In aviation, it is helpful for pilots to know the cloud celling, which is the distance between the ground and lowest cloud. The simplest way to measure this is by using a spotlight to shine a beam of light up at the clouds and measuring the angle between the ground ad where the beam hits the clouds. If the spotlight on the ground is 0.75 kg from the hangar door as shown in the image below, what is the cloud ceiling?

Each of the five friends needs to pay £12 to cover the cost of their own meals and contribute towards the birthday person's meal. Using mean allows us to distribute the cost equally among the friends, ensuring a fair division of expenses for the meal.

To determine how much each of the five friends needs to pay, we can use the concept of averages (mean) and divide the total cost by the number of people paying.

In this scenario, the total cost of the meal for 6 people is £12 per person. Since the other 5 friends have decided to pay for the birthday person's meal, they will collectively cover the cost of their own meals plus the birthday person's meal.

To calculate the total cost covered by the five friends, we can subtract the cost of one person's meal (since the birthday person's meal is being paid by the group) from the total cost. The cost of one person's meal is £12.

Total cost covered by the five friends = Total cost - Cost of one person's meal

= (£12 x 6) - £12

= £72 - £12

= £60

Now, to find out how much each of the five friends needs to pay, we divide the total cost covered by the five friends (£60) by the number of friends (5).

Amount each friend needs to pay = Total cost covered by the five friends / Number of friends

= £60 / 5

= £12

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It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Answer:

90.3125 m

Explanation:

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

a = 10m/s^2 (constant)

S=height

U=initial velocity

a= gravitational acceleration

t= time

s = 0 + 1/2 * 10 * 4.25 ^2

u is 0 because it is dropped without velocity

s =90.3125 m

A probabilistic assessment of seismic hazards requires an estimate of the maximum moment intensity M of the largest earthquake that is likely to occur in a given region. The answer is in the image.

Maximum magnitude, an important parameter in calculating seismic hazards, is also a controversial parameter. The choice of value can have a large impact on the final outcome of the results, but it is most likely the magnitude of an earthquake that has not yet occurred in the area under study.

The term size is defined as the number of flocks. For example, the magnitude can be used to describe the speed comparison between a car and a bicycle. It can also be used to describe how far an object has moved or how much an object has in relation to size.

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the kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20 m/s is  6,000 J. Option C is correct answer.

The kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20 m/s can be calculated by using the formula:

K.E = 1/2 mv²

Where, K.E = Kinetic energy of the objectm = Mass of the objectv = Velocity of the object

Putting the given values in the above formula:

K.E = 1/2 mv²K.E = 1/2 × 30 kg × (20 m/s)²K.E = 1/2 × 30 × 400K.E = 6000 joules

The correct answer is C.

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Answer:

Iron and lead generally have low specific heats.

Answer:

Well, you will get in trouble for plagiarism, and you will not get full credit or Maby none

Explanation:

Answer:

1: To make sure people don't think u plagiarized from another person

2: To show that you did good research by listing your sources

Answer:

meteor

Explanation:

A asteroid stays still and a meteor goes fast

Answer:

meteor

Explanation:

or some people call it a shooting star

The slope of the graph represent for ohmic material is as per the graph is that the current will increases with increase in potential.

Ohmic material is defined as anything that complies with Ohm's law, according to which the current flowing through a device is proportional to the applied voltage. Ohmic materials are those that adhere to Ohm's law.

Ohm's Law describes the relationship between voltage and current, and the resistance in a circuit is represented by the slope of the line from a graph of the two quantities. The slope of the distance-time plot, or the speed of the body, is the ratio of the distance traveled by the body to the time required for travel.

Thus, the slope of the graph represent for ohmic material is as per the graph is that the current will increases with increase in potential.

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The electric potential energy stored in the capacitor is 4.70 x 10^-8 Joules.

The electric potential energy stored in a capacitor is given by the formula:

U = (1/2) * C * V^2

where U is the potential energy in Joules, C is the capacitance in Farads, and V is the voltage across the capacitor in Volts.

In this case, we are given that the capacitor stores 9.40 x 10^-10 C of charge at 50.0 V. However, we are not given the capacitance value. Therefore, we cannot calculate the potential energy directly using the above formula.

To find the capacitance value, we can use the formula:

C = Q / V

where Q is the charge stored in the capacitor and V is the voltage across the capacitor.

Substituting the given values, we get:

C = 9.40 x 10^-10 / 50.0

= 1.88 x 10^-11 F

Now we can use the formula for electric potential energy to find the energy stored in the capacitor:

U = (1/2) * 1.88 x 10^-11 * (50.0)^2

= 4.70 x 10^-8 J

Therefore, the electric potential energy stored in the capacitor is 4.70 x 10^-8 Joules.

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I believe you are incorrect. A weathered mountain would appear more jagged.

I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.

If this is incorrect, please, don't refrain to tell me.

Explanation:

it has no energy when considered with respect to earth ,as it has neither height (i e potential energy) nor velocity (i.e kinetic energy).

Answer:

7,272.7m

Explanation:

Our knowns:

a = 5.2m/s^2

t = 32.8s

\(V_{i}\) = 0 m/s

Our Unknowns:

Distance (x): ?

We can use displacement formula:

ΔX = \(V_{i}\) t + \(\frac{1}{2}\) a \(t^{2}\)

Plug in knowns:

X = 0 + 1/2 ((5.2)(32.8))^2

Answer:

ΔX = 7272.7m

(I beleive this should be right) let me know if it helped!

The mass of the object experimented on in this situations is 20 kg. So, option(C) is correct.

A force is something that pushes or pulls an object. The objects' interactions with one another are what causes push and pull. Force can also be expressed using words like stretch and squeeze.

When an object with mass is pushed or pulled, it changes its velocity, according to the definition of force in physics.

The product of mass (m) and acceleration (a) represents the amount of force (F). Mathematically speaking, the force equation or formula can be written as follows:

F = ma

According to given parameters:

Force = 30 N; acceleration = 1.5 m/s². So, mass = 30/1.5 kg = 20 kg.

Force = 70 N; acceleration = 3.5 m/s². So, mass = 70/3.5 kg = 20 kg.

Force = 85 N; acceleration = 4.25 m/s². So, mass = 85/4.25 kg = 20 kg.

Hence, the mass of the object experimented on in this situations is 20 kg.

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1.00 kg of ice at -24.0°C is placed in contact with a 1.00 kg block of a metal at 5.00°C. They come to equilibrium at -8.88°C.

We can use the principle of conservation of heat to solve this problem. The heat lost by the metal must equal the heat gained by the ice.

The heat lost by the metal is given by

Q1 = m1c1ΔT1

Where m1 is the mass of the metal, c1 is its specific heat, and ΔT1 is the change in temperature.

The heat gained by the ice is given by

Q2 = m2c2ΔT2

Where m2 is the mass of the ice, c2 is its specific heat, and ΔT2 is the change in temperature.

Since the two objects come to thermal equilibrium, we can set Q1 equal to Q2

m1c1ΔT1 = m2c2ΔT2

Solving for c1, we get

c1 = m2c2ΔT2 / (m1ΔT1)

By putting these values we get

c1 = (1.00 kg)(2.06 kJ/kg·K)(-24.0°C - (-8.88°C)) / [(1.00 kg)(5.00°C - (-8.88°C))]

c1 = 0.902 kJ/kg·K

Hence, the specific heat of the metal is 0.902 kJ/kg·K.

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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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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 description that tells two processes that scientists think move Earth's lithospheric plates is convection in the mantle and pressure of magma on the edge of the plate.

The Earth's lithosphere is the rocky outer part of Earth which is made up of the brittle crust and the top part of the upper mantle.

The Earth's lithosphere deflects the convections and as the convections churn clockwise of anticlockwise, they drag the  lithosphere with it via friction an this is what is stipulated to cause tectonic plate movements.  

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Answer: convection in the mantle and pressure of magma on the edge of the plate

Explanation: I took the unit test

Answer:

a = (V2 - V1) / t = (0 - 14.6) / 3 = -4.87 m/s^2

It might be useful to convert 14.6 m/s to mph

14.6 m/s * 39.37 in/m = 575 in /s

575 in/s / 12 in/ft = 47.9 ft/sec

47.9 ft/s / 88 ft/sec *  60 mph = 32.6 mph

For the following are four electrical components:

a. For components A and B, both of which obey Ohm's law, the current-voltage characteristics would be a straight line passing through the origin. The slope of the line for component B would be steeper than that of component A, indicating higher resistance.

b. The shape of the characteristic for component C, the filament lamp, can be explained by its construction. A filament lamp consists of a filament made of a resistive material, typically tungsten, which heats up and emits light when an electric current passes through it.

c. The component chosen for D, which does not obey Ohm's law, could be a diode. A diode is a two-terminal electronic component that allows the current to flow in only one direction.

For the following are four electrical components:

a. Sketches of current-voltage characteristics:

For components A and B, both of which obey Ohm's law, the current-voltage characteristics would be a straight line passing through the origin. The slope of the line for component B would be steeper than that of component A, indicating higher resistance.

  Current (I)

     ^

     |          B

     |         /

     |        /

     |       /

     |      /

     |     /

     |    /

     |   /

     |  /

     | /

     |/

     +------------------> Voltage (V)

     Current (I)

     ^

     |          A

     |         /

     |        /

     |       /

     |      /

     |     /

     |    /

     |   /

     |  /

     | /

     |/

     +------------------> Voltage (V)

For component C, a filament lamp, the current-voltage characteristic would be a curve that is not linear. It would exhibit a non-linear increase in current with increasing voltage. At lower voltages, the lamp would have low resistance, but as the voltage increases, the resistance of the filament also increases due to the phenomenon of thermal self-regulation. This leads to a slower increase in current at higher voltages.

For component D, a component that does not obey Ohm's law, the current-voltage characteristic could be any non-linear curve depending on the specific component chosen. Examples of components that do not obey Ohm's law include diodes and transistors.

b. The shape of the characteristic for component C, the filament lamp, can be explained by its construction. A filament lamp consists of a filament made of a resistive material, typically tungsten, which heats up and emits light when an electric current passes through it. As the voltage across the filament increases, the temperature of the filament increases as well, causing its resistance to increase. This increase in resistance results in a slower increase in current with increasing voltage, leading to the characteristic non-linear curve observed.

c. The component chosen for D, which does not obey Ohm's law, could be a diode. A diode is a two-terminal electronic component that allows the current to flow in only one direction. It exhibits a non-linear current-voltage characteristic where it conducts current only when the voltage is above a certain threshold, known as the forward voltage. Below this threshold, the diode has a high resistance and blocks current flow in the reverse direction. The characteristic curve of a diode would show negligible current flow until the forward voltage is reached, after which it exhibits a rapid increase in current with a relatively constant voltage.

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Answer: Option A: electrical engineering.

Explanation: The field of electrical engineering, sometimes called electronics engineering, mostly outside the US, is a field of engineering that covers a number of subtopics that include large scale power systems, electronics, control systems, signal processing and telecommunications. In this case the engineer would specifically have to know about electronics (computer chips), signal processing (cell signals) and telecommunications.

The time taken for the group of people to pull the car, giving it a velocity of 1.5 m/s is 3.75 s

Momentum is simply defined as the product of mass and velocity i.e

Momentum = mass × velocity

To answer the question given, we'll begin by calculating the change in momentum. This can be obtained as follow:

Mass = 1500 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 1.5 m/s

Change in momentum = m(v – u)

Change in momentum = 1500 (1.5 – 0)

Change in momentum = 1500 × 1.5

Finally, we shall determine the time

Change in momentum = 2250 Kg•m/s

Force (F) = 600 N

Impulse = Ft = change in momentum

FT = change in momentum

600 × t = 2250

Divide both side by 600

t = 2250 / 600

Thus, the time required is 3.75 s

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Answer:

(a) Approximately \(0.335\; \rm m\).

(b) Approximately \(1.86\; \rm m\cdot s^{-1}\).

(c) Approximately \(0.707\; \rm m\).

(d) Approximately \(0.228\; \rm m\).

Explanation:

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy \(\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2\).

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of \(D\) and compressed the spring by the same distance.

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

\(\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2\).

Assume that \(g = 9.81\; \rm m \cdot s^{-2}\). In the equation above, all symbols other than \(D\) have known values:

Substitute in the known values to obtain an equation for \(D\) (where the unit of \(D\!\) is \(m\).)

\(3.178 = 2.69775\, D + 25\, D^2\).

\(2.69775\, D + 25\, D^2 + 3.178 = 0\).

Simplify and solve for \(D\). Note that \(D > 0\) because the energy lost to friction should be greater than zero.

\(D \approx 0.335\; \rm m\).

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

\(\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J\).

As the object moves to the left, part of that energy will be lost to friction:

\((\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J\).

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

\(2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J\).

Calculate the velocity corresponding to that kinetic energy:

\(\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}\).

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy (\(1.91\; \rm J\)) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is \(\mu \cdot m \cdot g\).

\(\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m\).

Similar to (a), solving (d) involves another quadratic equation about \(D\).

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) \(1.91\; \rm J\).

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

\(\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2\).

\(25\, D^2 + 2.69775\, D - 1.90811\approx 0\).

Again, \(D > 0\) because the energy lost to friction is greater than zero.

\(D \approx 0.228\; \rm m\).

The energy transferred between the object and the spring as a closed system, therefore, conserved are;

(a) The distance of compression, d ≈ 0.3354 meters

(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s

(c) The distance where the object comes to rest, D ≈ 0.7071 m

(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m

The reason the above values are correct are as follows;

The known parameters are;

Mass of the object, m₁ = 1.10 kg

Coefficient of friction, μ = 0.250

The initial speed of the object, \(v_i\) = 2.60 m/s

Force constant of the spring, K = 50.0 N/m

Distance the spring is compressed by the object = d

(a) Conservation of energy principle

\(Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2\)

Work done = Force × Distance

Friction force, \(F_f\) = W × μ

Weight, W = m·g

Weight = Mass × Acceleration

Energy transferred by object = Work done by spring + Work done by friction

\(Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718\)

Energy transferred by object = 3.718 J

\(Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2\)

\(Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2\)

\(W_{spring}\) = 25·d²

Work done by friction, \(W_{friction}\) = 1.10×9.81×0.250×d = 2.69775·d

Therefore;

3.718 = 25·d² + 2.69775·d

25·d² + 2.69775·d - 3.718 = 0

Solving gives

The distance of the compression d ≈ 0.3354 m

(b) The energy given by the spring = 25·d²

The work done by friction, \(W_{friction}\) = 2.69775·d

Kinetic energy given to object = 0.55·v²

0.55·v² = 25·d² - 2.69775·d

0.55·v² = 25×0.3354² - 2.69775×0.3354

∴ v = √(3.4682) = 1.8623

The velocity of the object at the un stretched position, v ≈ 1.8623 m/s

(c) The kinetic energy, K.E. of the object on the way left is given as follows;

K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J

The work done by friction before object comes to rest = 2.69775·D

\(D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m\)

The distance where the object comes to rest, D ≈ 0.7071 m

(d) The work done on spring, \(W_{spring}\) = 25·D'²

Work done on friction, \(W_{friction}\) = 2.69775·D'

Kinetic energy of object, K.E. ≈ 1.90751 J

K.E. = \(W_{spring}\) + \(W_{friction}\)

1.90751 ≈ 25·D'² + 2.6775·D'

25·D'² + 2.6775·D' - 1.90751 = 0

Solving with a graphing calculator gives;

D' ≈ 0.2278 m

The new value of the distance D = 0.2278 m

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The characteristic of a seismic wave that the motion shown by the sensor on the seismometer represents is the amplitude of the seismic wave.

Seismic waves are waves that transport energy away from the focus of an earthquake, and they cause the ground to shake. Seismometers are instruments that measure the shaking of the Earth caused by seismic waves. The black arrows in the picture show the relative motion of the sensor on the seismometer. The sensor moves back and forth as the ground shakes, and the amplitude of this motion represents the intensity of the seismic wave.The amplitude of a seismic wave is a measure of the strength or energy of the wave. It is the maximum displacement of the ground from its resting position during the passage of the wave. The greater the amplitude of the seismic wave, the stronger the shaking of the ground and the more damage it can cause. The amplitude of a seismic wave is usually measured in millimeters or centimeters.In summary, the motion shown by the sensor on the seismometer represents the amplitude of the seismic wave. This is an important characteristic of a seismic wave as it determines the intensity of the shaking and the potential damage that the earthquake can cause. Seismologists use this information to study earthquakes and to understand their effects on the Earth.

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Answer:

Power = 2000J/s

Explanation:

Power= Total Work done/Time

Given

mass = 100kg

distance(height) = 10m

time = 5s

acceleration due to gravity = 10m/s²

Work done = force × distance

Finding Force

Force = mass × acceleration due to gravity

Force = 100kg × 10m/s²

Force = 1,000N

Finding work done

Work done = Force × distance

Work done = 1,000N × 10m

Work done = 10,000J

Finding Power

Power = work done/time

Power = 10,000J/5s

Power = 2,000J/s

Therefore Power is 2,000J/s

FOREBRAIN: (front)

Includes:  cerebrum, thalamus, hypothalamus, pituitary gland, limbic system, and the olfactory bulb.

Example: THE PITUITARY GLAND carries out some of our most basic functions. This part of the brain controls one'sons metabolism, growth,reproduction, and it even affects the way we store energy from the foods we eat.

MIDBRAIN:(middle)

Includes: cranial nerve nuclei, tectum, tegmentum, colliculi, and crura cerebi.

Example: THE TECTUM; works as a neural processor. It controls swift eye movements, avoidance movements, and takes notice at your own or other's movements.

HINDBRAIN: (back)

Includes: pons, cerebellum, and medulla oblongata.

Example: THE CEREBELLUM, the part of the brain responsible for balance. Subtly it helps our bodies shift in ways to keep us balanced, spanning from posture to how we walk or stand.

Hopefully this helps, Have an amazing day

 ~JET