why is it considered that all electric charges are multiples of e
Answer:
electric charges are quantised
Explanation:
which means they exist as integral multiples q=ne . this was practically proved by the Millikan's oil drop experiment in which he let charged oil droplets as fine as mist fall under the effect of gravity, after measuring the charge on many drops the results showed that charges were multiples of a minimum value 1.6 x 10[tex]{-19}[/tex] which we now know as the charge on an electron or elementary charge.
A roller coaster is at a peak of 20m and has a mass of 900kg. What is the potential energy of the roller coaster?
O 100000 J
10000 J
O 9.8 J
O 176400 J
The potential energy of the roller coaster is 176,400 J (joules).
The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.
In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 [tex]m/s^2[/tex].
Using the formula, we can calculate the potential energy:
PE = mgh
= (900 kg)(9.8 [tex]m/s^2[/tex])(20 m)
= 176,400 J
Therefore, the potential energy of the roller coaster is 176,400 J (joules).
Know more about potential energy here:
https://brainly.com/question/21175118
#SPJ8
Two tiny objects with equal charges of 7.25 μC are placed at the two lower corners of a square with sides of 0.201 m, as shown.
Find the electric field at point B, midway between the upper left and right corners. If the direction of the electric field is upward, enter a positive value. If the direction of the electric field is downward, enter a negative value. answer in N/C
The electric field at point B, located at the midpoint between the upper left and right corners of the square, can be approximated as 3.244 x [tex]10^6[/tex] N/C in the upward direction.
To find the electric field at point B, we can consider the contributions from the two charges placed at the lower corners of the square. Since the charges are the same and the distance to point B is the same for both charges, the magnitudes of the electric fields produced by each charge will be equal.
First, let's calculate the magnitude of the electric field produced by one of the charges at point B using Coulomb's Law:
Electric field due to a point charge (E) = (k * q) / [tex]r^2[/tex]
Where:
- k is the electrostatic constant, approximately equal to 8.99 x 10^9 [tex]10^9[/tex]N [tex]m^2/C^2[/tex]
- q is the charge of the object, 7.25 μC (7.25 x [tex]10^-^6[/tex] C)
- r is the distance from the charge to point B, which is half the length of the square's side, 0.201 m / 2 = 0.1005 m
Plugging in the values, we have:
E = (8.99 x [tex]10^9 N m^2/C^2[/tex] * 7.25 x [tex]10^-^6[/tex] C) / (0.1005 [tex]m)^2[/tex]
E ≈ 1.622 x [tex]10^6[/tex] N/C
Since the electric fields from the two charges at the lower corners have equal magnitudes and point in the same direction (upward), the total electric field at point B is twice the magnitude of the individual electric field:
Electric field at point B = 2 * E ≈ 2 * 1.622 x [tex]10^6[/tex] N/C
Electric field at point B ≈ 3.244 x [tex]10^6[/tex] N/C
Therefore, the electric field at point B, midway between the upper left and right corners of the square, is approximately 3.244 x [tex]10^6[/tex]N/C upward.
For more such information on: electric field
https://brainly.com/question/19878202
#SPJ8
A rigid circular loop has a radius of 0.20 m and is in the xy-plane. The loop carries a
clockwise current I. The magnitude of the magnetic moment of the loop is 0.75 A.m2
.
A uniform external magnetic field, B = 0.20 T is directed parallel to the plane of the
loop. An external torque changes the orientation of the loop from one of lowest potential energy to one of highest potential energy. Calculate the work done by this external
torque.
A rigid circular loop having a radius of 0.20 m and is in the xy-plane, carries a uniform external magnetic field. B = 0.20 T and is directed parallel to the plane of the loop. The torque exerted on the loop is given by the formula as follows;τ = NABsin θ
Where τ is the torque, N is the number of turns, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the normal to the loop and the magnetic field direction. The direction of the torque is given by the right-hand rule and depends on the direction of the magnetic field.To obtain the magnitude of the torque, we need to evaluate all the terms in the formula, τ = NABsinθ.A = πr² = π (0.20 m)² = 0.1257 m²The angle between the normal to the loop and the magnetic field direction is 0 degrees since the field is parallel to the loop.θ = 0°Substituting the values in the formula, we have;τ = NABsinθ= N (0.1257 m²) (0.20 T) (sin 0°)τ = 0 NmThe torque exerted on the loop is zero since the angle between the normal to the loop and the magnetic field direction is zero. The torque only exists when there is an angle between the normal and the magnetic field direction. Therefore, there is no net force on the current loop.For such more question on torque
https://brainly.com/question/17512177
#SPJ8
A new ride being built at an amusement park includes a vertical drop of 79.8 meters. Starting from rest, the ride vertically drops that distance before the track curves forward. If friction is neglected, what would be the speed of the roller coaster at the bottom of the drop?
Answer:
Approximately [tex]39.6\; {\rm m\cdot s^{-1}}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Under the assumptions, vertical acceleration of the vehicle during the ride would be equal to the gravitational field strength: [tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Apply the following SUVAT equation to find the velocity of the vehicle at the bottom of the drop:
[tex]v^{2} - u^{2} = 2\, a\, x[/tex],
Where:
[tex]v[/tex] is the final velocity at the bottom of the drop;[tex]u[/tex] is the initial velocity at the beginning of the drop; [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] since the vehicle started from rest;[tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex] is the vertical acceleration of the vehicle during the drop;[tex]x = 79.8\; {\rm m}[/tex] is the vertical displacement of the vehicle during the drop.Rearrange this equation to find [tex]v[/tex]:
[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x} \\ &\approx \sqrt{0^{2} + 2\, (9.81)\, (79.8)} \; {\rm m\cdot s^{-1}} \\ &\approx 39.6\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the speed of this vehicle at the bottom of the drop would be approximately [tex]39.6\; {\rm m\cdot s^{-1}}[/tex].
determine energy of absorbed photon.
The energy of an absorbed photon is [tex]2.998 * 10^8[/tex] and the resulting unit for energy is Joules (J).
To determine the energy of an absorbed photon in electronvolts (eV), we can use the equation E = hv, where E represents energy, h is Planck's constant, and v is the frequency of the photon. By substituting the values and units correctly, we can calculate the energy in electronvolts.
To determine the energy of an absorbed photon, one can use the following formula:
energy of absorbed photon = Planck's constant x frequency of radiation
Where Planck's constant is equal to [tex]6.626 * 10^{-34[/tex] Joule seconds and frequency of radiation is measured in Hertz (Hz).
The energy of an absorbed photon can also be expressed using the wavelength of the radiation instead of its frequency.
The formula for this is:energy of absorbed photon = Planck's constant x speed of light / wavelength of radiation
Where the speed of light is equal to [tex]2.998 * 10^8[/tex] m/s.In both cases,
Know more about photon here:
https://brainly.com/question/31379966
#SPJ8
Two objects are held close together. When they are released, they move toward one another. Which conclusion is supported by this evidence? A) The objects are both positively charged. B) The objects are both negatively charged. C) The objects have no charge. D) The objects have opposite charges.
When two objects held close together and then released, they move toward each other. The phenomenon can be explained by the Coulomb's Law. Coulomb's Law is given by: $F=k\frac{q_{1}q_{2}}{r^2}$ where F is the electrostatic force, q1 and q2 are the magnitudes of the charges, r is the distance between the centers of the charged objects, and k is Coulomb's constant.
This law states that like charges repel and opposite charges attract. This means that when two charged objects of the same charge are held close together, they will repel one another, and when released, they will move away from each other. Therefore, option A (The objects are both positively charged) and B (The objects are both negatively charged) are ruled out. The only conclusion that is supported by the evidence is that the objects have opposite charges, which is option D. The objects attract each other because of the electrostatic force of attraction between opposite charges.In conclusion, when two objects held close together and then released, they move towards each other. This is due to the Coulomb's Law which states that like charges repel and opposite charges attract. Therefore, the conclusion that is supported by this evidence is that the objects have opposite charges.For such more question on Coulomb's Law
https://brainly.com/question/506926
#SPJ8
The student has a mass of 50.0 kg. What is her momentum at 2 s (in kgm/s)?
Answer:
50kgm/2
Explanation:
An electron is placed in a uniform electric field of strength 250 N/C. The electron is at rest at the origin of a coordinate system at t = 0 and the electric field is in the positive x-direction. Charge of electron is 1.602 × 10−19 C and mass of electron is 9.109 × 10−31 kg.
Problem 16.051
What is the x-coordinate of the electron at t = 4.00 ns? answer in mm
What is the y-coordinate of the electron at t = 4.00 ns? answer in mm
The x-coordinate of the electron at t = 4.00 ns is approximately 0.03515 mm, and the y-coordinate is 0 mm.
To determine the x-coordinate and y-coordinate of the electron at t = 4.00 ns, we need to use the equations of motion for an object experiencing constant acceleration in the presence of an electric field.
Given:
Electric field strength (E) = 250 N/C
Charge of the electron (q) = 1.602 ×[tex]10^-^1^9[/tex] C
Mass of the electron (m) = 9.109 × [tex]10^-^3^1[/tex]kg
Time (t) = 4.00 ns = 4.00 ×[tex]10^-^9[/tex] s
First, let's calculate the acceleration (a) experienced by the electron due to the electric field using Newton's second law, F = ma, where F is the force exerted on the electron.
F = qE
Substituting the values:
F = (1.602 ×[tex]10^-^1^9[/tex] C)(250 N/C) = 4.005 ×[tex]10^-^1^7[/tex] N
Using Newton's second law, we can find the acceleration:
F = ma
a = F/m = (4.005 × [tex]10^-^1^7[/tex]N) / (9.109 ×[tex]10^-^3^1[/tex]kg) ≈ 4.393 × [tex]10^1^3 m/s^2[/tex]
Next, we can calculate the x-coordinate using the equation of motion for linear motion with constant acceleration:
x = ut +[tex](1/2)at^2[/tex]
Since the electron is initially at rest (u = 0), the equation simplifies to:
x = [tex](1/2)at^2[/tex] = (1/2)(4.393 × [tex]10^1^3 m/s^2)[/tex](4.00 × [tex]10^-^9 s)^2[/tex] = 3.515 × [tex]10^-^5[/tex] m
To convert the result to millimeters (mm), we multiply by 1000:
x = 3.515 × [tex]10^-^5[/tex] m× 1000 = 0.03515 mm
Therefore, the x-coordinate of the electron at t = 4.00 ns is approximately 0.03515 mm.
Since the electric field is in the positive x-direction, the y-coordinate of the electron remains unchanged. Therefore, the y-coordinate at t = 4.00 ns is 0 mm.
For more such information on: electron
https://brainly.com/question/30128752
#SPJ8
A small sphere with charge 0.850 μC is placed at the center of a cube. What is the electric flux through one surface of the cube? answer in N⋅m2/C
The electric flux through the cube's surface is approximately 9.60 × 10⁵ N⋅m²/C.
To calculate the electric flux through one surface of a cube due to a charge at its center, we can apply Gauss's Law. Gauss's Law states that the electric flux (Φ) through a closed surface is equal to the enclosed charge (Q) divided by the electric constant (ε₀).
1. Calculate the electric flux:
The electric flux through one surface of the cube can be determined by considering a Gaussian surface that encloses the charge at the center. Since the cube is symmetrical, we choose a cube of the same size as our Gaussian surface.
The enclosed charge (Q) is the charge of the small sphere, which is 0.850 μC.
The electric constant (ε₀) is a fundamental constant equal to approximately 8.854 × 10⁻¹² N⋅m²/C².
Using Gauss's Law, the electric flux is given by:
Φ = Q / ε₀
Φ = (0.850 μC) / (8.854 × 10⁻¹² N⋅m²/C²)
2. Calculate the electric flux in N⋅m²/C:
Substituting the values into the equation:
Φ ≈ (0.850 × 10⁻⁶ C) / (8.854 × 10⁻¹² N⋅m²/C²)
Φ ≈ 9.60 × 10⁵ N⋅m²/C
Therefore, the electric flux through one surface of the cube due to the charge at its center is approximately 9.60 × 10⁵ N⋅m²/C.
For more such information on: electric flux
https://brainly.com/question/26289097
#SPJ8
Particles q₁, 92, and q3 are in a straight line.
Particles q₁ = -5.00 10-6 C,q2 = +2.50 10-6 C, and
q3= -2.50 10-6 C. Particles q₁ and q₂ are separated
by 0.500 m. Particles q2 and q3 are separated by
0.250 m. What is the net force on q2?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
When three particles q1, q2, and q3 are in a straight line, the force experienced by q2 depends on the distance between the particles, the magnitude of the charges of the particles, and the relative position of the particles. The electric force on a charge q1 from a charge q3, which is at a distance r3 from the first charge, is governed by Coulomb’s Law.
In mathematical notation, Coulomb’s Law is expressed as:F = kq1q3 / r3²where F is the electric force, q1 and q3 are the magnitudes of the charges, r3 is the distance between them, and k is the Coulomb’s Law constant which is equal to 8.99 × 109 N·m2 / C2. The direction of the electric force can be positive or negative. If q1 and q3 are of the same sign, the force between them is repulsive, while if they are of opposite sign, the force is attractive.The electric force experienced by q2 due to the other two particles is the vector sum of the forces F13 and F23, which are given by:F13 = kq1q3 / r13²andF23 = kq2q3 / r23²where r13 is the distance between q1 and q3, and r23 is the distance between q2 and q3. The net force on q2 is given by:Fnet = F13 + F23The magnitude and direction of Fnet depend on the charges of the particles and the distance between them. If the two charges are of the same sign, then the force is repulsive and the net force will be in the direction of the force between them. If the two charges are of opposite signs, then the force is attractive, and the net force will be in the direction of the force between them.For such more question on Coulomb’s Law
https://brainly.com/question/506926
#SPJ8
17. The time taken by a 100 g stone to reach the ground when dropped from the top of (A) your school building has been given. Write a suitable formula for the calculation of ght of your school building.
On a sunny spring day, Mr. Hane’s family went to a zoo in his town. His daughter looked at an elephant, and she found that the mass of the elephant is 2019 kg. What is the weight of the elephant in Newtons? Use the acceleration due to gravity (g) as 9.8 meters per second squared.
The weight of the elephant is approximately 19,780.2 Newtons, calculated by multiplying its mass (2019 kg) by the acceleration due to gravity (9.8 m/s²).
To calculate the weight of the elephant in Newtons, we need to use the formula:
Weight = Mass x Acceleration due to gravity
Given that the mass of the elephant is 2019 kg and the acceleration due to gravity is 9.8 m/s^2, we can substitute these values into the formula:
Weight = 2019 kg x 9.8 m/s^2
By multiplying these values, we find that the weight of the elephant is approximately 19,780.2 Newtons.
The weight of an object is the force with which it is attracted towards the center of the Earth due to gravity. The unit of weight is Newtons, and it represents the force required to support the object against gravity. In this case, the weight of the elephant tells us the force exerted by the Earth on the elephant, and it is approximately 19,780.2 Newtons.
Know more about gravity here:
https://brainly.com/question/940770
#SPJ8
Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.99 ✕ 105 m/s in the positive x-direction. Thousands of miles from Earth, they interact with Earth's magnetic field of magnitude 2.93 ✕ 10−8 T in the positive z-direction. Find the magnitude and direction of the magnetic force on a proton. Find the magnitude and direction of the magnetic force on an electron.
The magnitude of the magnetic force on a proton is 1.68 ✕ 10^−16 N, directed towards the negative y-direction. The magnitude of the magnetic force on an electron is also 1.68 ✕ 10^−16 N, but directed towards the positive y-direction.
Begin by using the formula for the magnetic force on a charged particle moving in a magnetic field:
F = q * v * B * sin(θ)
where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
For a proton, the charge is +e, where e is the elementary charge (1.6 ✕ 10^−19 C). The velocity of the proton is given as 3.99 ✕ 10^5 m/s in the positive x-direction. The magnetic field strength is 2.93 ✕ 10^−8 T in the positive z-direction.
The angle θ between the velocity vector and the magnetic field vector is 90 degrees since the velocity is perpendicular to the magnetic field.
Substituting the values into the formula, we get:
F_proton = (+e) * (3.99 ✕ 10^5 m/s) * (2.93 ✕ 10^−8 T) * sin(90°)
F_proton ≈ 1.68 ✕ 10^−16 N
Since the proton has a positive charge, the magnetic force is directed opposite to the direction of electron flow, which is in the negative y-direction.
Repeat the steps for an electron, but note that the charge is -e. The direction of the force on an electron will be opposite to that of a proton.
F_electron = (-e) * (3.99 ✕ 10^5 m/s) * (2.93 ✕ 10^−8 T) * sin(90°)
F_electron ≈ -1.68 ✕ 10^−16 N
The force on an electron is also 1.68 ✕ 10^−16 N, but directed towards the positive y-direction.
Remember to consider the proper units and take into account the directions of charge, velocity, and magnetic field when calculating the magnetic force on a charged particle.
For more such questions on magnetic force, click on:
https://brainly.com/question/13277365
#SPJ8
A roller coaster travels down the hill and hits a velocity of 10 m/s. What is the kinetic energy of the roller coaster if it has a mass of 500kg?
O 490000 J
25000 J
O 10 J
O 10000 J
Kinetic Energy is the energy that a moving object has by virtue of its motion. It is given by the formula:KE = 1/2mv²where m is the mass of the object and v is its velocity.
The mass of the roller coaster is given as 500 kg, and its velocity is 10 m/s,
The kinetic energy of the roller coaster can be calculated as follows:
[tex]KE = \frac{1}{2} mv²KE = \frac{1}{2} \times 500 kg x (10 m/s)²[/tex]
[tex]KE = \frac{1}{2} \times 500 kg \times 100 m^2/s^2KE = 50,000 J[/tex]
Therefore, the kinetic energy of the roller coaster is 50,000 J or 5 x 10⁴ J.
For such more question on Kinetic Energy
https://brainly.com/question/8101588
#SPJ8
A conducting sphere (radius a) is placed at the center of a conducting spherical shell (inner radius b and outer radius c). The conductors are in electrostatic equilibrium.
The sphere has a net charge of +5.10 μC and the shell has a net charge of +5.10 μC. What is the charge on the inner surface of the shell? answer in μC
The inner sphere has a net charge of +5.10 μC and the shell has a net charge of +5.10 μC. What is the charge on the outer surface of the shell? answer in μC
Charge on inner surface: -5.10 μC
Charge on outer surface: +10.20 μC
In electrostatic equilibrium, the charges on conductors distribute in such a way that the electric field inside them is zero. Given a conducting sphere with a net charge of +5.10 μC placed at the center of a conducting spherical shell with inner radius b and outer radius c, we can determine the charges on the inner and outer surfaces of the shell.
1. Charge on the inner surface of the shell:
Since the electric field inside the conductor is zero, the charges on the inner surface of the shell will redistribute in such a way that they cancel out the electric field from the sphere at the center.
Since the net charge on the sphere is +5.10 μC, the inner surface of the shell will have an equal and opposite charge to neutralize the electric field from the sphere. Therefore, the charge on the inner surface of the shell is -5.10 μC.
2. Charge on the outer surface of the shell:
The total charge on the shell is +5.10 μC, which means the sum of the charges on the inner and outer surfaces of the shell must equal this value.
Since the charge on the inner surface is -5.10 μC, the charge on the outer surface can be calculated as the difference:
Charge on the outer surface = Total charge on the shell - Charge on the inner surface
Charge on the outer surface = +5.10 μC - (-5.10 μC)
Charge on the outer surface = +10.20 μC
Therefore, the charge on the inner surface of the shell is -5.10 μC, and the charge on the outer surface of the shell is +10.20 μC.
For more such information on: Charge
https://brainly.com/question/2373424
#SPJ8
Oil-water separation is a very important process in petroleum engineering, as well as in industries that produces oily wastewater and other oil-water pollution. The basic process of oil-water separation is illustrated in the figures below. From left to right, the figures show a simple emulsion in which dispersed (Dutch: verspreidt) oil droplets in water coalesce (come together to form one mass) as time passes. Assume that the number of oil droplets decreases with time (in minutes) following the equation: ൌ െ0.02 a) Calculate how long it takes for half of the oil droplets to coalesce. (3p) b) How long will it take for practically all the oil droplets to coalesce? (2p) c) Provide a sketch of the number of oil droplets as function of time in the oily water. Add relevant values along the axis. (1p) To speed up the process of coalescence, a chemical component (surfactant) is added to the oily water. Assume that now the number of oil droplets decreases with time following the equation: ൌ െ0.02ଶ d) After adding the surfactant to the mixture, can the process still be considered a natural process? Explain. (1p) e) Show that, after adding the surfactant to the oil-water mixture, less time is needed for half of the oil droplets to coalesce. (
The time required for half of the oil droplets to coalesce is 34.65 minutes.b) Calculation to find how long it takes for practically all of the oil droplets to coalesce:To find the time it would take for practically all of the oil droplets to coalesce, we need to use the following formula and solve for time when n is equal to 0.0 = e^(-0.02t)-infinity = -0.02tNo oil droplets remain after an infinite amount of time. Therefore, it takes an infinite amount of time for all the oil droplets to coalesce.Answer: It takes an infinite amount of time for all the oil droplets to coalesce.
Sketch:We need to sketch the number of oil droplets as a function of time in the oily water. The axis should contain relevant values. As the oil droplets coalesce, the number of oil droplets decreases with time. This is shown by the line in the graph below.Answer: The figure is attached below. The x-axis shows the time in minutes, and the y-axis shows the number of oil droplets present. After adding the surfactant to the mixture, can the process still be considered a natural process? Explain. When a chemical component such as a surfactant is added to the oily water, the oil droplets are artificially made to coalesce. As a result, the process is no longer a natural one because it has been modified by human intervention. Answer: No, the process can no longer be considered natural, because a surfactant was added to the oily water to speed up the coalescence of the oil droplets. Show that less time is needed for half of the oil droplets to coalesce after adding the surfactant to the oil-water mixture. In order to find how long it takes for half of the oil droplets to coalesce after adding the surfactant, we will utilize the new equation that incorporates the surfactant.0.5 = e^(-0.02x)ln(0.5) = -0.02x17.32 minutes = xAnswer: After the addition of the surfactant, it only takes 17.32 minutes for half of the oil droplets to coalesce. This is less time than the original 34.65 minutes that it took before the surfactant was added.For such more question on coalesce
https://brainly.com/question/28150659
#SPJ8
Equation of path of projectile is y=x(1-x)
The equation of the path of a projectile is y = x(1 - x). Projectile motion is the movement of an object in a parabolic trajectory as a result of being propelled or released under the influence of gravity. A projectile, in simple terms, is any object that is launched into the air, such as a bullet, a baseball, or a rock. It's important to note that the parabolic trajectory is due to the force of gravity acting on the object.
To better understand projectile motion, we must first examine the horizontal and vertical components of motion. The horizontal component of motion is constant, indicating that there is no acceleration in that direction. The vertical component, on the other hand, has acceleration because of gravity. The parabolic trajectory is the result of these two components of motion.As the projectile is launched, it travels a certain distance horizontally before beginning to descend as a result of gravity's influence, resulting in a parabolic path.The general formula for the trajectory of a projectile in two dimensions is given by:y = xtanθ - (gx²) / 2(v₀cosθ)²Where:y is the vertical distance covered by the projectilex is the horizontal distance covered by the projectileθ is the angle of projectiong is the acceleration due to gravityv₀ is the initial velocity of the projectile In the case of the given equation, y = x(1 - x), the path of the projectile will be a parabolic trajectory with a vertex at x = 0.5 and y = 0.25. The equation represents the projectile's vertical distance, y, as a function of its horizontal distance, x. It is crucial to note that the projectile's initial velocity and angle of projection are not considered in this equation.For such more question on acceleration
https://brainly.com/question/460763
#SPJ8
A 50.0-kg wagon is pulled with a constant force of 380 N. Neglecting friction, the wagon's acceleration will be
A 50.0-kg wagon is pulled with a constant force of 380 N. Neglecting friction, the wagon's acceleration will be 7.6 m/s².
When a constant force acts on a wagon, it causes the wagon to accelerate We can calculate the wagon's acceleration using Newton's second law, which states that the force acting on an object is directly proportional to its acceleration, and the acceleration is inversely proportional to the mass of the object.A = F/mHere,A = AccelerationF = Force acting on the wagon m = mass of the wagon Substituting the given values, we getA = 380 N/50.0 kgA = 7.6 m/s²Therefore, the acceleration of the wagon is 7.6 m/s² when it is pulled with a constant force of 380 N, neglecting friction.For such more question on Newton's second law
https://brainly.com/question/25545050
#SPJ8
