A particle with positive charge q = 3 μC is at a distance of 1.5 cm from a long vertical wire that carries a current I = 5 A. The particle is traveling with a speed v = 1000 m/s perpendicular to the wire and is moving away from the wire. What are the magnitude and direction of the force on the particle?Group of answer choices1.0 x 10-8 N radially away from the wire1.0 x 10-8 N radially towards the wire2.0 x 10-8 N parallel to, but in the opposite direction of the current flow2.0 x 10-7 N parallel to, and in the direction of the current flow2.0 x 10-8 N radially away from the wire

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Answer:  two solid electrodes placed in an electrolyte connected together by an electrical conductor such as wire.

Explanation:

The two electrodes must be two different metals. The electrolyte can be acid solution, alkaline solution, salt solution or even a fruit such as orange or lemon.

If A 5kg objects is sliding across a floor at 10m/s.then the work done by friction to bring the 5 kg object to a stop is -250 Joules.

To calculate the work done by friction to bring the object to a stop, we need to determine the change in kinetic energy.

Given:

Mass of the object, m = 5 kg

Initial velocity, u = 10 m/s

Final velocity, v = 0 m/s (object comes to a stop)

The work done by friction can be calculated using the equation:

Work = Change in Kinetic Energy

The change in kinetic energy (ΔKE) can be calculated as:

ΔKE = (1/2) * m * (v^2 - u^2)

Plugging in the values:

ΔKE = (1/2) * 5 kg * (0 m/s)^2 - (10 m/s)^2

= (1/2) * 5 kg * (0 - 100 m^2/s^2)

= (1/2) * 5 kg * (-100 m^2/s^2)

= -250 J

The negative sign indicates that the work done by friction is in the opposite direction of the displacement of the object.

Therefore, the work done by friction to bring the 5 kg object to a stop is -250 Joules.

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The input work of a scissor jack is 8800

It is given that

Efficiency of the jack = 80%

output work = 8800

We have to find input work of a scissor jack

Force applied across a distance is called work. Examples of work include dragging down a captive helium balloon, driving a car up a hill, and lifting an object against the gravitational attraction of the Earth. Energy manifests mechanically as work. The joule (J), or newton-meter (N m), is the accepted unit of work.

Efficiency = work output / work input

work input = work output / efficiency

work input = 8800 / 80%

work input = 8800

Hence, input work of a scissor jack is 8800

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Answer:

Explanation: please see attached file I attached the answer to your question.

The angle between the magnetic field and the wire's velocity is 33.2 degrees.

Since the potential difference = 71mv = 71 *10 ^-3 V

The length is 12 cm = 0.12m

The magnetic field i.e. B = 0.27T

The speed or v = 4 m/s

here we assume \(\theta\) be the angle

So,

e = Bvl sin\(\theta\)

So,

\(Sin\theta\) = e/bvl

= 71*10^-3 / 0.27 *4*0.12

= 0.5478

= 33.2 degrees

Therefore, the angle should be 33.2 degrees

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The change in motion, visible effects of the collision, and temperature increase would all serve as evidence that energy was transferred during the collision between the cart and the track after the removal of electrostatic forces.

If the electrostatic forces on the cart and track were suddenly removed and the cart hits the track, there would be several pieces of evidence indicating that energy was transferred during the collision.Firstly, there would be a noticeable change in the motion of the cart. The cart would decelerate as it collides with the track, and its velocity would decrease due to the loss of energy. This change in motion demonstrates that energy has been transferred from the cart to the track.

Secondly, there may be visible effects of the collision, such as deformation or damage to the cart or the track. This deformation or damage occurs because the energy transferred during the collision is transformed into other forms, such as heat or sound energy. These visible effects provide evidence of energy transfer.Additionally, there could be an increase in the temperature of the cart and/or the track due to the conversion of kinetic energy into thermal energy upon impact. This temperature change would be another indication that energy has been transferred.

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Answer:

Therefore, the 10 kg mass should be placed at x = 5.7 m along the x-axis to achieve a center of mass located at y = 4.5 m.

Explanation:

To find the position along the x-axis where a 10 kg mass should be placed such that the center of mass is located at y = 4.5, we can use the formula for the center of mass:

x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)

Here, m1 and x1 represent the mass and position of the 8 kg mass, respectively. m2 is the mass of the 10 kg mass, and we need to find x2, its position.

Given:

m1 = 8 kg

x1 = 3 m

x_cm = unknown (to be found)

m2 = 10 kg

y_cm = 4.5 m

Since the center of mass is at y = 4.5, we only need to consider the y-coordinate when calculating the center of mass position along the x-axis.

