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

The force needed to accelerate a mass of 40 kg from velocity v₁ = (4î - 5) + 3k)m/s to v = (8î + 3) - 5k)m/s in 10 seconds is 12.4 N.

Start by calculating the change in velocity (Δv) experienced by the object. This can be done by subtracting the initial velocity v₁ from the final velocity v.

Δv = v - v₁ = ((8î + 3) - 5k) - ((4î - 5) + 3k)

= 8î + 3 - 5k - 4î + 5 - 3k

= 4î - 8k + 8

Next, calculate the acceleration (a) using the formula:

a = Δv / t

where t is the time interval, given as 10 seconds.

a = (4î - 8k + 8) / 10

= (0.4î - 0.8k + 0.8) m/s²

The force (F) required to accelerate the object can be found using Newton's second law:

F = m * a

where m is the mass, given as 40 kg.

F = (40 kg) * (0.4î - 0.8k + 0.8) m/s²

= (16î - 32k + 32) N

Simplify the expression to obtain the final answer:

F = 16î - 32k + 32 N

≈ 12.4 N

Therefore, the force needed to accelerate a mass of 40 kg from velocity v₁ = (4î - 5) + 3k)m/s to v = (8î + 3) - 5k)m/s in 10 seconds is approximately 12.4 N.

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The frictional force acting on a 20 kg block that is being pushed with 50 N of force and the weight of the block are calculated as follows:Frictional force:The frictional force is a force that opposes motion between two surfaces that are in contact. It's typically symbolized by the letter f.

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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).

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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.

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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.

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Answer:

50kgm/2

Explanation:

The alternating current (AC) in the primary circuit produces an induced current in the secondary current due to the phenomenon of electromagnetic induction.

However, a constant DC source will not induce a current in the secondary coil. When an alternating current (AC) flows through the primary coil of a transformer, it produces an alternating magnetic field around it. The magnetic field around the coil expands and contracts with each cycle of AC, inducing a voltage in the secondary coil, which is located near it.

This voltage induced in the secondary coil is alternating as well, with the same frequency as the primary AC, but with a different amplitude, which is determined by the transformer's turns ratio and impedance. However, when a direct current (DC) flows through the primary coil, it generates a static magnetic field around it that does not change with time.

The magnetic field does not expand or contract and does not induce any voltage or current in the secondary coil since electromagnetic induction needs a changing magnetic field. Therefore, a constant DC source will not induce a current in the secondary coil.

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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.

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Granite will heat up the quickest among dry soil, granite, water, and wet soil due to its lower specific heat capacity.

To determine which substance will heat up the quickest among dry soil, granite, water, and wet soil, we need to consider their specific heat capacities. The substance with the lowest specific heat capacity will heat up the quickest.

Specific heat capacity is defined as the amount of heat energy required to raise the temperature of a substance by a certain amount. Different substances have different specific heat capacities due to variations in their molecular structure and composition.

Water has a relatively high specific heat capacity of about 4.18 J/g°C, meaning it requires a significant amount of heat energy to raise its temperature. Dry soil and granite, on the other hand, have lower specific heat capacities compared to water.

Wet soil is a mixture of dry soil and water, and its specific heat capacity will lie between the values of dry soil and water. Since water has a higher specific heat capacity than dry soil, wet soil is expected to have a higher specific heat capacity than dry soil as well.

Therefore, based on the comparison of specific heat capacities, the substance that will heat up the quickest is granite. Granite has a lower specific heat capacity than water and wet soil, making it more susceptible to temperature changes. Dry soil and wet soil, including water, will heat up at a slower rate compared to granite.

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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,

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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.

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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:

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].

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.

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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

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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.

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.

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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.

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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.

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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.

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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.

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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.

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