The force of attraction that a -40.0 μC point charge exerts on a +108 μC point charge has magnitude 4.00 N. How far apart are these two charges? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

Answer:

  2.5 m/s²

Explanation:

a = ∆v/∆t = (15 m/s)/(6 s) = (15/6) m/s² = 2.5 m/s²

Answer:

  15√2 N

Explanation:

The acceleration is given by ...

  a = F/m = 5t/5 = t . . . . meters/second^2

The velocity is the integral of acceleration:

  v = ∫a·dt = (1/2)t^2

This will be 9 m/s when ...

  9 = (1/2)t^2

  t = √18 . . . . seconds

And the force at that time is ...

  F = 5(√18) = 15√2 . . . . newtons

Answer:

k = 65.3 N/m

Explanation:

Fs = Fg = mg = (1.0 kg)(-9.8 m/s²) = -9.8 N

x = 15 cm = 0.15m

Fs = -kx  (Hooke's Law)

-k = Fs/x = (-9.8 N)(0.15 m) = -65.3

k = 65.3 N/m

The boxes acceleration up the ramp is approximately 0.109 m/s².

The forces acting on the box can be resolved into two components: one parallel to the incline (F_parallel) and one perpendicular to the incline (F_perpendicular).

Given:

Mass of the box (m) = 40.7 kg

Coefficient of kinetic friction (μ) = 0.17

The force pulling the box (F_p) = 173 N

Incline angle (θ₁) = 14.5°

Force angle (θ₂) = 25.7°

First, we need to calculate the components of the force pulling the box:

F_parallel = F_p * sin(θ₂)

F_perpendicular = F_p * cos(θ₂)

Next, let's calculate the force of friction:

F_friction = μ * (mass of the box) * g, where g is the acceleration due to gravity (approximately 9.8 m/s²)

The force component parallel to the incline is opposed by the force of friction, so:

Net force parallel to the incline (F_net_parallel) = F_parallel - F_friction

Now, we can calculate the acceleration using Newton's second law:

F_net_parallel = (mass of the box) * acceleration

Rearranging the equation, we get:

acceleration = F_net_parallel / (mass of the box)

Now we can substitute the values into the equations and calculate the acceleration

F_parallel = 173 N * sin(25.7°) ≈ 73.88 N

F_perpendicular = 173 N * cos(25.7°) ≈ 154.37 N

F_friction = 0.17 * (40.7 kg) * 9.8 m/s² ≈ 69.44 N

F_net_parallel = F_parallel - F_friction ≈ 73.88 N - 69.44 N ≈ 4.44 N

acceleration = (4.44 N) / (40.7 kg) ≈ 0.109 m/s²

Therefore, the box's acceleration up the ramp is approximately 0.109 m/s².

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Answer:

The meaning of life is one of the most profound and enduring questions that has plagued humanity for centuries. Philosophers, scientists, and theologians have all attempted to answer this question in their own ways, yet it remains an enigma that has yet to be fully understood. At its core, the meaning of life is a subjective concept that is shaped by individual beliefs, experiences, and values. However, there are several common themes and ideas that have emerged from various attempts to answer this question. One of the most prominent views on the meaning of life is that it is to find purpose and fulfillment. This view suggests that we should strive to find something that gives our lives meaning, whether it be through our work, relationships, or personal pursuits. This idea is often associated with the concept of happiness, as many believe that true happiness can only be achieved by finding purpose and meaning in one's life. Another

If a wave has an amplitude of 0.0800 m and is moving 7.33 m/s. The

wavelength of the wave is 1.69m.

The wavelength of a wave can be determined using the equation:

Wavelength = velocity / frequency

To determine the frequency we need to calculate the reciprocal of the time it takes for one complete oscillation.

frequency = 1 / time

frequency = 1 / 0.230

frequency ≈ 4.35 Hz

Substitute the values into the wavelength equation:

wavelength = velocity / frequency

wavelength = 7.33 / 4.35

wavelength ≈ 1.69m

Therefore the wavelength of the wave is approximately 1.69 meters.

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Answer:

The apparent depth d = 19.8495 cm

Explanation:

The equation for apparent depth can be expressed as:

\(d = \dfrac{d_1} {\mu_1}+\dfrac {d_2}{\mu_2}\)

here;

\(d_1 = d_2 = 15 \ cm\)

\(\mu_1\) = refractive index in the first liquid = 1.75

\(\mu_2\) = refractive index in the second liqquid= 1.33

∴

\(d = \dfrac{15}{1.75}+\dfrac{15}{1.33}\)

\(d = 15( \dfrac{1}{1.75}+\dfrac{1}{1.33})\)

\(d = 15( 0.5714 +0.7519)\)

d = 15(1.3233 ) cm

d = 19.8495 cm

The initial volume of the liquid is 31.4 cm³. The volume coefficient of thermal expansion of the liquid is 0.002 per degree Celsius. A temperature increase of 109.5°C is needed for the liquid to rise to a height of 49cm.

The initial volume of the liquid can be found using the formula for the volume of a cylinder:

V = πr²h

where r is the radius (half the diameter), h is the height, and π is approximately 3.14. Plugging in the given values, we get:

V = π(0.5 cm)²(40 cm)

V = 31.4 cm³

The volume coefficient of thermal expansion (β) is defined as the fractional change in volume per degree Celsius change in temperature. It can be calculated using the formula:

β = ΔV/(VΔT)

where ΔV is the change in volume, V is the initial volume, and ΔT is the change in temperature. We can rearrange this formula to solve for ΔV:

ΔV = βVΔT

We know that ΔT = -10°C (a decrease of 10°C) and that the height decreased from 40cm to 37cm, or by 3cm. The change in volume can be found using the formula for the volume of a cylinder again, with the new height of 37cm:

ΔV = π(0.5 cm)²(40 cm - 37 cm)

ΔV = 0.59 cm³

Plugging in all the values, we get:

0.59 cm³ = β(31.4 cm³)(-10°C)

β = 0.002

To find the change in temperature needed for the liquid to rise to a height of 49cm, we can use the same formula as before, but solve for ΔT:

ΔT = ΔV/(βV)

We know that ΔV is the difference between the initial volume and the volume at the new height, which is:

ΔV = π(0.5 cm)²(49 cm - 40 cm)

ΔV = 6.86 cm³

Plugging in all the values, we get:

ΔT = 6.86 cm³/(0.002)(31.4 cm³)

ΔT = 109.5°C

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Answer: C

Explanation:

The x and y component of the ball's final velocity are respectively 7.35 m/s and  4.90 m/s.

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

Given that:

Mass of the ball: M = 6.35 kg.

Initial velocity of ball: U = 8.49 m/s.

Mass of the pin at rest: m = 1.59 kg.

Final velocity of pin: v = 20.1 m/s at a -77.0° angle.

Let the x and y component of the ball's final velocity are respectively V₁ m/s and  V₂ m/s.

Appling conservation of momentum along x axis:

MU + m.0 = MV₁ + mvcos(-77.0°)

⇒ V₁ = u - (m/M) v cos(-77.0°)

After putting the values we get:

V₁ = 7.35 m/s.

Appling conservation of momentum along y-axis:

M.0 + m.0 = MV₂ + mvsin(-77.0°)

⇒ V₂ = - (m/M) vsin(-77.0°)

After putting the values we get:

V₂ = 4.90 m/s.

Hence, the x and y component of the ball's final velocity are respectively 7.35 m/s and  4.90 m/s.

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Answer:

3g/mL

Explanation:

The density of a substance can be calculated using the formula;

Density = mass/volume

Where;

Density = g/mL

mass = grams (g)

volume = (mL)

According to this question, 45.0g of a sample placed in a graduated cylinder causes the water level to rise from

25.0mL to 40.0mL. This means that the volume of the sample is 40mL - 25mL = 15mL

Using D = m/v

D = 45/15

D = 3g/mL

Hence, the density of the sample is 3g/mL

The magnitude of the work done by friction in stopping the car 312500 J

v² = u² + 2as

0² = 25² + (2 × a × 10)

0 = 625 + 20a

Collect like terms

0 – 625 = 20a

–625 = 20a

Divide both side by 20

a = –625 / 20

a = –31.25 m/s²

NOTE: The negative sign indicates that the car is coming to rest (i.e decelerating)

F = ma

F = 1000 × 31.25

F = 31250 N

Wd = F × d

Wd = 31250 × 10

Wd = 312500 J

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Answer:

λ = 437 m.

Explanation:

       \(c = \lambda* f (1)\)

       \(\lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m (2)\)

Answer:

1.8 ms to the left

Explanation:

Answer:1.8 ms to the left

Explanation:

The formula for calculating attraction is F=GMmr².

Gravitational attraction is the force of attraction that occurs on all bodies

which have a mass as a result of the force of gravitational force.

The Force is directly proportional to the product of their masses  and

inversely proportional to the square of the distance between them.

Gravitational attraction =  F=GMmr².

where F is the Force, G is gravitational constant, M and m are masses and r²

is square of the distance between the masses.

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Answer:

To find the work done, we need to calculate the force applied to the block and the distance it moves.

The force applied to the block can be resolved into two components: one parallel to the floor and one perpendicular to the floor. The parallel component of the force is responsible for pushing the block forward, while the perpendicular component does not contribute to the work done.

The parallel component of the force is:

F_parallel = F * cos(30°) = F * √3/2

where F is the magnitude of the force applied.

The force of friction opposing the motion is:

F_friction = μ * F_norm

where μ is the coefficient of friction and F_norm is the normal force acting on the block, which is equal to the weight of the block since it is on a level floor:

F_norm = m * g = 25 kg * 9.81 m/s^2 = 245.25 N

where g is the acceleration due to gravity.

So the force of friction is:

F_friction = 0.4 * 245.25 N = 98.1 N

Since the block is moving at constant speed, the force applied must be equal and opposite to the force of friction:

F_parallel = F_friction

F * √3/2 = 98.1 N

F = 56.6 N

The work done by the force applied is:

W = F_parallel * d = 56.6 N * 5 m = 283 J

Therefore, the work done by the force applied is 283 J

An astronaut measure the period of a mass spring system on Earth.

The period of a mass spring system on the moon would be longer than the period on Earth. This is because the period of a mass spring system is dependent on the square root of the ratio of the mass to the spring constant, and the acceleration due to gravity. Since the acceleration due to gravity on the moon is only 1/6th of that on Earth, the restoring force on the mass will be weaker, resulting in a longer period. Therefore, the astronaut would measure a longer period for the same mass spring system on the moon than on Earth.

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In order to determine which orientation results in a larger depolarization factor for the similar dielectric ellipsoids placed in an electric field, we need to consider the shape and alignment of the ellipsoids with respect to the electric field.

The depolarization factor measures the reduction in the electric polarization of a material due to its shape and alignment in an electric field. It is influenced by the geometry of the material and how it interacts with the electric field.

Qualitatively, if the ellipsoids are aligned in such a way that their major axes are parallel to the electric field lines, the depolarization factor would be smaller. This is because the electric field would act along the long axis of the ellipsoid, resulting in less distortion of the polarized charges inside the material. The polarization would be more effectively aligned with the electric field, minimizing the depolarization effect.

On the other hand, if the ellipsoids are oriented such that their major axes are perpendicular or at an angle to the electric field lines, the depolarization factor would be larger. In this case, the electric field would act in a direction that is not aligned with the major axis of the ellipsoid, causing more distortion and misalignment of the polarized charges inside the material. This results in a larger depolarization effect.

Without a specific diagram or more information about the orientations shown in Figure P5.5, it is difficult to determine the exact orientation with the larger depolarization factor. However, based on the general understanding of the relationship between alignment and the depolarization effect, the orientation where the major axes of the ellipsoids are perpendicular or at an angle to the electric field lines is likely to result in a larger depolarization factor.

The equivalent capacitance of this system is approximately 8.0 µF.

The energy stored in 9.0 µF is 56250 J.

The potential difference across the 5 µF capacitor is given by 1.6 × 10⁻⁶ times the potential difference V₂.

To find the equivalent capacitance of the system, use the concept of capacitance in series and parallel combinations.

(a) Capacitors in series:

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄

Substituting the given values:

1/C_eq = 1/5.0 µF + 1/10.0 µF + 1/9.0 µF + 1/8.0 µF

Converting 8.0 µF to Farads:

1/C_eq = 1/5.0 µF + 1/10.0 µF + 1/9.0 F + 1/(8.0 × 10⁻⁶ µF)

Calculating the sum:

1/C_eq = 0.2 + 0.1 + 0.111 + 125000

Taking the reciprocal of both sides:

C_eq = 1 / (0.2 + 0.1 + 0.111 + 125000)

Calculating the value of C_eq:

C_eq ≈ 7.99999 µF or approximately 8.0 µF

(b) To calculate the energy stored in the 9.0 µF capacitor, use the formula:

E = (1/2) × C × V²

Substituting the given values:

E = (1/2) × 9.0 µF × (50.0 V)²

Calculating the energy:

E = (1/2) × 9.0 µF × 2500 V²

E = 56250 J

The energy stored in the 9.0 µF capacitor is 56250 J.

(c) To calculate the potential difference across the 5 µF capacitor, use the formula:

C₁ × V₁ = C₂ × V₂

Substituting the given values:

5.0 µF × V₁ = 8.0 µF × V₂

Converting 8.0 µF to Farads:

5.0 µF × V₁ = 8.0 × 10⁻⁶ µF × V₂

Simplifying:

V₁ = (8.0 × 10⁻⁶ µF / 5.0 µF) × V₂

V₁ = 1.6 × 10⁻⁶ × V₂

The potential difference across the 5 µF capacitor is given by 1.6 × 10⁻⁶ times the potential difference V₂.

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Answer:

mass

Explanation:

The value of g (acceleration due to gravitation) is independent of the mass of the object.

The current in the series circuit  across a 9.0 V potential difference is determined as 0.333 A.

The current in the series circuit is determined by applying ohms law as shown below;

V = IR

where;

R = 12 Ω  + 15 Ω  = 27 Ω

I = V/R

I = 9/27

I = 0.333 A

Thus, the current in the series circuit  across a 9.0 V potential difference is determined as 0.333 A.

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The work done in pulling the sled is 445,930.2 Joules.

The acceleration in the x direction, assuming that friction is negligible, is  20.85 m/s².

Time taken to pull the sled 185 m is 4.21s.

Work done is equal to product of force applied and distance moved.

Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.

Work = Force x Distance x cos(angle)

W= 1225 x 385 x cos 19°

W = 445,930.2 Joules

Thus, the  work done in pulling the sled is 445,930.2 Joules

From the Newton's second law of motion, we have'

F = ma

acceleration, a = 1225cos19° / ( 545 /9.81)

a = 20.85 m/s²

Thus, the acceleration in the x direction is  20.85 m/s²

Using the second equation of motion, we get

s = ut+ 1/2 at²

Substitute the values, we have

185 m = 0 +1/2 x 20.85 x t²

t = 4.21 s

Thus, the time taken to pull the sled 185 is 4.21s.

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The derivation of Green's function for the Schrödinger wave equation involves using the concept of superposition of solutions to find a particular solution for a given initial condition.

Green's function is a  fine tool that's used to  break  discriminational equations, including the Schrödinger equation. The idea behind Green's function is to find a  result to the  discriminational equation that satisfies a given  original condition.   In the  environment of the Schrödinger equation, Green's function represents the probability  breadth of chancing  a  flyspeck at a particular position at a particular time, given that it started from a known  original position and time.

The Green's function is  attained by  working the Schrödinger equation for a delta function source at the  original position.   To  decide Green's function, one can consider the Schrödinger equation with a delta function source at some  original position. This leads to the  result for the  surge function as a sum of two terms- a free  flyspeck term and a term due to the delta function source.

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The quantity of heat energy that must be transferred to 2.25 kg of brass to raise its temperature from 20 °C to 240 °C is 195030 J

First, we shall list out the given parameters from the question. This is shown below:

The quantity of heat energy that must be transferred can be obtained as follow:

Q = MCΔT

= 2.25 × 394 × 220

= 195030 J

Thus, we can conclude quantity of heat energy that must be transferred is 195030 J

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The energy transferred in Joules by the radio when it has been switched on for 2 minutes would be 6000 Joules.

Power is defined as the rate of energy transfer or the rate at which work is done, and is given by the equation:

Power = Energy transferred / Time

Rearranging the equation to solve for energy transferred, we get:

Energy transferred = Power x Time

We are given:

Power = 50 W

Time = 2 minutes = 120 seconds

Therefore, the energy transferred by the radio when it has been switched on for 2 minutes is:

Energy transferred = Power x Time = 50 W x 120 s = 6000 J

In other words, the energy transferred by the radio is 6000 Joules.

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Scientists typically collect ozone samples from the stratosphere, which is the atmospheric layer located between about 10 and 50 kilometers (6 to 30 miles) above the Earth's surface.

The stratosphere is the layer of the Earth's atmosphere located above the troposphere and below the mesosphere. It extends from about 10 kilometers (6 miles) to about 50 kilometers (30 miles) above the Earth's surface.

The stratosphere is characterized by a gradual increase in temperature with altitude, due to the absorption of ultraviolet radiation by ozone in the stratosphere.

This is where most of the Earth's ozone is found and where the ozone layer is located. The ozone layer absorbs harmful ultraviolet (UV) radiation from the sun, protecting life on Earth from its harmful effects. Scientists collect air samples from this layer using airplanes or weather balloons equipped with specialized instruments.

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Answer:

The type of medium affects a sound wave as sound travels with the help of the vibration in particles.

Explanation:

As different mediums have different amount and size of particles, for example, the speed of sound is faster through solid than liquid as solids have closely packed particles whereas liquids are loosely packed.

The speed of sound in a given medium is determined by its density and stiffness (or compressibility in the case of gases).The speed of sound increases with the rigidity (or lack of compressibility) of the medium. The speed of sound decreases with increasing medium density.

Any material or area through which a wave is transmitted is referred to as a medium. Four variables impact a wave's speed: wavelength, frequency, medium, and temperature. The wavelength and frequency are multiplied to determine the wave speed (speed = l × f).

Therefore, The rate at which energy is transferred through a medium depends on the amplitude of the vibrations of its constituent particles; the higher this rate, the more powerful the sound wave.

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Answer:

the intercept is the orgin (0,0)

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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