a) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric force that q1 and q2 exert on a third charge q3 = +2.00 nC on the x -axis at x = 4.00 cm is in the +x-direction and has magnitude 49.0 μN. Find the x-component of the force that q1 exerts on q3.


b) Find the x-component of the force that q2 exerts on q3.


c) Find the charge q2

The velocity in which this second particle finds itself at a distance r2 = 25,0 cm from point P is 17.22 m/s.

The electric force between the two particles is calculated from Coulomb's law of electrostatic force.

F = kq₁q₂/r²

where;

F = (9 x 10⁹ x 3.10 x 10⁻⁶ x 3.10 x 10⁻⁶) / (0.09 m)²

F = 10.68 N

The velocity in which this second particle finds itself at a distance r2 = 25,0 cm from point P is calculated by applying the principle of conservation of energy.

W = ΔK.E

where;

Fr = ¹/₂mv²

2Fr = mv²

v² = 2Fr/m

v = √(2Fr/m)

v = √(2 x 10.68 x 0.25 / 1.8 x 10⁻²)

v = 17.22 m/s

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The centripetal force is obtained by its formulas is,\(\rm F_C= \frac{mv^2}{r}\)

The acceleration needed to move a body in a curved way is understood as centripetal acceleration.

The direction of centripetal acceleration is always in the path of the center of the course. The total acceleration is the result of tangential and centripetal acceleration.

The formula for the centripetal force is ;

\(\rm F_C= \frac{mv^2}{r}\)

Where,

r is the radius

v is the linear speed

m is the mass of an object

Hence, the centripetal force obtained by its formulas as,\(\rm F_C= \frac{mv^2}{r}\)

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The amount of work done by the gas is proportional to the pressure and the change in volume, as well as the efficiency of the process. If the pressure and volume are known, the work done by the gas can be calculated by multiplying these values by the efficiency of the process.

The amount of work done by a gas when it expands is proportional to the change in volume, pressure, and temperature. According to the first law of thermodynamics, the energy of a closed system is conserved, so the work done by the expanding gas is equal to the energy transferred from the gas to the environment in the form of work. Therefore, the work done by the gas is equal to the change in energy of the system. Assume that the process is 10% efficient. Then, only 10% of the energy available to the system is converted into work. This means that the remaining 90% of the energy is lost to the environment in the form of heat. As a result, the amount of work done by the gas expanding into the atmosphere is given by the formula

W = E x η, where W is the work done by the gas, E is the energy available to the system, and η is the efficiency of the process. The energy available to the system is determined by the difference between the internal energy of the gas before and after the expansion. The internal energy of a gas is determined by its temperature, pressure, and volume.

Assuming that the temperature and pressure are constant, the change in internal energy is proportional to the change in volume. Therefore, the energy available to the system is equal to the product of the pressure and the change in volume: E = P x ΔV, where P is the pressure of the gas and ΔV is the change in volume during the expansion. Substituting this equation into the formula for work, we get W = P x ΔV x η.

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Cheese is made of casein protein, which is originally made from milk and is high in protein. The texture of the cheese is determined by the quality of the milk, so cow's milk cheese differs from goat's milk cheese.

Diary products are made from milk, such as cheese, curd, yogurt, etc., but all of them have different nutrients. The quality of these dairy products depend upon the quality of the milk, as some milking animals have a higher concentration of fats in their milk than other animals. These dairy products are used in different industrial sectors, such as for making ice cream, chocolate, and different food products.

Hence, cheese is made up of casein proteins that are present in the milk.

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The given statement is true.

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Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

The equivalent resistance of this combination of resistors is 2.9Ω.

The combined resistance in such arrangement of resistors is provided by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

here.

Req means  the equivalent resistance and R1, R2, R3

.Rn means the resistance of individual resistors interlinked in parallel.

Also,

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

So,

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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The acceleration of skateboard on road A will be 7.74 \(m/s^{2}\)

The acceleration of skateboard on road B will be 3.87 \(m/s^{2}\)

The acceleration of skateboard on road C will be 9.68 \(m/s^{2}\)

force = mass * acceleration

mass1 = 62 kg

force1 = 480 N

acceleration1 = force / mass

                     = 480 / 62 = 7.74 \(m/s^{2}\)

mass2 = 62 kg

force2  = 240 N

acceleration 2  = 240 / 62 = 3.87  \(m/s^{2}\)

mass 3 = 62 kg

force 3  = 600 N

acceleration 3  = 600 / 62  = 9.68  \(m/s^{2}\)

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At the given erro in angle, the error in the measurement of sin 90 degrees would be 0.001.

The percentage error of any measurement is obtained from the ratio of the error to the actual measurement.

The error of sin 90 degrees is calculated as follows;

sin 90 = 1

error in measurement = sin(90 - 0.5)

error in measurement = sin(89.5) = 0.999

Error in sin 90 degrees = 1 - 0.999

Error in sin 90 degrees = 0.001

Thus, at the given erro in angle, the error in sin 90 degrees would be 0.001.

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Answer: A literature review consists of an overview, a summary, and an evaluation (“critique”) of the current state of knowledge about a specific area of research.

Explanation:

Answer:

16 meters

Explanation:

When the rope is suddenly cut the object moving tangent at the circle. In that moment the gravity act in the object making it falls.

First we need to find how much time de object take to reach at the ground.

g=acceleration of gravity=10m/s²

v= vertical velocity =0m/s

h=vertical altitude =20m

We will find t such that h(t)=0

\(0=20-5t^{2} \\\\5t^{2} =20\\\\t^{2} =4\\\\t=2s\)

v=horizontal velocity

D(t=2) is the horizontal distancetravelled by the object:

\(D(2)=8*2\\\\D(2)=16m\)

Correct question is;

A tiny water droplet of radius 0.010 cm descends through air from a high building .Calculate its terminal velocity . Given that η of air = 19 × 10^(-6) kg/m.s and density of water ρ = 1000kg/ms

Answer:

1.146 m/s

Explanation:

We are given;

Radius; r = 0.010 cm = 0.01 × 10^(-2) m

η = 19 × 10^(-6) kg/m.s

ρ = 1000 kg/ms

The formula for the terminal velocity is given by;

V_t = 2r²ρg/9η

g is acceleration due to gravity = 9.8 m/s²

Thus;

V_t = (2 × (0.01 × 10^(-2))² × 1000 × 9.8)/(9 × 19 × 10^(-6))

V_t = 1.146 m/s

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

F is an applied force

r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

This is the moment of inertia.

Thank you!

The Phase difference between points A and B is zero of a circular wave.

                    =  5 pi/2-5pi/2=0

b) What is the phase difference between points C and D?

  Phase difference between points C and D will be

                         = 9pi/2- pi/2=4pi

c) What is the phase difference between points E and F?

    Phase difference between E and F will be

                     = 5pi/2 - 2pi/2= pi

The phase difference of a sine wave can be defined as "the time interval by which one wave leads or lags another", and phase difference is not only a property of one wave, but also the relative property of two or more waves. am. This is also called the "phase angle".

The phase difference between two sound waves of the same frequency passing through a fixed position is given by the time difference between the same position within the wave cycle of the two sounds.

Phase difference is the difference in phase angle between two sine waves or phasors. In a three-phase system, the phase difference between conductors is one-third of the period.

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We will have that each 10-decible the increment is by a factor of 10, so:

So, it is 10 000 times louder.

Answer:

85

Explanation:

soln

given that;

frequency=17Hz

wavelength=5m

speed?

formula for wavelength is;

wavelength= speed/frequency

then ; making v the subject formula

we have that v=wavelength*frequency

v=17*5=>85ms

The rate of heat loss, in watts, does this achieve is 37.66 W

It leads in cooling since water absorbs heat equivalent to mass times latent heat of evaporation to get converted into vapor .

So,

latent heat of evaporation of water = 2260 x 10³ J / kg or 2260 J / g

Now

in the evaporation of 1 g of water , heat lost = 2260 J

And,

heat lost per minute = 2260 J

So,

heat lost per second = 2260 / 60

= 37.66 J /s

= 37.66 W

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Before the ball hits the ground, it is 2.8 seconds in the air. The result is obtained by using the equations in uniformly accelerated straight motion.

The equations apply in uniformly accelerated straight motion in vertical dimension are

v₁ = v₀ + gt

v₁² = v₀² + 2gh

h = v₀t + ½ gt²

Where

A student standing on the ground throws a ball straight up with

Find the time it takes for the ball to reach the ground!

We use g = 9.8 m/s². See the illustration picture in the attachment!

