Answer:
n = 1 10¹⁰ electron
Explanation:
For this exercise we will use Coulomb's law
F = k q₁ q₂ / r²
where it indicates that the value of the force is 7.61 10⁻²¹ N, the distance between the objects is r = 1.94 10⁻³ m and the charge on the objects has the same charge q₁ = q₂ = q,
F = k q² / r²
q = [tex]\sqrt{ \frac{ k \ q^2}{k} }[/tex]
let's calculate
q = [tex]\sqrt{ \frac{ 7.61 \ 10^{-21} \ (1.93 \ 10^{-3})^2 }{9 \ 10^{9} } }[/tex]
q = √(3.182 10⁻¹⁸)
q = 1.783 10⁻⁹ C
the charge of an electron is q₀ = 1.6 10⁻¹⁹ C, therefore
q = n q₀
n = q / q₀
n = 1,78 10⁻⁹ / 1.6 10⁻¹⁰
n = 1.06 10¹⁰ electron
the number of electrons must be integer
n = 1 10¹⁰ electron
Coulomb's Law is a fundamental principle in physics that describes the electrostatic interaction between charged particles. It took approximately 3.62 electrons to produce the charge on one of the objects.
According to Coulomb's Law, the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
To solve this problem, we can use Coulomb's law, which relates the electrostatic force between two charged objects to the charges and the distance between them. Coulomb's law is given by:
[tex]F = k * (|q1 * q2|) / r^2[/tex]
Where:
F is the electrostatic force,
k is the electrostatic constant [tex](k = 8.99 x 10^9 Nm^2/C^2),[/tex]
q1 and q2 are the charges on the objects, and
r is the distance between the objects.
Given that the force F is [tex]7.61 x 10^{-21} N[/tex], and the distance r is [tex]7.61 x 10^{-21} N[/tex]m, and assuming the charges on both objects are equal (let's call it q), we can rewrite Coulomb's law as:
[tex]F = k * (q^2) / r^2[/tex]
Rearranging the equation, we get:
[tex]q^2 = (F * r^2) / k[/tex]
Substituting the given values:
[tex]q^2 = (7.61 x 10^{-21} N) * (1.94 x 10^{-3} m)^2 / (8.99 x 10^9 Nm^2/C^2)[/tex]
Calculating this expression:
[tex]q^2 = 3.352 x 10^{-37} C^2[/tex]
Taking the square root of both sides:
[tex]q = \sqrt {3.352 x 10^{-37} C^2}\\q = 5.79 x 10^{(-19)}C[/tex]
Now, we know that the elementary charge of an electron is approximately[tex]1.6 x 10^{(-19)} C[/tex]. Since each electron carries a charge of[tex]-1.6 x 10^{-19} C[/tex], we can calculate the number of electrons:
Number of electrons = q / (elementary charge)
Number of electrons =[tex](5.79 x 10^{-19} C) / (1.6 x 10^{-19} C)[/tex]
Number of electrons = 3.62
Therefore, it took approximately 3.62 electrons to produce the charge on one of the objects.
For more details regarding Coulomb's law, visit:
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The International Space Station has a mass of 419,455 kg. When it orbits the
Earth at an altitude of 400,000 m, what is the approximate gravitational force
on the Station due to Earth's gravity? (Recall that Earth has a radius of 6.37 x
106 m, it has a mass of 5.97 x 1024 kg, and G = 6.67 x 10-11 N·m²/kg2.)
O A. 3.9 ~ 106N
B. 2.0 x 106N
O C. 3.6 x 106N
O D. 2.8 x 106N
Answer:
[tex]F=3.6\times 10^6\ N[/tex]
Explanation:
Given that,
The mass of International Space Station, m = 419,455 kg
It orbits around the Earth at an altitude of, h = 400,000 m
The radius of Earth, [tex]r=6.37\times 10^6\ m[/tex]
The mass of the Earth, [tex]m_e=5.97 \times 10^{24}\ kg[/tex]
We need to find the approximate gravitational force on the Station due to Earth's gravity. The formula for the gravitational force between the satellite and the Earth is given by :
[tex]F=G\dfrac{mm_e}{(r+h)^2}\\\\F=6.67\times 10^{-11}\times \dfrac{419455\times 5.97 \times 10^{24}}{(6.37 \times 10^6+400000 )^2}\\\\F=3.6\times 10^6\ N[/tex]
So, the required force on the station due to Earth's gravity is [tex]3.6\times 10^6\ N[/tex].
Based on the information presented in the introduction of this problem, what is a sound wave? Based on the information presented in the introduction of this problem, what is a sound wave? Propagation of sound particles that are different from the particles that comprise the medium Propagation of energy that does not require a medium Propagation of pressure fluctuations in a medium
Answer:
sound is a wave that carries energy and the particles are oscillating around their equilibrium positions.
Explanation:
When the air particles are subjected to a periodic force they oscillate around their equilibrium possessions, this oscillation can propagate in the air creating a sound wave.
Therefore the sound wave is a wave that propagates energy, but the particles have only one oscillation around their equilibrium position.
These waves can be mathematically analyzed as displacement waves and since the oscillation is in the same line of the movement of the wave we can also analyze it as a pressure wave
In summary, sound is a wave that carries energy and the particles are oscillating around their equilibrium positions.
A transverse, sinusoidal wave travels in a string and can be described by the function: y(x,t)=0.87 sin(21x−4.9t). What is the speed of this wave?A transverse, sinusoidal wave travels in a string and can be described by the function: . What is the speed of this wave?
A. 0.18 m/s
B. 9.2 m/s
C. 4.3 m/s
D. 0.23 m/s
Answer:
D. 0.23m/s
Explanation:
Given the equation of a wave expressed as;
y(x,t)=0.87 sin(21x−4.9t)
Comparing with the standard equation of a wave;
y(x,t)= Asin(2πx/λ−2πft) where;
f is the frequency
λ is the wavelength
On comparing;
2πx/λ = 21x
2π/λ = 21
λ = 2π/21
λ = 2(3.14)/21
λ = 6.28/21
λ = 0.299m
Get the frequency f;
2πft = 4.9t
2πf = 4.9
f = 4.9/2π
f = 4.9/2(3.14)
f = 4.9/6.28
f = 0.78Hz
Get the speed of the wave
Speed = frequency * wavelength
Speed = 0.299 * 0.78
Speed = 0.233m/s
Hence the speed of the wave is 0.23m/s
prove the identity
Trigonometry grade 10
Answer:
and is in photo given.I didn't get time to type.
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 47.0 cm . The explorer finds that the pendulum completes 91.0 full swing cycles in a time of 125 s. What is the magnitude of the gravitational acceleration on this planet?
Answer:
[tex]g=9.83\ m/s^2[/tex]
Explanation:
Given that,
The length of a simple pendulum, l = 47 cm = 0.47 m
The pendulum completes 91.0 full swing cycles in a time of 125 s.
The time period of a simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{L}{g}} \\\\or\\\\g=\dfrac{4\pi^2L}{T^2}[/tex]
Also,
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{t}{n}[/tex]
t is 125 s and n = 91 cycles
Substitute all the values to find g.
[tex]g=\dfrac{4\pi^2\times 0.47}{(\dfrac{125}{91})^2}\\\\g=9.83\ m/s^2[/tex]
So, the gravitational acceleration on this planet is [tex]9.83\ m/s^2[/tex].
First aid for a strain or a sprain
A) the FITT formula.
B) ineffective.
C) the RICE formula. .
D) the ABC formula
Answer: C) the RICE formula
Explanation:
Particularly for the treatment of injury associated with the sprain or strain one can get relieve in pain and swelling on the damaged area by promoting flexibility and healing of that area utilizing the RICE formula. R- rest, I- Ice, C- Compression, and E- Elevation. Resting and restricting any activity can help in enhancing the healing process. The application of ice will reduce pain and swelling. The injured area can be compressed by using a tight bandage. This will also help in reducing the swelling. Keeping the injured body part at an elevation like on a pillow can also reduce the swelling.
HIGHLIGHT the correct answer: a.) As the skater goes up the hill, his kinetic energy: increases / decreases / stays constant b.) As he goes up the hill, his potential energy: increases / decreases / stays constant c.) As he goes up the hill, his mechanical energy: increases / decreases / stays constant d.) As he goes down the hill, his kinetic energy: increases / decreases / stays constant e.) As he goes down the hill, his potential energy: increases / decreases / stays constant f.) As he goes down the hill, his mechanical energy: increases / decreases / stays constant
Answer:
a) Kinetic energy decreases, b) Potential energy increases, c) Mechanical energy stays constant, d) Kinetic energy increases, e) Kinetic energy increases, f) Mechanical energy stays constant.
Explanation:
a) Let suppose that the skater is a conservative system. If he goes up the hill, gravitational potential energy is increased at the expense of kinetic energy. In a nutshell, his kinetic energy decreases.
b) His potential energy increases.
c) Mechanical energy is the sum of gravitational potential and kinetic energies, since skater is conservative, then mechanical energy stays constant.
d) If he goes down the hill, his kinetic energy is increased at the expense of gravitational potential energy. In a nutshell, his kinetic energy increases.
e) His potential energy decreases.
f) Mechanical energy is the sum of gravitational potential and kinetic energies, since skater is conservative, then mechanical energy stays constant.
Answer:
The answer would be what the other guy said
Explanation:
How much energy is needed to heat 1kg of copper by 20 degrees Celsius
Step by step explanation
Answer:
Q = 0.0077 J
Explanation:
Given that,
Mass of copper, m = 1 kg = 0.001 g
The change in temperature = 20°C
We need to find the energy needed to heat copper. The formula is given by :
[tex]Q=mc\Delta T[/tex]
c is the specific heat of copper, c = 0.385 J/g°C
Put values in the above formula
[tex]Q=0.001\ g\times 0.385\ J/g^{\circ} C\times 20^{\circ} C\\\\Q=0.0077\ J[/tex]
So, the energy needed to heat 1 kg of copper is 0.0077 J.
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass?
Answer:
The quarterback applies a force of 8.08 N to the ball during the pass.
Explanation:
To find the force we need to calculate the acceleration first so we can use the following equation:
[tex] F = ma [/tex]
Now, we can calculate the acceleration as follows:
[tex] x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2} [/tex]
Since the ball starts from rest [tex]v_{0} = 0[/tex] and [tex]x_{0} = 0[/tex]:
[tex] a = \frac{2*x_{f}}{t^{2}} = \frac{2*1.909 m}{(0.439 s)^{2}} = 19.81 m/s^{2} [/tex]
Now we can find the force:
[tex] F = 0.408 kg*19.81 m/s^{2} = 8.08 N [/tex]
Therefore, the quarterback applies a force of 8.08 N to the ball during the pass.
I hope it helps you!
The dotted lines and arrows represent
Why is temperature scalar?
A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that is stationary. The carts stick together after the collision. What is the final kinetic energy of the system?
Answer:
0.006075Joules
Explanation:
The final kinetic energy of the system is expressed as;
KE = 1/2(m1+m2)v²
m1 and m2 are the masses of the two bodies
v is the final velocity of the bodies after collision
get the final velocity using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
0.12(0.45) + 0/12(0) = (0.12+0.12)v
0.054 = 0.24v
v = 0.054/0.24
v = 0.225m/s
Get the final kinetic energy;
KE = 1/2(m1+m2)v
KE = 1/2(0.12+0.12)(0.225)²
KE = 1/2(0.24)(0.050625)
KE = 0.12*0.050625
KE = 0.006075Joules
Hence the final kinetic energy of the system is 0.006075Joules
A ball is thrown straight upward with an initial speed v0. When it reaches the top of its flight at height h, a second ball is thrown straight upward with the same initial speed. Do the balls cross paths at height 1/2h, above 1/2h, or below 1/2h
Answer:
The balls cross paths above 1/2h
Explanation:
Let's analyze the picture when the first ball is at the top of its flight and the second ball starts to flight.
When the second ball is thrown it has the maximum speed and the first ball starts with an initial velocity equal to zero. We know that they will cross at the same time, hence the second ball will travel a longer distance, due to its velocity, than the first ball.
Therefore, the correct answer is above the midpoint of h (1/2h).
I hope it helps you!
A burning candle provides :
a.radiant energy
b.solar energy
c.chemical potential energy
d.thermal energy
A student must perform an experiment in which two objects travel toward each other and collide so that the data collected can be used to show that the collision is elastic within the acceptable range of experimental uncertainty. Which of the following measuring tools, when used together, can the student use to verify that the collision is elastic?
A. A motion detector.
B. A meterstick.
C. A balance.
D. A stopwatch.
Answer:
A. A motion detector.
C. A balance.
Explanation:
An elastic collision can be defined as a type of collision in which the total kinetic energy possessed by the colliding bodies remain the same.
This ultimately implies that, in an elastic collision the maximum amount of kinetic energy possessed by the colliding bodies is conserved (kinetic energy is not lost) after the collision.
In this scenario, a student must perform an experiment in which two objects travel toward each other and collide so that the data collected can be used to show that the collision is elastic within the acceptable range of experimental uncertainty.
The measuring tools, which when used together, can be used by the student to verify that the collision is elastic are a motion detector and a balance.
Basically, the motion detector is an electronic device that can be used to determine the initial and final velocity of the objects while the balance would be used to take a measurement of the mass possessed by the objects.
The image below represents molecules in which state of matter? *
da answer is liquiddddddddd
How many times more acidic is solution A with a pH of 3.4 than solution B with a pH of 8.4?
Solution A is _________ times more acidic than solution B.
(Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to the nearest tenth as needed.)
Answer:
Solution A is 1 x 10⁵ times more acidic than solution B
Explanation:
The pH scale is a logarithmic scale used to express the acidity or basicity of a solution.
The values are written from 1 - 14 with 1 being the most acidic and 14 a basic solution.
Each interval is ten folds more concentrated than the next because it is a logarithmic scale.
Since the pH of A = 3.4
pH of B = 8.4
The difference = 8.4 - 3.4 = 5.0
So, Solution A is 1 x 10⁵ times more acidic than solution B
Can someone help me with this
Answer:
Explanation:
try the fourth one :)
A piece of metal with a mass of 1.9 kg, and specific heat of 200J/kg°C, and an initial temperature of 100°C is dropped into an insulated jar that contains liquid with a mass of 3.9 kg, a specific heat of 1000J/kg°C, and initial temperature of 0°C. The piece of metal is removed after 7 s, at which time its temperature is 20°C.
Required:
Find the temperature of the liquid after the metal is removed. Neglect any effects of heat transfer to the air or to the insulated jar. Answer in units of ∘C.
Answer:
7.79 °C
Explanation:
Given that
Mass of the metal, m = 1.9 kg
Specific heat of the metal, c(m) = 200 J/kgC
Initial temperature, T(0) = 100 °
Mass of liquid, M = 3.9 kg
Specific heat of liquid, c(l) = 1000 J/kgC
Initial temperature, T = 0 °C
Time taken, t = 7 s
Final temperature = 20 °C
To solve the question, we use the formula the sum of the heats gained is zero.
massmetal * cmetal (20-100) + massliquid * cliquid (T.f-0)=0
On solving for T.f
1.9 * 200 * (20 - 100) + 3.9 * 1000 * (t.f - 0) = 0
380 * -80 + 3900 * t.f = 0
-30400 + 3900t.f = 0
3900t.f = 30400
t.f = 30400 / 3900
t.f = 7.79 °C
write the realtionship between the density of a liquid and its upthrust? clarify
Answer:
B = ρ g V_liquid
the thrust is proportional to the density of the liquid
Explanation:
The density of a liquid is defined as the relationship between the mass and the volume of the liquid
ρ = m / V
The upward push of the liquid is given by the principle of Archimedes Archimedes establishes that the push is equal to the weight of the dislodged liquid
B = W_liquid
B = m _liquid g
we substitute mass for density
B = ρ g V_liquid
therefore we see that the thrust is proportional to the density of the liquid
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 s after it leaves the thrower’s hand. Ignore air resistance.
(a) What is the initial speed of the egg?
(b) How high does it rise above its starting point?
(c) What is the magnitude of its velocity at the highest point?
(d) What are the magnitude and direction of its acceleration at the highest point?
(e) Sketch ay−t,vy−ta y −t,v y −t, and y-t graphs for the motion of the egg.
Answer:
1. 18.5m/s
2. 17.5 m
3. 0 at its highest point
4. Direction is downwards
Explanation:
1. This egg is thrown vertically from a height
Yo = 0. This egg then falls to the point y = -30.0 at t = 5seconds
Y-Yo = V0t - 1/2gt²
-30-0 = V0(5)-1/2(9.8)(5²)
-30 = 5v0 - 4.9x25
-30 = 5V0 - 122.5
-30+122.4 = 5v0
V0 = 92.5/5
= 18.5m/s
this is the initial speed of the egg
2. When the egg is at a maximum height it would have a velocity equal to 0
V² = V0² - 2*g*y
V = 0, V0 = 18.5, g = 9.8
0 = 18.5²-2x9.8*y
342.25-19.6y = 0
342.25 = 19.6y
Divide through by 19.6
Y = 342.25/19.6
Y = 17.5m
this value is how high it rises above starting point
3.
The magnitude of velocity is = 0 at its highest point
4.
This egg falls under gravity. Therefore the acceleration due to gravity has a constant magnitude and direction. Magnitude = 9.8m/s and it's direction is downwards.
5. Please check attachment for graph
Which object has the most momentum?
A toy car rolling along the floor
A child peddling their bike as fast as possible
&o
A large moving truck stopped at a red light
A 20 kg cart with frictionless bearings is initially at rest on a horizontal tabletop. The cart is connected by a light string to a hanging 5 kg block. The string passes over an ideal pulley. The system is released from rest and both objects have an acceleration of magnitude a 5 kg
(a) Draw the forces (NOT components) that act on each object.
(b) Calculate the acceleration ay of the system of masses once the system is released.
Answer:
[tex]a=1.96m/s^2[/tex]
Explanation:
From the question we are told that
Mass of cart =20kg
Mass of block =5kg
Magnitude of acceleration M_a= 5 kg
Generally Force on the 20kg mass in the [tex]\bar X[/tex] is mathematically given as
[tex]\sum F=ma\\t=(20kg)a[/tex]
Generally Force on the 2kg in the[tex]\bar Y[/tex] mass is mathematically given as
[tex]\sum F=ma\\(5kg)(9.8m/s^2)-(5kg)a[/tex]
Therefore
[tex](20kg)a=(5kg)(9.8m/s^2)-(5kg)a[/tex]
[tex]a=\frac{5kg)(9.8m/s^2)-(5kg)}{25}[/tex]
[tex]a=1.96m/s^2[/tex]
What do you think is the hardest problem for Divine Command Theory to overcome, and why?
Explanation:
One problem with opting for number 1 in the above dilemma is that it becomes difficult if not impossible to conceive of God as morally good, because if the standards of moral goodness are set by God's commands, then the claim “God is morally good” is equivalent to “God obeys His own commands”.
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Helge, Steve, and Heidi are sitting on a sled on a slope covered with a hard snow. The sled is stationary. The friends have different suggestions for how to make the sled start moving:
Helge: If one of us gets off, the sled will start moving.
Steve: We should invite another person to join us, and then the sled will start moving.
Heidi: We should get off the sled, polish the bottom of the sled to make it smoother, and sit back down on it. The sled will then start moving.
Who suggests the correct solution? Select the correct equation for acceleration ax along the incline. Assume the x-axis is directed down along the incline, and the y-axis is directed normal to the incline. Let μs and θ denote the coefficient of static friction and the angle the incline makes with the horizontal, respectively.
Answer:
Heldi is right
Explanation:
It will not move even if more people or one person gets off it will stay the same til a force pushes down on it.
An elevator travels up and down. The vertical position of the elevator in meters over time is shown below
what is the displacement of the elevator between 8s and 20s?
What is the distance traveled by the elevator between 8s and 20s
Answer:
The displacement is
-27m
The distance traveled is
27m
Dont listen to the other guys they are wrong
Explanation:
Well you see the elevator starts at 12m and 8s and moves down to 0mat 16s, traveling at a distance of 12m
The Elevator starts at 0m at 16s and moves down to -15m at 20s traveling a distance of 15m
So afterwards what you do is find the distance between 8 and 20 so do this equation
12m + 15m = 27m
So that gives you
-27
27
HOPE THIS HELPS!!!
The displacement and distance of the elevator between 8s and 20s will be -27 meters and 27 meters respectively.
What is distance?Distance is a numerical representation of the distance between two objects or locations.
Distance can refer to a physical length or an estimate based on other factors in physics or common use. |AB| is a symbol for the distance between two points A and B.
From the given graph, it is observed that;
Distance is the sum of the physical length travelled. First, the elevator travels the 15 m down and the 12-meter up. The distance travelled will be;
d= 15 +12 = 27 m
Displacement is the net distance travelled or the shortest distance. The value of the displacement can be negative. The displacement is found as;
D= -15-12 =-27 m
Hence, the displacement and distance of the elevator between 8s and 20s will be -27 meters and 27 meters respectively.
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A particular race car can cover a quarter-mile track (402 m) in 6.40 s starting from a standstill.
Find its acceleration.
If the combined mass of the driver and race car is 485 kg, what horizontal force must the road exert on the tires?
Answer:
9520N
Explanation:
For the first part of this problem:
Given parameters:
Distance traveled = 402m
Time taken = 6.4s
Initial velocity = 0m/s
Unknown:
Find the acceleration = ?
Solution:
To solve this problem, we use on of the kinematics equation:
S = ut + [tex]\frac{1}{2}[/tex]at²
S is the distance moved
u is the initial velocity
t is the time taken
a is the acceleration
402 = (0 x 6.4) + ([tex]\frac{1}{2}[/tex] x a x 6.4²)
402 = 20.48a
a = 19.63m/s²
Second problem:
Given parameters:
Mass of the driver and race car combined = 485kg
So;
Horizontal force = mass x acceleration
Horizontal force = 485 x 19.63 = 9520N
Answer:7.1*10^3
Explanation:
You exert a 100-N force on a pulley system to lift 300-N. What is the mechanical advantage of this system? How many sections of rope support weight?
Answer:
Mechanical advantage = 3
Explanation:
You exert a 100-N force on a pulley system to lift 300-N.
The mechanical advantage of the system is given by the ratio of output force to the input force.
Here, output force = 300 N and input force = 100 N
Mechanical advantage,
[tex]m=\dfrac{300}{100}\\\\m=3[/tex]
Mechanical advantage is 3 it means that there are 3 sections of rope support. Hence, this is the required solution.
In the diagram, the block is sliding at a constant velocity down a rough incline, which makes a 45 degree angle to the horizontal. Which of the following statements is NOT true?
A) The magnitude of the Net Force is equal to the magnitude of Force C.
B) The magnitude of Force A is equal to the magnitude of Force B
C) The magnitude of Force B is equal to the magnitude of Force D
D) The coefficient of friction between the block and the ramp is equal to 1.
Answer:
I haven't done physics for a while, but with the information given I don't understand how D could be correct, so I think D is your anwser
Explanation:
To say that the rights to life, liberty, and
property are unalienable means that they
