A block with mass 4kg is suspended from a vertical Spring. What is the spring constant if the ebrignation of the spring is 10cm? Use g=10

Programmable Array Logic (PAL) has a programmable AND array and a fixed OR array.

Programmable Array Logic (PAL)  is a logic device.

Hence PAL has programmable AND array and a fixed OR array.

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The Title of the interviews Questionnaire is : Community Members' Right to Safe and Healthy Living Questionnaire. The Questionnaire is attached.

A tool for research known as a questionnaire comprises a series of questions formulated to extract data from individuals or a collective of individuals. A systematic approach in acquiring data, which permits researchers to obtain uniform feedback and perspectives from respondents, is termed as structured data collection.

Questionnaires have multiple applications such as conducting surveys, holding interviews, performing assessments, and carrying out evaluations.

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The resistance indicated by the platinum thermometer at 200°C is 1.648 times the reference resistance Ro at 0°C.

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Answer:

The atomic and molecular interactions unveil the bulk properties of matter in our environment by ways of the fact that everything in the whole universe is made of either atoms, molecules or even ions

How matter is made up of atoms and molecules?

It has been proven practically everything in the whole universe is matter and everything which interact with matter is also matter. This explains to us the reasons why matter could be atoms, molecules or ions.

That being said, some substances (matter) is made up of atoms of elements, some made up of molecules or atoms and molecules and others ions or both. However, matter is anything which has mass and occupies space.

In conclusion, we can now conclude from the explanation above that the properties of matter are as a result of the interaction which exists between matter at the atomic and molecular level.

This is the pressure at the top of the container due to the height of the water. The pressure at the bottom of the container due to the height of the water is 0 Pa since the datum was set at the bottom.

Pressure is a physical quantity used to measure the force applied by an object to another object or by an object to a surface. It can be expressed as the force per unit area and is typically measured in units such as pounds per square inch (psi) or pascals (Pa). Pressure can be applied to gases, liquids, and solids, and is generated by the weight of the atmosphere, the force of gravity, and by the movement of air or liquids. Pressure can also be created by mechanical devices such as pumps and compressors, as well as by chemical reactions.

ρ = density of water = 1000 kg/m3

g = acceleration due to gravity = 9.81 m/s2

h = height of the container = 7.7 in = 0.198 m

Using the equation ρgh = density x gravity x height, we can calculate the pressure at the top of the container to be:

ρgh = (1000 kg/m3)(9.81 m/s2)(0.198 m) = 1960.38 Pa

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Answer:

a) a= 19 m/s2, b ) F = F = 38 N

c)  the correct one is B

Explanation:

1) To solve this exercise, let's use the concept of torque, let's start by looking for the mass of the mass of the head, which is 65 of 65 kg

         m= 0.06  65 = 3.9 kg

torque is defined

        τ = F x R

the bold indicate vectors

        τ = F R sin θ

the force is the weight of the head and the distance R = 0.20 m

        τ = mg R sin θ

        τ = 3.9 9.8 0.2 sin 50

        τ = 5.86 N m

Newton's second law expression for rotational motion is

         τ = I α

In general, the head is approximated by a sphere, so the moment of foolishness with respect to the base of the neck (turning point) using the parallel axis theorem is

         I = I_sphere + m d²

Neck height data is not indicated, so we will assume that it is zero (d = 0)

         I = 2/5 m R²

angular and linear acceleration are related

        a = α R

        α = a / R

let's replace

         τ = 2/5 mR²  (a / R)

        a = 5/2 τ / mR

let's calculate

         a = 5/2 5.86 / (3.9 0.2)

         a = 18.8 m / s²

         a= 19 m/s²

2) as the neck and the head are related, the forces are action and reaction, therefore the neck force is

         F = m g

         F = 3.9 9.8 = 38.2 N

          F = 38 N

3) He asks us about the efficiency of the headrest, after reviewing the different alternatives, the correct one is B

Yes. A headrest ensures that both the head and the body experience the same forward force at approximately the same time.

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field  is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

\(\Delta A = \frac{1}{2} \Delta \theta R^2\)

The  formula for the induced emf is

\(E = \frac{\Delta \phi}{\Delta t}\)

\(\phi = \texttt {magnetic flux}\)

\(E=\frac{\Delta (BA) }{\Delta t}\)

\(=B\frac{\Delta A}{\Delta t}\)

B is the magnetic field strength

substitute

\(\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A\)

\(E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega\)

The magnetic field of the earth is oriented at 14.42

\(\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5\)

we plug in the values in the equation above

so, the induce EMF will be

\(E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega\)

\(=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V\)

Explanation:

The kinetic energy of the marble is converted to elastic potential energy in the rubber ball. When the marble hits the rubber ball, the ball compresses and stores energy in its structure, which is released as it expands back to its original shape. This stored energy is known as elastic potential energy, which is a form of potential energy that depends on the deformation of an object. In this case, the rubber ball deforms when it is hit by the marble, and the deformation results in the storage of energy.

Answer:

Energy, in physics, the capacity for doing work. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another.

Explanation:

Hope this helps!

Explanation:

Hypothesis: The gecko tape will exhibit stronger adhesive properties and superior sticking ability compared to regular duct tape.

To test this hypothesis, a possible experimental setup could involve the following steps:

1. Materials: Gather the gecko tape and the regular duct tape as the comparison tape.

2. Test Surface: Prepare a clean and smooth surface, such as a glass or metal panel, to which the tapes can be applied.

3. Application Procedure: Cut equal-sized strips of gecko tape and duct tape. Apply one strip of each tape to the test surface with the same pressure and technique.

4. Adhesive Strength Measurement: Use a force gauge or a testing device to measure the force required to remove each tape from the test surface. Repeat this process multiple times to ensure accuracy.

5. Data Collection: Record and compare the adhesive strength values (force required to remove the tapes) for the gecko tape and the duct tape.

6. Statistical Analysis: Use appropriate statistical methods to analyze the data and determine if there is a significant difference in adhesive strength between the two tapes.

7. Conclusion: Based on the results, evaluate the hypothesis by determining whether the gecko tape demonstrates superior sticking ability compared to the regular duct tape.

By comparing the adhesive strength of the gecko tape and the regular duct tape, this experiment aims to provide measurable and specific evidence to support or refute the hypothesis regarding the functional effectiveness of the gecko tape.

Given parameters:

First velocity  = 2.50m/s

Time of travel = 3s

Second velocity  = 1.50m/s

Unknown:

The displacement during the first interval = ?

Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.

    Velocity  = \(\frac{Displacement}{Time taken}\)

So;

      Displacement  = Velocity x Time taken

Now input the parameter for the first velocity and time of travel;

      Displacement  = 2.5 x 3  = 7.5m

The displacement id 7.5m

a. The free-body diagrams of the 8.0 kg box and the 2.0 kg mass is attached below.

bi) the acceleration of the system is 0.4 m/s^2

bii)  the tension in the string is 0.8 N.

Net force = ma

Net force = T - F_friction

where T is the tension force and F_friction is the frictional force.

Substituting the given values, we have:

T - F_friction = ma

T - 6.0 N = (8.0 kg + 2.0 kg) a (since the tension force acts on both masses)

T - 6.0 N = 10.0 kg a

(ii) To find the tension in the string, we can apply Newton's second law of motion to the 2.0 kg mass:

Net force = ma

where m is the mass of the mass and a is the acceleration of the system. The net force on the mass is the tension force:

Net force = T

Substituting the given values, we get:

T = ma

T = 2.0 kg x a

To solve for T, we need to find the acceleration of the system.

To find both a and T, we can use the equations we derived above and solve them simultaneously:

T - 6.0 N = 10.0 kg a (equation 1)

T = 2.0 kg x a (equation 2)

Substituting equation 2 into equation 1, we get:

2.0 kg x a - 6.0 N = 10.0 kg a

Solving for a, we get:

a = 0.4 m/s^2

Substituting a into equation 2, we get:

T = 2.0 kg x 0.4 m/s^2 = 0.8 N

Therefore, the acceleration of the system is 0.4 m/s^2 and the tension in the string is 0.8 N. of the system is 0.4 m/s^2 and the tension in the string is 0.8 N.

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The distance formula used for a falling object is y = 1/2gt^2 where g is gravity which is 9.80m/s^2 and t is the time it falls ( reaction time).

Y = 1/2(9.8)0.2^2

Y = 0.196

It travels 0.196 meters

The velocity of the car 1 can be seen from the calculation as 20 m/s West

A coordinate system or point of view used to observe motion is known as a frame of reference. It can be used as a guide when describing an object's position, speed, and acceleration. Different frames of reference may result in various motion observations.

The relative velocity is the velocity of an object or observer as observed from a particular frame of reference.

We can see that the velocity of the car 1 is;

45 m/s - 25 m/s

= 20 m/s West

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The density of the second liquid is 0.812 g/cm³

To calculate the density of the second liquid, we need to use the principle of displacement. The mass of the liquid can be found by subtracting the mass of the empty density bottle from the mass of the bottle filled with the liquid. Therefore, the mass of the liquid is:

mass of liquid = mass of bottle + liquid - mass of empty bottle

mass of liquid = 39.84g + x - 18.00g

where x is the mass of the liquid.

We can now use the density formula, which is:

density = mass/volume

The volume of the liquid is equal to the volume of the density bottle that is filled with the liquid, which can be calculated by subtracting the volume of the empty bottle from the volume of the bottle filled with the liquid. Therefore, the volume of the liquid is:

volume of liquid = volume of bottle filled with liquid - volume of empty bottle

We can now substitute this expression into the density formula to get:

density of liquid = mass of liquid / (volume of bottle filled with liquid - volume of empty bottle)

We know that the density of water is 1000 kg/m³, which is equal to 1 g/cm³. We can use this to find the volume of the liquid by dividing the mass of water by its density:

volume of water = mass of water / density of water

volume of water = 44.00g / 1 g/cm³

volume of water = 44.00 cm³

Now, we can calculate the volume of the density bottle filled with the second liquid by using the principle of displacement:

volume of bottle filled with liquid = volume of water - volume of liquid

volume of bottle filled with liquid = 44.00 cm³ - (39.84g - 18.00g) / 1 g/cm³

volume of bottle filled with liquid = 44.00 cm³ - 21.84 cm³

volume of bottle filled with liquid = 22.16 cm³

Finally, we can substitute these values into the density formula to get:

density of liquid = x / 22.16 cm³

Solving for x, we get:

x = density of liquid x 22.16 cm³

Substituting x back into the mass equation, we get:

mass of liquid = 39.84g + (density of liquid x 22.16 cm³) - 18.00g

Solving for the density of the liquid, we get:

density of liquid = (mass of liquid - 21.84g) / 22.16 cm³

Substituting the given values, we get:

density of liquid = (39.84g - 21.84g) / 22.16 cm³ = 0.812 g/cm³

In conclusion, the density of the second liquid is 0.812 g/cm³. This value is less than the density of water, which means that the second liquid is less dense than water and will float on top of water.

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The speed of the mass when it hits the ground from the given height is \(\sqrt{2gh}\).

The given parameters:

The final velocity of the ball when it hits the ground is calculated as follows;

\(v_f^2 = v_0_y^2 + 2gh\\\\\)

where;

\(v_f^2 = 0 + 2gh\\\\v_f^2 = 2gh\\\\v_f = \sqrt{2gh}\)

Thus, the speed of the mass when it hits the ground from the given height is \(\sqrt{2gh}\).

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The voltage between the plates is 3.9 × \(10^-^3\) V.

The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential.  The greater the change in voltage per unit distance, the greater the electric field.

The kinetic energy of the electrons = 4.1 × \(10^-^1^5\) J

Charge of the electron = 1.602 × \(10^-^1^9\) coulomb

Using,

     ΔU = q × ΔV

4.1 × \(10^-^1^5\) = 1.602 × \(10^-^1^9\) × ΔV

      ΔV  = 3.9 × \(10^-^3\) V

Therefore, the voltage between the plates is 3.9 × \(10^-^3\) V.

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Answer:

This problem involves the concept of a random walk, which is a mathematical model of a path consisting of a succession of random steps.

The question asks for the distance, R(n), from the center of a star after n steps of a photon, assuming a 2D random walk.

The random walk in two dimensions has a step length of A_i and the direction of the steps is uniformly distributed in [0, 2π). The change in position after each step can be written in Cartesian coordinates (Δx, Δy), where Δx = A_i cos(θ_i) and Δy = A_i sin(θ_i).

The displacement from the center after n steps is given by the vector sum of all the individual steps. This vector sum can be written in terms of its Cartesian coordinates, (X, Y), where X = Σ Δx and Y = Σ Δy. This sum over n random vectors is itself a random variable. The net displacement R(n) from the center of the star after n steps is given by the magnitude of the net displacement vector:

R(n) = √(X² + Y²)

Because each step is independent and has a random direction, the expected value of the cosine and sine for any step is zero. This means that the expected values of X and Y are both zero.

However, the mean square displacement is not zero. Because the steps are independent, the mean square displacement in each direction is additive. For a 2D random walk:

<X²> = Σ <(Δx)²> = n <(A cos θ)²> = n A²/2

<Y²> = Σ <(Δy)²> = n <(A sin θ)²> = n A²/2

Because <X²> = <Y²>, we can write:

<R²> = <X²> + <Y²> = n A²

So, the root mean square distance (the square root of the mean square displacement) after n steps is:

R(n) = √(<R²>) = √(n) * A

Therefore, the distance R(n) that the photon is expected to be from the center of the star after n steps grows as the square root of the number of steps, with each step having a length A. Please note that this result holds for a 2D random walk. A real photon in a star would be performing a 3D random walk, which would have slightly different characteristics.

There are only two types of charges which are the positive and the negative charges.

These two charges (positve and negative), are created in pairs, they are not formed alone.

Like charges repel while unlike charges attract each other.

Charge can be said to be created from nothing.

There is no actual positive charge without a negative charge.

The two charges, positive and negative, are created in pairs.

Therefore, the given statement "the two types of charges, positive and negative, are not created alone, but in pairs" is True,

ANSWER:

True.

Answer:

True

Explanation:

(a) The ball goes up to the height of 31.89 m. (b) The ball stays for 5.1 s in the air. (c) The acceleration due to gravity and wind resistance can affect the estimation.

Acceleration owing to gravity is the term used to describe the rate at which a body's velocity changes as a result of the earth's gravitational pull. In general, it is assumed that the acceleration caused by gravity is in the downward direction.

The acceleration caused by gravity has been calculated as, however as it changes from location to location, it may have an impact on the estimation.

You may have thought that the wind has no impact, but it can actually generate drag and even cause the ball to shift course.

Therefore, (a) The ball goes up to the height of 31.89 m. (b) The ball stays for 5.1 s in the air. (c) The acceleration due to gravity and wind resistance can affect the estimation.

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The magnetic flux Φ of a field B on a loop with area A and normal vector n is:

There θ is the angle between the normal vector n and the field B.

If the normal to the plane of the loop is parallel to the field, then θ=0º. In that case, the magnetic flux is simply the product of the field and the area of the loop, since cos(θ)=0:

The area of a circle with radius r is:

Replace the expression for the area in terms of r and substitute B=1.194T and r=9.792*10^(-2)m to find the magnetic flux:

Therefore, the magnetic flux if the normal is parallel to the field, is 0.03597 weber.

The acceleration of the jet when it speeds up = 7.5 m/s²

The initial speed, u = 100 m/s

Distance travelled, s = 2. km

s = 2 x 1000m

s = 2000 m

Final speed, v = 200 m/s

The acceleration is found by using the equation of motion below

Substitute v = 200 m/s, u = 100 m/s, s = 2000 m

The acceleration of the jet when it speeds up = 7.5 m/s²

When materials vibrate, waves are created that travel through the substance, and this energy is what we hear as sound. Earthquakes are earth vibrations that cause the (potential) energy held within rocks to be released (as a result of their pressure-generating relative positions). Seismic waves are produced by earthquakes.

How do sound waves and earthquakes compare?

The waves lose energy as they move through the air with sound or through the ground with shaking during an earthquake. Therefore, a band can be heard louder close to the stage than farther away, and an earthquake can be felt more strongly close to the fault than farther away.

In actuality, sound in the air cannot match how quickly earthquake waves move. In rock, the compressional or "P" wave of an earthquake moves at the In actuality, sound in the air cannot match how quickly earthquake waves move. The speed of a P wave is typically 10,000 mph. The speed of sound through air is roughly 750 mph.

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Since the car is accelerating at a constant rate, the distance it travels during each second of its motion will be directly proportional to the time it has been accelerating.

In the first second, the car moved a distance of 2 meters, and in the second second, it will move twice the distance of the first second, so the car will move additional distance of 2*2 = 4 meters during the second second of its motion.

The distance traveled during the second second of its motion is 1/2 * 2 = 1 meters.

A car that accelerates at a constant rate will move a distance equal to the initial velocity multiplied by time plus 1/2 the acceleration multiplied by the square of time. Since the car starts from rest, the initial velocity is zero.

Therefore, the distance traveled during the second second is 1/2 * acceleration \(* (time)^2 = 1/2 * a * t^2 = 1/2 * a * 1^2 = 1/2 * a\) Since the car moved 2.0 meters in the first second, it means the acceleration is\(2m/s^2\), and the distance traveled during the second second is 1/2 * 2 = 1 meters.

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Answer:

The traffic cop needs 23.3 seconds to pass the automobile.

When your velocity (not speed) changes, you are accelerating. A automobile moving at a steady 100 km/h in a straight line has no acceleration. Average acceleration is equal to (change in velocity) / (duration). The car's acceleration is zero because its change in velocity is also zero.

\(d1 = v1*t1 = 35.0 m/s * 1 s = 35.0 m\)

\(d = d1 = 35.0 m\)

\(d2 = v2*t + (1/2)at^2\)

\(d2 = (1/2)at^2\)

\(v2*t + (1/2)at^2 = (1/2)at^2\)

\(v2*t = (1/2)at^2\)

Solving for t, we get:

\(t = (2v2/a) = (235.0 m/s)/3.0 m/s^2 = 23.3 s\) (rounded to 2 decimal places)

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The acceleration of the car, when it is made to stop is -13.33 m/s². The correct option is B

Acceleration is given by the ratio of resultant or total force acting on any object and the its mass.

It can also be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

A car came to a stop from a speed of 28 m/s in a time of 2.1 seconds.

The final velocity will be zero.

So, the acceleration is

a = 0-28/2.1

a = -13.33 m/s²

Thus, the acceleration of the car is -13.33 m/s².

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Answer:

D. Conclusion.

Explanation: