Answer:
462000J
Explanation:
Quantity of heat= mass x specific heat capacity of iron x change in temp
specific heat capacity of iron is 462J/Kg/K
change in temp = 250-50= 200°C
200°C is equivalent to 200K since 1°C is 1K
Q= mct
= 5x462x200
= 462000J
Suppose that the sound level of a conversation is initially at an angry 71 dB and then drops to a soothing 54 dB. Assuming that the frequency of the sound is 504 Hz, determine the (a) initial and (b) final sound intensities and the (c) initial and (d) final sound wave amplitudes. Assume the speed of sound is 346 m/s and the air density is 1.21 kg/m3.
Answer:
Explanation:
In the decibel scale , intensity of sound changes logarithmically as follows
[tex]10log\frac{I}{I_0} =[/tex] Value in decibel scale , the value of I₀ = 10⁻¹² W /m².
Putting the values
[tex]10log\frac{I}{10^{-12}} = 71[/tex]
[tex]log\frac{I}{10^{-12}} = 7.1[/tex]
[tex]\frac{I}{10^{-12}} = 10^{7.1}[/tex]
[tex]I= 10^{-4.9}[/tex] W/m²
Similarly for 54 dB sound intensity can be given as follows
I = 10⁻¹² x [tex]10^{5.4}[/tex]
[tex]I= 10^{-6.6 }[/tex] W / m²
For intensity of sound the relation is as follows
I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .
Putting the given values for 71 dB
[tex]I= 10^{-4.9}[/tex] = 2π² x 504²xA²x 1.21 x 346
A² = 60.03 x 10⁻¹⁶
A = 7.74 x 10⁻⁸ m
For 54 dB sound
[tex]10^{-6.6}[/tex] = 2π² x 504²xA²x 1.21 x 346
A² = 1.1978 x 10⁻¹⁶
A = 1.1 x 10⁻⁸ m
What is the ball's acceleration?
Answer:
g, downward
Explanation:
It is given that, a baseball is thrown straight upward. The force acting on the stone is force of gravity. It is moving under the action of gravity. We know that the force of gravity always acts in a downward direction.
At the highest point, the velocity of the stone will be equal to 0. It will move will constant acceleration equal to g and it always acts in downward direction.
Hence, the correct option is (e) "downward direction".
(a) In electron-volts, how much work does an ideal battery with a 17.0 V emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If 3.88 × 1018 electrons pass through each second, what is the power of the battery?
Answer:
(a) 17.0eV
(b) 10.55W
Explanation:
(a) The amount of work done (W) on an electron by an ideal battery of emf value of V as it moves from the positive to the negative terminal is given by;
W = q x V --------(i)
Where;
q = charge on the electron = 1e
From the question;
V = 17.0 V
Substitute the values of q and V into equation (i) as follows;
W = 1e x 17.0
W = 17.0eV
Therefore, the work done in electron volts is 17.0
(b) The power (P) of the battery as some electrons (n) pass through it at time t, is given as;
P = (n q V) / t --------------(ii)
Where;
n = number of electrons = 3.88 x 10¹⁸
t = 1s
q = 1.6 x 10⁻¹⁹C
V = 17.0V
Substitute these values into equation (ii) as follows;
P = (3.88 x 10¹⁸ x 1.6 x 10⁻¹⁹ x 17.0) / 1
P = 10.55W
Therefore the power of the battery is 10.55W
we can catch a rolling ball but not a flying bullet?? give reason
Answer:
yeah this statement is tru
Explanation:
it is because the speed of the bullet is more than the speed of rolling ball .so from this reason we cannot catch a bullet.
A bullet will be moving much faster than a rolling ball. Even seeing a bullet in flight requires it to be extremely low velocity.
Determine whether or not each of the following statements is true. If a statement is true, prove it. If a statement is false, provide a counterexample and explain how it constitutes a counter-example.
A capacitor consists of two flat, metal plates with unequal areas. Each of the plates starts neutral, and then each plate is connected to a dierent terminal of a battery. After some time, the plates will have excess charge on them, and the magnitude of the excess charge on one plate will equal the magnitude of the excess charge on the other plate.
A. True
B. False
If a wire carries current, then it has a net non-zero charge in it.
A. True
B. False
Answer:
a) True b) True
Explanation:
a) a capacitor is made up of two flat plates and each one has a charge of the same sign, therefore the statement is true
b) the current is the flow of electrons per unit of time, therefore the charge is not zero, therefore the statement is True
An arrow is launched from P with a speed Vi = 25m / s. Knowing that the target Q is 10 m high, and the arrow reaches it as shown in the figure, we are asked to determine the distance X. (g = 10m / s2).
Answer:
20 m
Explanation:
Given in the y direction:
Δy = 10 m
v₀ = 25 m/s sin 37° = 15.0 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
10 m = (15.0 m/s) t + ½ (-10 m/s²) t²
10 = 15t − 5t²
2 = 3t − t²
t² − 3t + 2 = 0
(t − 1) (t − 2) = 0
t = 1 or 2
Since the projectile reaches Q before it reaches the peak, we want the lesser time, so t = 1.
Given in the x direction:
v₀ = 25 m/s cos 37° = 20.0 m/s
a = 0 m/s²
t = 1 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20.0 m/s) (1 s) + ½ (0 m/s²) (1 s)²
Δx = 20 m
Choose the correct answer/s
refer to the attachment for the explanation
Radar uses radio waves of a wavelength of 2.5 m . The time interval for one radiation pulse is 100 times larger than the time of one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect
Answer:
The minimum distance to the object that the radar can detect is 124.995 m
Explanation:
Here, we are to calculate the shortest distance to an object the radar in the question can detect.
Mathematically;
v = c/λ
Where v is the frequency, c is the speed of light and λ is the wave length
Thus;
v = (3 * 10^8)/2.5 = 1.2 * 10^8 Hz
Mathematically, the time period
t = 1/v = 1/(1.2 * 10^8) = 0.000000008333 = 8.333 * 10^-9 sec
From the question, we are told that the transmitting time is 100 times a single oscillation
Transmitting time = 100 * one oscillation
Hence Transmitting time = 100 * 8.33 * 10^-9 = 8.33 * 10-7
Mathematically;
Minimum distance =( Transmitting time * speed of light)/2 =
(8.33 * 10^-7 * 3 * 10^8)/2 = 124.995 m
Runner 1 has a velocity of 10 m/s west. Runner 2 has a velocity of 7 m/s east. From the frame of reference of runner 2, what is the velocity of runner 1? A.17 m/s east. B.3 m/s east. C.17 m/s west. D.3 m/s west
Answer:
17 m/s west
Explanation:
Runner 1 has velocity = 10 m/s west
runner 2 has velocity = 7 m/s east
From the frame of reference of runner 2, we can imagine runner 2 as standing still, and runner 1 moving away from him, towards the west with their combined velocity of
velocity = 10 m/s + 7 m/s = 17 m/s west
Answer:
17 m/s west
Explanation:
Hope this helps!
Q. A train accelerates from 36 km/h to 54 km/h in 10 sec. (i) Acceleration (ii) The distance travelled by car.
u=10m/s
v=15m/s
acceleration=
v_u/ t
5/10
0.5
define the term 'one solar mass.'
Answer:
the mass of one sun
Explanation:
its really heavy
A duck is 12 m from the edge of a pond. A student stands in the middle of the pond
and creates ripples that travel past the duck and towards the edge of the pond. The
ripples are produced uniformly at 2 ripples per second. The student determines that
the ripples take 3.0 seconds after they pass the duck to reach the edge of the pond.
Determine the wavelength of the ripples.
Answer:
The wavelength is 2 meters
Explanation:
The relationship between the frequency, the speed and the wavelength is given by the relation;
v = f × λ
The given parameters are;
The distance of the duck from the edge of the pond = 12 m
The number of ripples produced per second = Frequency, f = 2 Hz
The time it takes the ripple to reach the edge of the pond after travelling past the duck = 3 seconds
Therefore, speed of the wave, v = Distance/time = 12 m/(3 s) = 4 m/s
The wavelength, λ, is therefore;
λ = v/f = (4 m/s)/(2 Hz) = 2 meters.
4.
An "extreme" pogo stick utilizes a spring whose uncompressed length is 46 cm and whose force constant is 1.4 x 104 N/m. A 60-kg person is jumping on the pogo stick,
compressing the spring to a length of only 5.0 cm at the bottom of their jump. Which is the upward acceleration of the person at the moment the spring reaches its greatest
compression at the bottom of their jump?
6 m 2
Answer:
a = 85.9 m / s²
Explanation:
For this exercise we can use Newton's second law in the most compressed part
F - W = m a
force is the spring elastic force
F = - k Δx
k Δx - m g = m a
a = k/m Δx - g
Δx = x₀ -[tex]x_{f}[/tex]
ΔX = 46 - 5 = 41cm (1m / 100cm) = 0.41 m
let's calculate
a = 1.4 10⁴/60 0.41 - 9.8
a = 85.9 m / s²
Define centre of gravity and centre of Bouyancy
Answer:
Center of Gravity is the point in a body where the gravitational force may be taken to act. Center of Buoyancy is the center of gravity for the volume of water which a hull displaces.
A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?
Answer:
ugmd = 1/2 kx²
d = (1/2 kx²) / (ugm)
= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)
= 7.4 m
ugmd = 1/2 mv²
v = √2ugd
= √(2(0.23)(9.81 m/s²)(7.4 m)
= 5.8 m/s
Explanation:
The box's launch speed is 5.8 m/s.
What is energy?Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
Given in question spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor,
ugmd = 1/2 kx²
d = (1/2 kx²) / (ugm)
= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)
= 7.4 m
ugmd = 1/2 mv²
v = √2ugd
= √(2(0.23)(9.81 m/s²)(7.4 m)
= 5.8 m/s
The box's launch speed is 5.8 m/s.
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3.) [15 points] A physics teacher is on the west side of a small lake and wants to swim across and up at a point directly across from his starting point. He notices that there is a current in the lake and
that a leaf floating by him travels 4.2m [S] In 5.0s. He is able to swim 1.9 m/s in calm water,
(a) What direction will he have to swim in order to arrive at a point directly across from his position?
Answer:
The teacher should swim in a direction 29.24° North of East
Explanation:
Given that the there is a water current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;
The speed of the water current is given by the speed of the floating leaf traveling with the water current
Distance traveled by the leaf = 4.2 m South
Time of travel of the leaf = 5.0 s
Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current
Swimming peed of the teacher, v = 1.9 m/s
To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.
Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim
Therefore;
1.9 × sin(θ) = 0.84
sin(θ) = 0.84/1.9 = 0.44
θ = 26.24°
That is the teacher should swim in a direction 29.24° North of East.
To cross the lake the teacher has to swim in a direction 29.24° North of the East
Finding the direction of speed required:
The speed of the water current can be derived from the speed of the floating leaf :
The distance traveled by the leaf L = 4.2 m South
Time taken T = 5s
So, the speed of the leaf is:
u = 4.2/5
u = 0.84 m/s South
So, the speed of the current is 0.84 m/s South
Now, it is given that the speed of the teacher is, v = 1.9 m/s East
To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.
Let, the speed of the teacher makes an angle of θ from the EAST.
So, the North component is given by:
v(north) = vsinθ
it must be equal to the speed of the current:
vsinθ = u
1.9 × sinθ = 0.84
sinθ = 0.84/1.9
sinθ = 0.44
θ = 26.24°
The teacher should swim in a direction 29.24° North of East.
Learn more about vector components:
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Earthquakes at fault lines in Earth's crust create seismic waves, which are longitudinal (P-waves) or transverse (S-waves). The P-waves have a speed of about 9 km/s. Estimate the average bulk modulus of Earth's crust given that the density of rock is about 2500 kg/m3
Answer:
[tex]B=2.025\times 10^{11}\ Pa[/tex]
Explanation:
It is given that,
Speed of P- waves, v = 9 km/s = 9000 m/s
The density of rock is about [tex]2500\ kg/m^3[/tex]
We need to find the average bulk modulus of Earth's crust. Let it is given by B. So,
[tex]v=\sqrt{\dfrac{B}{d}} \\\\B=v^2d\\\\B=(9000)^2\times 2500\\\\B=2.025\times 10^{11}\ Pa[/tex]
So, the bulk modulus of the Earth's crust is [tex]2.025\times 10^{11}\ Pa[/tex].
Unpolarized light, with an intensity of I0, is incident on an ideal polarizer. A second ideal polarizer is immediately behind the first and its axis of polarization is oriented at an angle of 60° relative to the first polarizer’s. How much of the light will be transmitted through the system?
Answer:
The light transmitted through the system will be 0.125*I₀.
Explanation:
The light transmitted through the system can be found using Malus Law:
[tex] I = I_{0}cos^{2}(\theta) [/tex] (1)
Where:
I: is the intensity of the light transmitted
I₀: is the initial intensity
θ is the angle relative to the first polarizer’s = 60°
Because the light transmitted by the first polarizer is dropped by half, the equation (1) results as:
[tex] I = \frac{I_{0}}{2}cos^{2}(\theta) [/tex]
[tex] I = \frac{I_{0}}{2}cos^{2}(60) [/tex]
[tex] I = 0.125I_{0} = \frac{1}{8}I_{0} [/tex]
Therefore, the light transmitted through the system will be 0.125*I₀.
I hope it helps you!
(blank)a0 is undeniably accepted by scientists all over the world as the primary language of science.
Answer:
English.
Explanation:
English is undeniably accepted by scientists all over the world as the primary language of science. It is a language spoken all over the world and is also the language of science because it is an official language and can be understood by almost everyone.
Answer:
[tex]\boxed{\mathrm{English}}[/tex]
Explanation:
English is undeniably accepted by scientists all over the world as the primary language of science. The scientific methods and experiments are done by communicating and writing in English. Scientists use English to communicate and share knowledge as English is understood by people from around the world.
When we double the distance between a source of light and the
surface on which it falls, the amount of illumination on the surface
decreases to
(what fraction) of the
original illumination.
The amount of lighting on a surface drops to (1/4) of the initial illumination when the distance between a light source and the surface it falls on is doubled.
What is illumination?The amount of light or luminous flux that falls on a surface is known as illumination. It is expressed as lux or lumens per square meter.
The square of the distance has an inverse relationship with the light intensity;
[tex]\rm I = \frac{1}{r^2}[/tex]
Where,
I is the light intensity
r is the distance
Let r is the distance and I is the sound intensity for case 2;
r' = 2r
[tex]\rm I' = \frac{1}{(2r)^2} \\\\ I' = \frac{1}{4r^2} \\\\ I' = \frac{I}{4}[/tex]
When we double the distance between a source of light and the surface on which it falls, the amount of illumination on the surface decreases to(1/2) of the original illumination.
Hence the value of the fraction is 1/4.
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The plates of a parallel-plate capacitor are maintained with a constant voltage by a battery as they are pushed together, without touching. How is the amount of charge on the plates affected during this process
Answer:
The amount of charge increases.
Explanation:
For a capacitor that has its plates still connected to a battery, the voltage on the capacitor remains constant. If the distance between the two plates of this capacitor is decreased, the capacitance will increase, and then the charge on the plate will increase in accordance with the relationship
Q =CV
From this relationship, if V is held constant, then Q has to increase for C to increase also.
A train travels 120 km at a speed of 60 km/h, makes a stop for 0.5 h, and then travels the next 180 km at a speed of 90 km/h. What is the average speed of the train for this trip? 9th grade level pls FFFFFFFFFFFAAAAAAAAAAASSSSSSSSSSSSSTTTTTTTTTTTTT
Answer:
average speed = 66.67 km/h
Explanation:
In order to find the average speed of the train, you need to calculate the total distance traveled, divided by the time it took to cover that distance. So for the total distance:
Distance= 120 km + 180 km = 300 km
For the total time we need to add three different quantities, two of which we need to derived based on the information provided:
time for first part of the trip:
[tex]time_1=\frac{D_1}{v_1} =\frac{120}{60} \,h= 2\,h[/tex]
for the time of the stop:
[tex]time_2=0.5\,\,h[/tex]
for the last part of the trip:
[tex]time_3=\frac{180}{90} \,h= 2 \,\,h[/tex]
Which gives a total of 4.5 hours
Then, the average speed is: 300/4.5 km/h = 66.67 km/h
The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is the kinetic energy of the protons when they are ejected from the cyclotron
Answer:
1.92 x 10⁻¹²J
Explanation:
The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e
F = ma
Where;
a = [tex]\frac{v^2}{r}[/tex] [v = linear velocity, r = radius of circular path]
=> F = m[tex]\frac{v^2}{r}[/tex] ------------(i)
We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;
F = q v B sin θ ----------(ii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = the angle between the velocity and the magnetic field.
Combining equations (i) and (ii) gives
m[tex]\frac{v^2}{r}[/tex] = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]
=> m[tex]\frac{v^2}{r}[/tex] = q v B sin 90°
=> m[tex]\frac{v^2}{r}[/tex] = q v B
Divide both side by v;
=> m[tex]\frac{v}{r}[/tex] = qB
Make v subject of the formula
v = [tex]\frac{qBr}{m}[/tex]
From the question;
B = 1.25T
m = mass of proton = 1.67 x 10⁻²⁷kg
r = 0.40m
q = charge of a proton = 1.6 x 10⁻¹⁹C
Substitute these values into equation(iii) as follows;
v = [tex]\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}[/tex]
v = 4.79 x 10⁷m/s
Now, the kinetic energy, K, is given by;
K = [tex]\frac{1}{2}[/tex]mv²
m = mass of proton
v = velocity of the proton as calculated above
K = [tex]\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )[/tex]
K = 1.92 x 10⁻¹²J
The kinetic energy is 1.92 x 10⁻¹²J
Suppose a 50.0 g block of silver (specific heat = 0.2350 J/g·°C) at 100.°C is placed in contact with a 50.0 g block of iron (specific heat = 0.4494 J/g·°C) at 0.00°C, and the two blocks are insulated from the rest of the universe. The final temperature of the two blocks will be:
Answer:
34.34 °C
Explanation:
From the question,
Heat lost by the silver block = heat gained by the iron block.
cm(x-y) = c'm'(y-z)................... Equation 1
Where c = specific heat capacity of the silver block, m = mass of the silver block, c' = specific heat capacity of the iron, m' = mass of the iron. x = initial temperature of the silver block, z = initial temperature of the iron, y = final temperature of the mixture.
make y the subject of the equation
y = (cmx+c'm'z)/(cm+c'm')............... Equation 2
Given: c = 50 g, c = 0.2350 J/g·°C, x = 100°C, m' = 50 g, c' = 0.4494 J/g.°C, z = 0°C
Substitute these values into equation 2
y = [(50×0.2350×100)+(50×0.4494×0)]/[(50×0.2350)+(50+0.4494)]
y = 1175/(11.75+22.47)
y = 1175/34.22
y = 34.34 °C
car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car makes it one-quarter of the way around the circle before it skids off the track.From these data, determine the coefficient of static friction between the car and the track
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
Find the speed of the mobile after 4 seconds
Answer:
27 m/s
Explanation:
V_final = v_initial + a*t
V_final = 15 + 3 * 4
V_final = 27 m/s
You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a battery. The first piece has a length L and cross-sectional area A. The second is twice as long as the first, but has the same thickness. The third is the same length as the first, but has twice the cross-sectional area. Rank the wires in order of which carries the most current (has the lowest resistance) when connected to batteries with the same voltage difference.
A. Most Current
B. Wire of length L and area 2A
C. Wire of length L and area A
D. Wire of length 2L and area A
E. Least Curren
Answer:
Explanation:
Resistance of first wire
R₁ = ρL / A where ρ is specific resistance
for second wire the resistance
R₂ = ρ2L / A = 2 ρL / A
For third wire
R₃ = ρL /2 A
The resistance is in the order
R₂ > R₁ > R₃
So , current will be maximum in R₃
I₃ > I₁ > I₂
Where I₁ , I₂ and I₃ is current in first , second and third wire .
A car travels from city A to city B separated by 282 km and the journey takes approximately 6 hours and 32 minutes. If they come back in 8 hours. What is your average speed throughout the journey? What is its average speed?
Answer:
38.8 km/hr
Explanation:
The total distance is 2 × 282 km = 564 km.
The total time is 6 hr 32 min + 8 hr = 14 hr 32 min = 14.533 hr.
The average speed is:
(564 km) / (14.533 hr)
38.8 km/hr
FORENSIC SCIENCE, PLEASE HELP ASAP Investigators find a hair clasped in the fist of a dead victim. Upon close examination, they realize there is follicular tissue attached to that hair. What could investigators use to prove that the hair in the victim's fist belonged to a specific suspect? A) The medulla. B) Nuclear DNA. C) Mitochondrial DNA. D) Physical comparison.
Answer:
C) Mitochondrial DNA
Explanation:
Mitochondrial DNA can help forensic scientists in recovering the follicular tissue that is attached to the hair, they can find out the specific suspect from that hair.
Mitochondrial DNA is maternally inherited and enables forensic scientists to trace maternal lineage.
Answer:
Mitochondrial DNA
Explanation:
Mitochondrial DNA should be used to prove that the hair in the victim's fist belonged to a specific suspect. Mitochondrial DNA enables scientists to trace maternal lineage from which it can be known that who is the suspect.
what is the electric force between two points charges when q1=-4e, q2 = +3e, and r = 0.05 m?
Answer:
Option A. 1.1×10¯²⁴ N
Explanation:
The following data were obtained from the question:
Charge 1 (q1) = - 4e
Charge 2 (q2) = + 3e
Distance apart (r) = 0.05 m
Electric field constant (K) = 9×10⁹ N•m²/C²
Electron (e) = 1.6×10¯¹⁹ C.
Electric Force (F) =..?
Next, we shall determine the value of the two charges.
This is illustrated below:
Charge 1 (q1) = - 4e
Charge 1 (- q1) = 4e
Electron (e) = 1.6×10¯¹⁹ C.
Charge 1 (- q1) = 4 × 1.6×10¯¹⁹ C.
Charge 1 (- q1) = 6.4×10¯¹⁹ C.
Charge 2 (q2) = + 3e
Electron (e) = 1.6×10¯¹⁹ C.
Charge 2 (q2) = 3 × 1.6×10¯¹⁹ C.
Charge 2 (q2) = 4.8×10¯¹⁹ C.
Finally, we shall determine the value of the electric force. This can be obtained as shown below:
Charge 1 (- q1) = 6.4×10¯¹⁹ C.
Charge 2 (q2) = 4.8×10¯¹⁹ C.
Electric field constant (K) = 9×10⁹ N•m²/C².
Distance apart (r) = 0.05 m
Electric Force (F) =..?
F = Kq1q2 /r²
F = (9×10⁹× 6.4×10¯¹⁹× 4.8×10¯¹⁹)/(0.05)²
F = 1.1×10¯²⁴ N
Therefore, the electric force between the two point charge is 1.1×10¯²⁴ N