To solve for x2, we can rearrange the formula as follows:

x2 = (x_cm * (m1 + m2) - m1 * x1) / m2

Substituting the given values:

x2 = (x_cm * (8 kg + 10 kg) - 8 kg * 3 m) / 10 kg

Simplifying:

x2 = (x_cm * 18 kg - 24 kg*m) / 10 kg

Now, we can set the y-coordinate of the center of mass equal to 4.5 m and solve for x_cm:

4.5 m = (8 kg * 3 m + 10 kg * x2) / (8 kg + 10 kg)

Simplifying:

4.5 m = (24 kg + 10 kg * x2) / 18 kg

Multiplying both sides by 18 kg:

81 kg*m = 24 kg + 10 kg * x2

Subtracting 24 kg from both sides:

10 kg * x2 = 81 kg*m - 24 kg

Dividing both sides by 10 kg:

x2 = (81 kg*m - 24 kg) / 10 kg

Simplifying:

x2 = 8.1 m - 2.4 m

x2 = 5.7 m

(brainlest?) ples:(

Answer:

the 10 kg mass should be placed at x = -2.4 m to achieve a center of mass at y = 4.5 m.

Explanation:

To find the position along the x-axis where the 10 kg mass should be placed so that the center of mass is located at y = 4.5, we can use the principle of the center of mass.

The center of mass of a system is given by the equation:

x_cm = (m1x1 + m2x2) / (m1 + m2),

where x_cm is the x-coordinate of the center of mass, m1 and m2 are the masses, and x1 and x2 are the positions along the x-axis.

Given:

m1 = 8 kg,

x1 = 3 m,

m2 = 10 kg,

y_cm = 4.5 m.

To solve for x2, we need to find the x-coordinate of the center of mass (x_cm) by using the y-coordinate:

y_cm = (m1y1 + m2y2) / (m1 + m2),

where y1 and y2 are the positions along the y-axis.

Rearranging the equation and substituting the given values:

4.5 = (83 + 10y2) / (8 + 10).

Simplifying the equation:

4.5 = (24 + 10*y2) / 18.

Multiplying both sides by 18:

81 = 24 + 10*y2.

Rearranging the equation:

10*y2 = 81 - 24,

10*y2 = 57.

Dividing both sides by 10:

y2 = 5.7.

Therefore, the y-coordinate of the 10 kg mass should be 5.7 m.

To find the x-coordinate of the 10 kg mass, we can use the equation for the center of mass:

x_cm = (m1x1 + m2x2) / (m1 + m2).

Substituting the given values:

x_cm = (83 + 10x2) / (8 + 10).

Since the center of mass is at x_cm = 0 (the origin), we can solve for x2:

0 = (83 + 10x2) / (8 + 10).

Rearranging the equation:

83 + 10x2 = 0.

24 + 10*x2 = 0.

10*x2 = -24.

Dividing both sides by 10:

x2 = -2.4.

The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.

To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.

The force between two charged particles can be calculated using Coulomb's law:

F = k * |q₁ * q₂| / r²

Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

First, let's calculate the force exerted on q₂ by q₁:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²

F₁₂ ≈ 252000 N

The force is positive because q₁ and q₂ have opposite charges.

Next, let's calculate the force exerted on q₂ by q₃:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²

F₂₃ ≈ -63000 N

The force is negative because q₂ and q₃ have the same charge.

Finally, we can find the net force on q₂ by summing the individual forces:

Net force = F₁₂ + F₂₃

Net force = 252000 N + (-63000 N)

Net force ≈ 189000 N

The net force on particle q₂ is approximately 189000 N.

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Answer:

Electrons in different metals

Explanation:

When you heat an atom, some of its electrons are "excited* to higher energy levels. When an electron drops from one level to a lower energy level, it emits a quantum of energy.  The different mix of energy differences for each atom produces different colours. Each metal gives a characteristic flame emission spectrum.

The result of the ratio of the displacement module of the second body at the point of meeting is 0.5.

To determine the ratio of the displacement of the first body to the displacement of the second body at the moment of their meeting, use the equation of motion:

d = vt + 1/2at²

where d is the displacement, v is the initial velocity, t is the time, and a is the acceleration.

Since the bodies are moving simultaneously towards each other, then assume that their initial velocities are zero. Also, at the moment of their meeting, their displacement will be the same, d₁ = d₂.

Assume that the time at which they meet is t, then:

d₁ = 1/2 * 2.4t²

And the equation for the displacement of the second body:

d₂ = 1/2 * 4.8t²

If d₁ = d₂

then, 1/2 * 2.4t² = 1/2 * 4.8t²

Solving this equation for t and substituting it into the equation for d₁ or d₂, the ratio of the displacement of the first body to the displacement of the second body: d₁/d₂ = 2.4/4.8 = 0.5 or 1/2

So, at the moment of their meeting, the displacement of the first body is half of the displacement of the second body.

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Answer:

b

Explanation:

Answer:

\(F=3.6\times 10^6\ N\)

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, \(r=6.37\times 10^6\ m\)

The mass of the Earth, \(m_e=5.97 \times 10^{24}\ kg\)

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 :

\(F=G\dfrac{mm_e}{(r+h)^2}\\\\F=6.67\times 10^{-11}\times \dfrac{419455\times 5.97 \times 10^{24}}{(6.37 \times 10^6+400000 )^2}\\\\F=3.6\times 10^6\ N\)

So, the required force on the station due to Earth's gravity is \(3.6\times 10^6\ N\).

The force acting on the electron is 1.92 x 10^-17 N.

The problem states that an electron of charge 1.6 x 10^-19 is located in a uniform electric field of 120 Vm^-1, and it asks us to determine the force acting on it.

We can use Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. If the charges are of opposite signs, the force is attractive, while if the charges are of the same sign, the force is repulsive.

The formula for Coulomb's law is F = kq1q2/r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

Since the electron has a charge of 1.6 x 10^-19 C, and the electric field strength is 120 Vm^-1, we can use the equation F = qE to find the force acting on it.

F = qE = (1.6 x 10^-19 C)(120 Vm^-1) = 1.92 x 10^-17 N.

Therefore, the force acting on the electron is 1.92 x 10^-17 N.

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The compressive force that would be enough to produce a fracture is 4000 N.

We know that the femur is one of the most important bones that we have in the human body. In this case, we have been told that In the human femur, bone tissue is strongest in resisting compressive force, approximately half as strong in resisting tensile force, and only about one- fifth as strong in resisting shear force.

Then we know that;

Tensile force = 8000

The compressive force would be half of this magnitude as such;

Compressive force = 4000 N

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Answer:

Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons.

Answer:

thomsons experiment he used ray tubes that showed all electrons contain negatively cahrged particles

Answer:

A. The long, medium, and short straws all mixed together and placed

in the box

With the use of first and third equation of motion, the distance covered by a T-Rex is 30.51 m

When a body is in linear motion, the body is moving in a straight line. some of the parameters to consider are:

Given that a T-Rex move from 0 m/s to 9 m/s in 6.78 seconds, the distance covered can be found by calculating the acceleration.

Let us use equation 1

V = U + at

9 = 0 + 6.78a

a = 9 / 6.78

a = 1.33 m/\(s^{2}\)

Now let us use equation 3

\(v^{2}\) = \(u^{2}\) + 2as

\(9^{2}\) = 2 x 1.33 x S

81 = 2.655S

S = 81/2.655

S = 30.51 m

Therefore, the distance covered by a T-Rex is 30.51 m.

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An airplane pilot sets a compass course due west and maintains an airspeed of 225  \(\frac{km}{h}\) . After 0.500 hours of flight, she reaches a town 121 kilometers to the west and 14 kilometers to the south of her starting place. Her velocity at this moment is 240  \(\frac{km}{h}\) .

Using south as the negative x direction and west as the positive

positive y direction as the direction.

The aircraft's speed is= 220 \(\frac{km}{h}\)due west.

Then \(v_{x}\) = 220  \(\frac{km}{h}\)and \(v_{y}\)=0

The position of the aircraft at time t = 0.500h is 120 km west and 20 km south.

then \(S_{x}\) = 120 km and \(S_{y}\) = 20 km

Then

\(v_{x}\)= \(\frac{S_{x} }{t}=\frac{120}{0.500}\) = 240   \(\frac{km}{h}\)

\(v_{y}\)=\(\frac{S_{y} }{t}=\frac{20}{0.500}\)= 40 \(\frac{km}{h}\)

Wind velocity is a 40  \(\frac{km}{h}\) due south.

Then \(v_{windx}\)=0 and \(v_{windy}\) = 40 km/h

Then  \(v_{x}\)=\(v_{aero x} +v_{wind x}\)

\(v_{aerox}=240-0\)

\(v_{aerox}=240\)  \(\frac{km}{h}\)

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Answer:

28.1 degrees

Explanation:

Use formula: \(N_1Sin(Angle_1) = N_2Sin(Angle_2)\)

* use angles given

Answer:

Net force = 11340 N

Explanation:

Given that,

Mass of an automobile, m = 2100 kg

Acceleration of the automobile, a = 5.4 m/s²

We need to find the net force required for the automobile. The net force is the product of mass and acceleration. It can be given by the formula as follows :

\(F=ma\\\\F=2100\ kg\times 5.4\ m/s^2\\\\F=11340\ N\)

So, the net force is 11340 N.

\(\blue{\bold{\underline{\underline{Answer:}}}}\)

\(\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}\)

\(\green{\underline{\bold{Given :}}} \\ \tt: \implies Constant \: force(F) = 8 \: N \\ \\ \tt: \implies Displacement(s) = 3 \: m \\ \\ \red{\underline{\bold{To \: Find : }}} \\ \tt: \implies Work \: Done(W.D) = ?\)

• According to given question :

\( \green{ \star} \tt \: \theta \: = 0 \degree \: \: \: \: (Angle \: between \: force \: and \: displacement) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: Done = FS \: cos \: \theta \\ \\ \tt: \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\ \\ \green{ \circ} \tt \: cos \: 0 \degree = 1 \\ \\ \tt: \implies Work \: Done =24 \times 1 \\ \\ \green{\tt: \implies Work \: Done =24 \: J}\)

If I were an astronaut, the first job I would do on the International Space Station (ISS) would be to familiarize myself with the station and its systems.

I would need to learn how to operate the various equipment and how to maintain the station in good working order. I would also need to learn the procedures for conducting experiments and for performing spacewalks.

Once I had a good understanding of the station and its systems, I would begin working on my assigned tasks. These tasks could include conducting experiments, performing maintenance, or teaching other astronauts new skills. I would also take the opportunity to conduct research on my own and to learn more about the space environment.

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Answer:................

Explanation:.............

The electric field is -641.25 volt, potential energy is -1.76x 10⁻³ J.

What is electric field ?

The term "electric field" or "electrostatic field" refers to the area surrounding an electric charge where stress or an electric force occur. A significant amount of tension might be produced in the area if the charge is of a big magnitude. The letter E is used to denote the electric field. Newtons per coulomb, or volts per metre, is the SI unit for the electric field.

What is potential energy ?

Potential energy is energy that is preserved or stored in a material or an item. The item or substance's position, organization, or condition determines the amount of stored energy. Consider it as energy with the "potential" to accomplish work.

Va = kq1/r1 + Kq2/r2

Va = 9 x 10/0.05 (2.3 x 10⁻⁹ – 6x 10⁹)

Va = -666 volt

Vb = 9x10⁹ (2.3x10⁻⁹/.08 – 6x10⁻⁹/.06)

Vb = -641.25 volt

Work done by electric field from b to a

W e = q(Vb-Va)  

W e = 3x10⁻⁹ (-641.25+666)

W e = 74.25x10⁻⁹ J

W e = qEd = 30x 10⁻⁹x 3.2x10⁴x.63

W e = 6.05x10⁻⁴ J

W e = -qEd cos 45

W e = -30x10⁻⁹x3.2x10⁴x2.6 cos 45

W e = -1.76x 10⁻³ J

Therefore, electric field is -641.25 volt, potential energy is -1.76x 10⁻³ J.

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Answer:

The net force on the desk is zero.

Explanation:

Find the time it takes to hit the  bottom....then multiply this time by the horizontal velocity .......

Time to hit bottom :

       d = 1/2 at^2

        9 m  = 1/2 (9.81 m/s^2) (t^2)    shows  t = 1.35 s

Now the horizontal displacement is

      x = rate * time = 24.40 m/s * 1.35 s = 33.1 m

We are given:

Frequency of the wave = 10 Hz

Speed of the wave = 25 m/s

We can use the formula:

speed = frequency x wavelength

Rearranging the formula, we get:

wavelength = speed / frequency

Substituting the given values, we get:

wavelength = 25 m/s / 10 Hz

wavelength = 2.5 m

Therefore, the wavelength of the wave is 2.5 m.

Answer:

B = 0.62 T

Explanation:

       \(B= \mu_{o} *n*I (1)\)

The average speed of the motorist in the second half of the journey is determined as 7 km/h.

The average speed of an object is the ratio of total distance to total time of motion.

21 = (35 + x) / 2

35 + x = 42

x = 42 - 35

x = 7 km/h

Thus, the average speed of the motorist in the second half of the journey is determined as 7 km/h.

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