.7 × 10¹⁵ Hz
Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz
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Read the text below. Each sentence is about one, two or no energy at all. (5 points) Name the type (s) of energy for each sentence, or leave the space blank (if in the sentence no energy is mentioned). Artan decided to paint the house. He moved the furniture, climbed the stairs, and began work. After two hours he took a break, ate lunch and turned on the radio to listen to some music. When done, turn on a heater to allow the paint to dry as quickly as possible. At dinner everything had ended. a) ............................................................................................................................................ b) ............................................................................................................................................ c) ............................................................................................................................................ d) ............................................................................................................................................ e) ............................................................................................................................................
A professional golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00410 s. After the collision, the ball leaves the club at a speed of 32.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?
The magnitude of the average force exerted on the ball by the club is 429.27 N.
What is force?Force can be defined as the product of mass and acceleration
To calculate the force exerted on the ball by the club, we use the formula below.
Formula:
F = m(v-u)/t............ Equation 1Where:
F = Force exerted on the ballm = mass of the ballv = Final velocityu = initial velocityt = timeFrom the question,
Given:
m = 55 g = 0.055 kgu = 0 m/sv = 32 m/st = 0.0041 sSubstitute these values into equation 1
F = 0.055(32-0)/0.0041F = 429.27 NHence, The magnitude of the average force exerted on the ball by the club is 429.27 N.
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5 waves with a length of 4m hit the shore every 2 seconds, what is the frequency?
The frequency of the 5 waves with a length of 4m hit the shore every 2 seconds is 2.5 Hz.
What is frequency?This is the number of cycles completed by a wave in one second. The s.i
unit of frequency is Hert (Hz).
From the question, to calculate the frequency of 5 waves with length of 4 m that hit the shores every 2 seconds, we use the formula below.
Formula:
F = n/t........... Equation 1Where:
n = Number of waveF = Frequencyt = timeFrom the question,
Given:
n = 5 waves t = 2 secondsSubstitute these values into equation 1
F = 5/2F = 2.5 Hz.Hence, The frequency of the wave is 2.5 Hz.
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define capacitance of a capacitor and state its SIbunit
Answer: what I know that the unit is farad
Explanation:
An object of mass m is oscillating with a period T. The position x of the object as a function of time is given by the equation x(t)=Acosωt . The maximum net force exerted on the object while it is oscillating has a magnitude F. Which of the following expressions is correct for the maximum speed of the object during its motion?
What the equation given?
x(t)=Acos[tex]\omega[/tex]tMaximum velocity occurs at the equilibrium position
So
x=0Now
x(0)=Acos0[/tex]x(0)=ANow
As we know the formula
[tex]\\ \rm\rightarrowtail V_max=A\omega[/tex]
These expressions can be used
The maximum speed of the oscillating object will be given by [tex]V_{max}=Aw[/tex]
What is oscillation?An oscillation is defined as the repitative periodic motion of any object about its mean or equilibrium position.
The given equation is as follows:
x(t)=Acost
Maximum velocity occurs at the equilibrium position x=0
x(0)=Acos0
x(0)=A
Hence the maximum velocity will be [tex]V_{max}=Aw[/tex]
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The force of a hammer drives a nail into wood. This is an example of?
A. An unbalanced force.
B. Gravitational force.
C. Friction.
D. Balanced forces.
Answer:
A. An unbalanced force.
Explanation:
There needs to be a net force in order for the nail to be driven into presumably the wall. Without the net force then the hammer and nail wouldn't move.
A school bus and a small Toyota Prius collide in a head-on collision
Which vehicle pushed with a greater force? Explain
Which vehicle has a greater mass?
Which vehicle has a greater acceleration after the moment of impact? Explain why
Hi there!
Part 1:
According to Newton's Third Law, every action has an EQUAL and OPPOSITE reaction.
For a collision, the objects involved exert an EQUAL and OPPOSITE impulse on each other, which means the forces exerted are equal as well.
Therefore, the school bus and Toyota Prius exert an EQUAL FORCE on each other.
Part 2:
The school bus has a greater mass.
Part 3:
Using Newton's Second Law:
[tex]\Sigma F = ma[/tex]
Using this relationship, the smaller the mass, the larger the acceleration. Since the forces are EQUAL, the Toyota Prius will experience a larger acceleration because it has a smaller mass.
At an altitude of 1.3x10^7 m above the surface of the earth an incoming meteor mass of 1x10^6 kg has a speed of 6.5x10^3 m/s. What would be the speed just before impact with the surface of earth?Ignore air resistance.
Show all steps.
Answer:
Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.
Explanation:
Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].
Let [tex]m[/tex] denote the mass of the meteor.
Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.
Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.
Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.
The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:
At an altitude of [tex]1.3 \times 10^{7}\; {\rm m}[/tex] about the surface of the earth, the meteor would be approximately [tex]r_{0} \approx 6.371 \times 10^{6}\; {\rm m} + 1.3 \times 10^{7}\; {\rm m} \approx 1.9 \times 10^{7}\; {\rm m}[/tex] away from the surface of planet earth. The meteor would be only [tex]r_{1} \approx 6.371 \times 10^{6}\; {\rm m}[/tex] away from the center of planet earth just before landing.Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.
During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.
Initial [tex]\text{KE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].
Initial [tex]\text{GPE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].
(Note the negative sign in front of the fraction.)
Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:
[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].
[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:
[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].
[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].
Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:
[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].
Substitute in the values and evaluate:
[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].
(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.
a 1. You found that the MCB was tied with a thread and the thread was fixed with a nail on the wall in your friend's house. i. Is it good idea to do this? ii. What could be the possible hazard of this? iii. What should have done to keep the circuit safe?
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
What is a miniature circuit breaker?A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.
This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
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To travel at a constant speed, a car engine provides 24 KW of useful power. The driving force on the car is 600 N. At what speed does it travel?
[tex]\text{Given that,}\\ \\\text{Power,} ~P=24~ KW = 24000~ W\\\\\text{Force,} ~F=600~N\\\\\text{We know that,}\\\\P=Fv\\\\\implies v = \dfrac{P}{F} = \dfrac{24000}{600}=40~ ms^{-1}\\\\\\\text{It travels at 40 m/s}[/tex]
how r u
________________.
A wire is attached to the ceiling so that the current flows south to north. A student is standing directly below the wire facing north. What is the direction of the B-field (caused by the current in the wire) at this observation point
Answer:
If one wraps the fingers around the wire and points the thumb in the direction of the "conventional" current the fingers will point towards the North pole - the direction of the B-field.
In this case the B-field is pointed "West".
a 2.99 kg sphere makes a perfectly inelastic collision with a second sphere that is intially at rest. the composite moves with a speed equal to one third the original speed of the 2.99kg. what is the mass of the second sphere?
Answer:
5.98 kg
Explanation:
To solve this problem, let use the Linear Momentum Conservation Law:
Before collision: [tex]\sum p=p_{1}+p_{2}=m_{1}v_{1}+m_{2}v_{2}=(2.99v_{1})+0=2.99v_{1}[/tex]
After collision: [tex]\sum p'=p_{1}'+p_{2}'=(2.99+m_{2})(v_{1}/3)[/tex]
So, we obtain:
[tex]\sum p =\sum p' \rightarrow 2.99v_{1}=(2.99+m_{2})(v_{1}/3) \rightarrow 8.97 = 2.99 + m_{2}[/tex]
[tex]m_{2}=8.97-2.99=5.98 kg[/tex]
An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?
Answer:
See below.
Explanation:
According to the question, we know that,
work done is given by, [tex]W=qV[/tex]
and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]
therefore equating both the equations we get,
[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]
m= mass of electron = [tex]9.1*10^{-31} kg[/tex]
q= charge on an electron = [tex]1.6*10^{-19} C[/tex]
v= speed of electron= 700000m/s
substituting the values in the above equation, we get
[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]
(1). the potential difference that stopped the electron is 1.39 volts.
now the kinetic energy equation is : 2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]
or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]
(2). the initial kinetic energy of the electron is 1.39eV.
Ways in which a teacher plays a role in the literacy development of the learners
Answer:
encourage all attempts at reading, writing, and speaking
Explanation: