A Galapagos turtle travels 7 cm/s. How long would it take the turtle to travel 90 m (that's
the length of the science hallway

The order of events when a neuron fires an action potential is as follows:

1. Depolarization: The neuron's membrane potential is changed from its resting potential to a more positive voltage.

2. Threshold: The membrane potential reaches a critical level, known as the threshold potential, which triggers the action potential.

3. Action potential: An all-or-none electrical impulse is generated and propagates along the axon.

4. Repolarization: The membrane potential returns to its resting potential.

The Porsche taycan turbo S can accelerate at 35.71 m / s² and the Tesla P100D can accelerate at 42.35 m / s². Tesla P100D will be faster after 2 seconds.

a = ( v - u ) / t

a = Acceleration

v = Final velocity

u = Initial velocity

t = Time

For Porsche taycan turbo S,

u = 0

v = 100 km / h

t = 2.8 s

a = ( 100 - 0 ) / 2.8

a = 35.71 m / s²

For Tesla P100D,

u = 0

v = 60 mi / h = 96.56 km / h

t = 2.28 s

a = ( 96.56 - 0 ) / 2.28

a = 42.35 m / s²

At t = 2 sec,

\(v_{P}\) = \(a_{P}\) * t

\(v_{P}\) = 35.71 * 2

\(v_{P}\) = 71.42 m / s

\(v_{T}\) = \(a_{T}\) * t

\(v_{T}\) = 42.35 * 2

\(v_{T}\) = 84.7 m / s

\(v_{T}\) > \(v_{P}\)

Therefore,

(a) The Porsche taycan turbo S can accelerate at 35.71 m / s² and the Tesla P100D can accelerate at 42.35 m / s²

(b) Tesla P100D will be faster after 2 seconds.

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Answer:

128.205.1

Explanation:

Answer:

Most commonly anti-freeze is used in the motor engine, but water is sometimes used because it has a high thermal conductivity.

Explanation:

Answer:

P = 30000 [kg*m/s]

Explanation:

Momentum in physics is defined as the product of force in the given time interval. In this way, it can be expressed by means of the following equation.

\(P=F*t\)

where:

P = momentum [kg*m/s]

F = force = 6*10⁴ [N]

t = 0.5 [s]

Now replacing:

\(P=6*10^{4}*0.5\\P=30000[kg*m/s]\)

Answer:4.62

Explanation:

Acellus

Answer:

Explanation:

The equation for Kinetic Energy is

\(KE=\frac{1}{2}mv^2\) and we know everything but the mass of this ball. Filling in:

\(3.00=\frac{1}{2}m(5.00)^2\). Get everything isolated from the :

\(m=\frac{2(3.00)}{(5.00)^2}\) which tells us that we need 3 significant digits in our answer.

m = .240 kg since mass is always in kg i the velocity is in meters/sec

Answer:Earth-orbiting astronauts are weightless for the same reasons that riders of a free-falling amusement park ride or a free-falling elevator are weightless. They are weightless because there is no external contact force pushing or pulling upon their body. In each case, gravity is the only force acting upon their body.

Explanation:

Answer:

Friction is another force that energy gets "wasted" on in the dragging. Therefore, the machine gives you less mechanical advantage when too much friction is involved because it wastes energy. It also makes it less efficient!

The centripetal acceleration of the professor holding onto the end of the minute hand of a clock atop City Hall is 0.00133 m/s².

Centripetal acceleration is the inward force acting on a body moving in a circular path that changes the direction of the velocity of the body and constantly pulls it toward the center of the circle.To determine the professor's centripetal acceleration, we use the formula;

`a= (v²)/r`

Where `a` is the centripetal acceleration, `v` is the velocity, and `r` is the radius. We have the length of the minute hand which is the radius of the circle.

So,`r = 4 m`We need to find the velocity which is given by the formula:

`v= (2πr)/T`

Where `π` is pi (3.14), `r` is the radius, and `T` is the time taken for one complete rotation which is 60 minutes since it's the minute hand.

Therefore;`v = (2 x 3.14 x 4 m) / (60 min x 60 s / 1 min)``v = 0.42 m/s`Substitute `v` and `r` into `a = (v²)/r` to get:`a = (0.42 m/s)² / 4 m``a = 0.00133 m/s²`

Therefore, the centripetal acceleration of the professor holding onto the end of the minute hand of a clock atop City Hall is 0.00133 m/s².

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The resistance of the wire is R1

Let L be the length of the wire and d be the diameter

Then resistance

R1 = ρL/A

Where ρ is the resistivity of the material

A is the crossectional area = πd2/4

As the material is not changed, ρ does not changes

new length L' = L/2

new diameter d' = d/2

Therefore the new resistance is

R2 = ρL'/A'

R2 = ρ(L/2)/(A/4)                (A' = π(d/2)2/4)

R2 = 2 ρL/A

R2 = 2R1

R2/R1 = 2

What is resistance?

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The answers are:

(a) The length of the string is 1 meter.

(b) The frequency of the wave is 303 Hz.

To find the length of the string, we can use the formula for wave velocity:

v = λf,

where v is the velocity of the wave, λ is the wavelength, and f is the frequency.

Given that the velocity of the wave on the string is V = 303 m/s, and the length of the string is L, we can equate the velocity to the product of the wavelength and frequency:

V = λf.

Since L represents the length of the string and is equal to one wavelength (L = λ), we can rewrite the equation as:

V = Lf.

We can solve this equation for L:

L = V / f.

Since we don't have the frequency given directly, we need to find it using other information. However, in part (b), we are asked to find the frequency when L is one wavelength. This means that the frequency will be equal to the velocity of the wave, as the velocity equals the product of the wavelength and the frequency.

Therefore, in this case, the frequency is f = V.

Now we can substitute this value into the equation for L:

L = V / f = V / V = 1 meter.

Hence, the length of the string is 1 meter.

If L is one wavelength, the frequency of the wave on the string is equal to the velocity of the wave, which is V = 303 m/s. Thus, the frequency is 303 Hz.

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After a great many contacts with the charged ball, the charge on the rod will be arranged with negative charge on end A and positive charge on end B when the charged ball is far away.

When the charged ball is brought in contact with the rod, some of the charge transfers from the ball to the rod. If the ball has a positive charge, electrons will flow from the rod to the ball, resulting in an excess of positive charge on the rod. Conversely, if the ball has a negative charge, electrons will transfer from the ball to the rod, leaving an excess of negative charge on the rod.

After numerous contacts between the charged ball and the rod, the charge will eventually distribute evenly along the length of the rod. This is because the excess charge on the rod will repel similar charges and attract opposite charges. As a result, the negative charge will accumulate on end A of the rod, while the positive charge will accumulate on end B.

When the charged ball is far away, the excess charge on the rod will remain distributed in this manner. The negative charge on end A and the positive charge on end B will persist, creating a charge separation along the rod. This charge arrangement is a result of the redistribution of charges due to the electrostatic forces between them.

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We need to locate the places on the graph where the distance is not changing, suggesting that Valentina was not moving, in order to calculate how long she was stationary for. By examining the graph, we can tell that Valentina was still for 15 minutes, from 17:30 to 17:45.

The particle's acceleration is represented by the gradient of the line of the speed-time graph (for straight lines acceleration is constant). Line gradient = change in velocity time = v ut = a. Line gradient is equal to the change in velocity over time, or v u t, or a. The distance traveled by a particle is indicated by the area under the speed-time graph.

On a speed-time graph, we can see how the speed of a moving object changes with time. The steeper the graph, the greater the acceleration. A horizontal line means the object is moving at a constant speed.

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Answer:

the book is nice.

Explanation:

because the book is nice.

(a) The budget constraint can be written as: C = wL + V, where C is consumption, w is the real wage, L is leisure, and V is non-labor market income.

(b) With ẞ = 1/2, V = 100, and w = 200, John's optimal supply of labor cannot be determined without information about his preferences for leisure and consumption. The utility function only represents his preferences, but we need additional information to determine the specific amount of labor he would choose to supply.

(c) John's total income is the sum of his labor income and non-labor market income: Total income = Labor income + Non-labor income = wL + V. Without knowing the specific value of L, we cannot calculate the total income.

(d) Similarly, without knowing John's preferences for leisure and consumption, we cannot determine the specific level of consumption he would choose.

(e) In the case where John is subject to a 10% income tax on labor income only, his new optimal supply of labor would depend on the tax rate's impact on his preferences and the trade-off between leisure and consumption. Without further information on his preferences and the specific tax structure, we cannot determine the new optimal supply of labor.

Additional information about John's preferences for leisure and consumption, as well as the specific tax structure, is necessary to calculate his optimal labor supply, total income, consumption, and the impact of the income tax on his labor supply.

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Answer:

Gravitational potential energy

Explanation:

PE = m*g*h

m = mass

g = 9.81m/s², acceleration of gravity

h = height

So m and g stay the same but the height can change so the higher you go the higher the Gravitational potential energy.

Placing a soccer ball on the top shelf increases the Gravitational potential energy

In a gravitational field the energy possessed by an object due to a change in its position called as gravitational potential energy, it is an energy that is related to gravitational force or to gravity.

When a mass of the body  (m) is moved from infinity to a point inside the gravitational influence of a source mass (M) in absence of acceleration and the amount of work done in displacing it into the source field is stored in the form of potential energy denoted with the symbol Ug.

when external force is applied, then the  position of the body changes potential energy is equal to the amount of work done on the body by the forces.

The gravitational influence on a body at infinity is zero, then the  potential energy is zero, PE = m*g*h where m = mass, g = 9.81m/s², acceleration of gravity, h = height

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Answer:

The Answer is D!

Explanation:

I checked it on Khan Academy.

The correct free body diagram of the hanging carrot is option B, where, the forces from two ropes and the net force acting upward and the gravitational force acting downwards.

A free body diagram is a representation of a body with different kinds of forces acting upon it with respect to its position and direction.

There are different kinds of forces namely magnetic force, tension force, nuclear force, gravitational force and the normal force  etc. The force acting on a massive moving body will be the product of its mass and acceleration.

The direction of each force acting upon a body will be different and the displacement also. Here, the carrot hanged by the two ropes will experience forces from both ropes which hangs it down but the force acting is upwards and thus the net force  is upward.

The gravitational force Fg acting on a body is always downward and thus the correct diagram depicting the free body of the carrot with forces is option B.

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Answer:

The static "frictional force" and the "kinetic friction" will act on the box.

When  A force is applied to a box to move it to the right across the kitchen floor, static friction force comes into play before start moving. When the box starts moving, dynamic friction force comes into play.

The resistance provided by surfaces in touch as they move past one another is known as friction.

To walk without slipping, traction is provided by friction. In most situations, friction is advantageous. They do, however, give a strong amount of opposition to the move.

When  A force is applied to a box to move it to the right across the kitchen floor, static friction force comes into play until it starts moving. When the box starts moving, dynamic friction force comes into play.

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The nurse should insert the needle at a 45-degree angle to ensure proper medication administration into the subcutaneous tissue.

When administering a subcutaneous injection, the nurse inserts the needle at a 45-degree angle to ensure proper medication administration. This angle allows for even distribution of the medication in the subcutaneous tissue, facilitating absorption and therapeutic effect. The subcutaneous tissue has a rich blood supply, and the medication is absorbed through capillaries in this layer. The 45-degree angle ensures that the medication reaches the appropriate depth without hitting muscle or bone. It also balances effective delivery with patient comfort, minimizing pain and tissue trauma. Inserting the needle at a steeper angle could cause the medication to be delivered too deeply, potentially causing discomfort or injury. Proper injection techniques, including selecting the correct needle length and angle, are crucial for safe and accurate administration of subcutaneous medications.

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Answer:

2.45 m/s²

Explanation:

From the question,

On the Earth

W = mg.................. Equation 1

Where W = weight of the alien on the earth, m = mass of the alien on the earth, g = acceleration due to gravity of the earth.

Make m the subject of equation 1

m = W/g................... Equation 2

Given: W = 3200 N

Constant: 9.8 m/s²

Substitute these value into equation 2

m = 3200/9.8

m = 326.5 kg.

Similarly,

On planet 9,

W' = mg'............... Equation 3

Where W' = weight of the alien on planet 9, g' = acceleration due to gravity on planet 9.

make g' the subject of the equation

g' = W'/m............ Equation 4

Given: W' = 800 N

Substitute into equation 4

g' = 800/326.5

g' = 2.45 m/s²

Answer :

The answer is clearly C

Explanation:

Because the only way currents move are to the side

The amount of kinetic energy dissipated in the perfectly inelastic head-on collision between the two cars is 107,726.0 J.

To solve this problem, we need to calculate the kinetic energy of each car before the collision, and then add them together to find the total kinetic energy. We can then use the principle of conservation of momentum to calculate the final velocity of the combined mass after the collision, assuming that the collision is perfectly inelastic.

First, we need to convert the speeds from km/h to m/s, since kinetic energy is measured in joules, which is a unit of energy in the metric system.

The speed of the first car traveling east is:

v1 = 106.0 km/h = 29.4 m/s

The speed of the second car traveling west is:

v2 = 75.0 km/h = 20.8 m/s

The mass of each car is:

m = 60.0 kg

The mass of each driver is:

md = 50.0 kg

The total kinetic energy of the first car and driver is:

KE1 = (1/2)mv1^2 + (1/2)mdv1^2

KE1 = (1/2)(60.0 kg)(29.4 m/s)^2 + (1/2)(50.0 kg)(29.4 m/s)^2

KE1 = 94,908.0 J

The total kinetic energy of the second car and driver is:

KE2 = (1/2)mv2^2 + (1/2)mdv2^2

KE2 = (1/2)(60.0 kg)(20.8 m/s)^2 + (1/2)(50.0 kg)(20.8 m/s)^2

KE2 = 48,727.0 J

The total kinetic energy before the collision is:

KEtotal = KE1 + KE2

KEtotal = 94,908.0 J + 48,727.0 J

KEtotal = 143,635.0 J

Now, we can use the principle of conservation of momentum to find the final velocity of the combined mass after the collision. Assuming that the collision is perfectly inelastic, we can assume that the two cars stick together after the collision.

The total momentum before the collision is:

p = (m + md)v1 - (m + md)v2

p = (60.0 kg + 50.0 kg)(29.4 m/s) - (60.0 kg + 50.0 kg)(20.8 m/s)

p = 1,760.0 kg·m/s

The final velocity of the combined mass after the collision is:

vfinal = p/(m + md + m + md)

vfinal = 1,760.0 kg·m/s / (60.0 kg + 50.0 kg + 60.0 kg + 50.0 kg)

vfinal = 9.8 m/s

Finally, we can calculate the amount of kinetic energy dissipated in the collision by subtracting the final kinetic energy from the initial kinetic energy:

KEdissipated = KEtotal - (1/2)(m + md + m + md)vfinal^2

KEdissipated = 143,635.0 J - (1/2)(60.0 kg + 50.0 kg + 60.0 kg + 50.0 kg)(9.8 m/s)^2

KEdissipated = 107,726.0 J

Therefore, the amount of kinetic energy dissipated in the perfectly inelastic head-on collision between the two cars is 107,726.0 J.

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The force constant of the spring is 200.696 N/m & The height the object achieves when the spring releases its energy is 2.5087 m

The spring constant is the force needed to stretch or compress a spring, divided by the compressive or expansive distance. It's used to determine stability or instability in the spring, and therefore the system it's intended for. we know,

F = kx

Therefore,

k = F/x

We also know that the force being exerted on the spring is equal to the mass of the object. Hence, F = mg = 5.13 * 9.8 N = 50.174 N and we know compression due to the mass is 0.25m. Therefore,

K = 50.174/0.25 N/m

K = 200.696 N/m

Therefore, The Spring Constant is 200.696 N/m

On release, the spring potential energy gets converted to kinetic energy. Hence, on release, the height attained by the object is given by:

h = \(1/2 kx^{2}\)

We know that k=200.696 N/m and x=0.25 m. Therefore the height is:

h = \(1/2 (200.696 N/m)(0.25 m)^{2}\)

h = 2.5087 m

Therefore, the force constant of the spring is 200.696 N/m & The height the object achieves when the spring releases its energy is 2.5087 m

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Answer:

The body must have to be situated at a height 5.10 m.

Explanation:

The body of mass 10 kg must be situated at a height of approximately 51.02 meters above the ground.

The conservation of energy principle states that the total energy of a system is constant. In this case, we can equate the potential energy of the first body to the kinetic energy of the second body, as follows:

Potential energy = Kinetic energy

\(mgh = 1/2 mv^2\)

where m is the mass of each body, g is the acceleration due to gravity, h is the height above the ground, and v is the velocity of the second body.

Substituting the given values, we get:

\(10 kg * g * h = 1/2 * 10 kg * (10 m/s)^2\)

Simplifying this equation, we get:

\(h = (1/2 * v^2) / g\\h = (1/2 * 10 m/s * 10 m/s) / 9.81 m/s^2\\h = 51.02 meters\)

Therefore, the body of mass 10 kg must be situated at a height of approximately 51.02 meters above the ground to have potential energy equal in value to the kinetic energy possessed by the other body of mass 10 kg moving with a velocity of 10 m/s.

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The collision occurred at a height that is 8/9 of the height of the building. This problem involves the concepts of kinematics and conservation of energy.

Let's assume that the height of the building is h, and the speed of ball b just before the collision is v. Since ball a is dropped from rest, its initial velocity is 0.

We know that at the time of collision, the velocity of ball a is three times the velocity of ball b, and their directions are opposite. Therefore, the velocity of ball a is -3v, and the velocity of ball b is v.

Let's find the time it takes for each ball to reach the collision point.

For ball a, using the equation of motion for constant acceleration:

h = (1/2)at²

where a is the acceleration due to gravity, which is approximately 9.8 m/s², and t is the time it takes for ball a to reach the collision point.

Solving for t, we get:

t = √(2h/9.8)

For ball b, using the equation of motion for constant acceleration:

v = u + at

where u is the initial velocity, which is upward for ball b, and a is the acceleration due to gravity. We know that v = -3v at the collision point, so:

-3v = u - 9.8t

Solving for t, we get:

t = (4/3)v/9.8

Now we can equate the two expressions for t and solve for h:

√(2h/9.8) = (4/3)v/9.8

h = (8/9)v²

Therefore, the collision occurred at a height that is 8/9 of the height of the building.

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Answer:

Reaction force. An object at rest on a surface experiences reaction force.

Tension. An object that is being stretched experiences a tension force.

Friction. Two objects sliding past each other experience friction forces.

Air resistance.

Explanation:

Answer with Explanation:

Conntact forces: Frictional force, Muscular force etc

Non-contact : gravitational, electrostatic, and magnetic

I hope   im right!!

Answer:

3.038 m/s^2

Explanation:

What we have

m = 6.0 kg

\(\mu_{k}\) = 0.20

\(\mu_{s}\) = 0.40

\(F_{applied}\) = 30 N

Since the situation is that the box has already been pushed, lets get the total net force of both components.

∑Fy = \(ma_{y}\) = 0 (bc the normal and weight cancel out each other)

∑Fx = \(F_{applied}\) - \(f_{k}\) = \(ma_{x}\) (We are going to use this equation to find acceleration)

To find the friction force

\(f_{k}\) = \(\mu_{k}\) * N

\(f_{k}\) = .20 * 6.0 kg * 9.81 m/s^2 = 11.772 N

Now we can use ∑Fx equation to actually find the acceleration of the box!

∑Fx = \(F_{applied}\) - \(f_{k}\) = \(ma_{x}\)

Plug it in

30 N - 11.772 N = 6 kg * a

Do the algebra...

a = \(\frac{30 N - 11.772 N}{6 kg}\) = 3.038 m/s^2