The ball will go upward and stop at a certain height with v₁ = 0. The time needed is

v₁ = v₀ - gt₁

0 = 13.0 - 9.8t

13.0 = 9.8t

t₁ = 13.0/9.8

t₁ = 1.3 s

The height above the hand when the ball stops is

v₁² = v₀² - 2gh₂

0 = 13.0² - 2(9.8)h₂

13.0² = 2(9.8)h₂

169 = 19.6h₂

h₂ = 8,62 m

The ball stops at a height of

h₃ = h₁ + h₂

h₃ = 2.50 + 8.62

h₃ = 11.12 m

The ball goes downward and reach the ground. Initial velocity in this condition is v₁ = 0. The time needed is

h₃ = v₁t + ½ gt₂²

11.12 = 0 + ½ (9.8)t₂²

11.12 = 4.9t²

t₂² = 2.27

t₂ ≈ 1.5 s

The time that the ball in the air is

t = t₁ + t₂

t = 1.3 + 1.5

t = 2.8 s

Hence, the ball is in the air for 2.8 seconds.

Your question is incomplete, but most probably your full question was

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the hand is 2.50 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way).

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A heat engine, symbolized by a circle in this illustration, uses some of the heat transfer to produce work.

The most effective heat engine that is theoretically possible—i.e., any machine that can transform thermal energy (heating) into mechanical energy—is the Carnot engine (work).The heat and cold reservoirs are the names given to the hot and cold items. Only the hot temperatures' absolute values affect efficiency. A portion of the heat QH that is extracted from the high-temperature source TH in the case of a heat engine is transformed into work. The ratio of net work completed to heat absorbed throughout one full cycle is known as a heat engine's thermal efficiency.

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Complete Question

A proton of mass m​p​​= 1.67×10​−27​​ kg and a charge of q​p​​= 1.60×10​−19​​ C is moving through vacuum at a constant velocity of 10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E = 3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region. Answer is to be in units of meters

Answer:

    \(s = 0.039 \ m\)

Explanation:

From the question we are told that

    The  mass of the proton is  \(m = 1.67 *10^{-27} \ g\)

    The charge of on the proton is \(q = 1.60 *10^{-19} \ C\)

      The speed of the proton is  \(v = 10000 \ m/s\)

     The magnitude of the electric field is  \(E = 3.62*10^{3 } \ N/C\)

       The width covered by the electric field    \(d = 5mm = 5 *10^{-3} \ m\)      

Generally the acceleration of the proton due to the electric toward the south  (at the point where the force on the proton is equal to the electric force due to the electric field) is  mathematically represented as

       \(a = \frac{q* E}{m}\)  

Substituting values

       \(a = \frac{1.60*10^{-19 } * 3.26 *10^{3}}{ 1.67*10^{-27}}\)

      \(a = 3.12*10^{11} \ m/s^2\)

Generally the time it will take the proton to cross the electric field is  mathematically represented as

      \(t = \frac{d}{v}\)

Substituting values

      \(t = \frac{5 *10^{-3}}{10000}\)

     \(t = 5 *10^{-7} \ s\)

Generally the the distance covered by the proton toward the south is  

       \(s = ut + \frac{1}{2} * a*t^2\)

   Here  u = 0  m/s  this  because before the proton entered the electric field region the it velocity towards the south is  zero

     So

       \(s = \frac{1}{2} * a*t^2\)

Substituting values

      \(s = \frac{1}{2} * 3.12 *10^{11}*(5 *10^{-7})^2\)

      \(s = 0.039 \ m\)

Answer:

It is the carrier of genetic information.

Plate boundaries represent the parts of the Earth's crust where plates come in contact with one another. There are three types of plate boundaries based on the movement and interaction of the plates involved. These are: Divergent Plate Boundaries, Convergent Plate Boundaries, and Transform Plate Boundaries.

Divergent Plate Boundaries
At divergent plate boundaries, two plates move away from each other as magma rises to the surface and creates new crustal material. Examples of divergent plate boundaries include the Mid-Atlantic Ridge, the East Pacific Rise, and the African Rift Valley.

Convergent Plate Boundaries
At convergent plate boundaries, two plates move toward each other and eventually collide. Depending on the type of plate involved, different types of interactions can occur. The three types of convergent plate boundaries are oceanic-continental, oceanic-oceanic, and continental-continental. An example of oceanic-continental convergence is the Pacific Northwest region of the United States. An example of oceanic-oceanic convergence is the Japanese Islands, and an example of continental-continental convergence is the Himalayas.

Transform Plate Boundaries
At transform plate boundaries, two plates move past each other in a horizontal direction. These boundaries are characterized by faults and earthquakes, such as the San Andreas Fault in California.

To create an illustration that represents each type of plate movement, you can draw a diagram that shows the direction of plate movement, the type of boundary, and any notable geological features associated with that type of boundary.

For example, a divergent plate boundary illustration could include a depiction of magma rising to the surface and creating new crustal material, while a transform plate boundary illustration could include a fault line and a depiction of the earthquakes that occur along that boundary.

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(a) The new angular speed of the student is 2.34 rad/s, and (b) The kinetic energy of the student before the objects are pulled in is 8.22 J, and the kinetic energy of the student after the objects are pulled in is  24.8 J.

Angular speed, also known as rotational speed, is the measure of how fast an object rotates or revolves around a central point or axis, and it is measured in radians per second (rad/s) or degrees per second (°/s). It is a scalar quantity that describes the magnitude of the rotational velocity of an object.

(a) To solve for the new angular speed of the student, we can use the conservation of angular momentum. Initially, the student, stool, and objects have an angular momentum given by:

L = Iω

where I is the moment of inertia and ω is the angular speed. Since there is no external torque acting on the system, the angular momentum is conserved. Therefore, we can write:

I1ω1 = I2ω2

where I1 and ω1 are the initial moment of inertia and angular speed, and I2 and ω2 are the final moment of inertia and angular speed.

At the initial state, the moment of inertia of the student plus stool and objects is given by:

I1 = I_student + I_objects

where I_student is the moment of inertia of the student and stool and I_objects is the moment of inertia of the objects. The moment of inertia of the objects can be approximated as:

I_objects ≈ 2mr²

where m is the mass of one object and r is the distance from the axis of rotation to the object. Substituting the given values, we get:

I1 = 3.0 kg·m² + 2(3.6 kg)(1.0 m)² ≈ 29.04 kg·m²

At the final state, the moment of inertia is given by:

I2 = I_student + 2mr²

where r is the new distance from the axis of rotation to the objects. Substituting the given values, we get:

I2 = 3.0 kg·m² + 2(3.6 kg)(0.43 m)² ≈ 8.97 kg·m²

Substituting the known values and solving for ω2, we get:

I1ω1 = I2ω2

(29.04 kg·m²)(0.75 rad/s) = (8.97 kg·m²)ω2

ω2 ≈ 2.34 rad/s

So, the new angular speed of the student is approximately 2.34 rad/s.

(b) To solve for the kinetic energy of the student before and after the objects are pulled in, we can use the formula:

KE = (1/2)Iω²

where KE is the kinetic energy, I is the moment of inertia, and ω is the angular speed.

At the initial state, the kinetic energy is given by:

KE1 = (1/2)I1ω1² ≈ 8.22 J

Substituting the known values, we get:

KE1 ≈ (1/2)(29.04 kg·m²)(0.75 rad/s)² ≈ 8.22 J

At the final state, the kinetic energy is given by:

KE2 = (1/2)I2ω2²

Substituting the known values, we get:

KE2 ≈ (1/2)(8.97 kg·m²)(2.34 rad/s)² ≈ 24.8 J

So, the kinetic energy of the student before the objects are pulled in is approximately 8.22 J, and the kinetic energy of the student after the objects are pulled in is approximately 24.8 J.

Therefore, (a) The student's new angular speed is 2.34 rad/s, and (b) the student's kinetic energy before the objects are drawn in is around 8.22 J, and the student's kinetic energy after the things are pushed in is roughly 24.8 J.

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Answer: the answer is (C)

Explanation: Bar graphs show comparisons using number comparisons

Answer:

d

Explanation:

anser: 27s PoP

Explanation:

1 is the force, so the equal antiforce will be 1. weight of car divided by force from engine

34 pop / s = 129

27pOp

Answer:

what are the choices

Explanation:

Answer:

Landforms are a result of a combination of constructive and destructive forces. Collection and analysis of data indicates that constructive forces include crustal deformation, faulting, volcanic eruption and deposition of sediment, while destructive forces include weathering and erosion.

Explanation: