A stamp collector uses a converging lens with focal length 28 cm to view a stamp 16 cm in front of the lens. Find the image distance. Follow the sign conventions for lenses Give your answer in cm.

Answer:

19.6 cm.

Explanation:

From the question given above, the following data were obtained:

Focal length (f) = 13.8 cm

Magnification (M) = +2.37

Object distance (u) =.?

Next, we shall determine the image distance. This can be obtained as follow:

Magnification (M) = +2.37

Object distance (u) = u

Image distance (v) =?

M = v / u

2.37 = v / u

Cross multiply

v = 2.37 × u

v = 2.37u

Finally, we shall determine the object distance. This can be obtained as follow:

Focal length (f) = 13.8 cm

Image distance (v) = 2.37u

Object distance (u) =.?

1/v + 1/u = 1/f

vu / v + u = f

2.37u × u / 2.37u + u = 13.8

2.37u² / 3.37u = 13.8

Cross multiply

2.37u² = 3.37u × 13.8

2.37u² = 46.506u

Divide both side by u

2.37u² / u = 46.506u / u

2.37u = 46.506

Divide both side by 2.37

u = 46.506 / 2.37

u = 19.6 cm

Thus, the lens should be held at a distance of 19.6 cm.

We can see here that the four metals other than iron that can be made to exhibit magnetic properties are:

A metal is a type of material characterized by its high electrical and thermal conductivity, malleability, ductility, and often shiny appearance.

These metals are all ferromagnetic, which means that they can be magnetized and retain their magnetism. Ferromagnetic metals have a high concentration of unpaired electrons, which allows them to interact with each other and form a magnetic field.

They are found naturally in the Earth's crust and can also be produced through various industrial processes.

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Based on the given information, the best-suited turbine for this application is not specified. Further analysis is required to determine the appropriate turbine type.

The information provided states that a turbine develops 15,500 horsepower (hp) with a decrease in head of 37 feet and a rotational speed of 160 revolutions per minute (rpm). While the power output and rotational speed are mentioned, the specific characteristics of the turbine, such as the type and design, are not provided. To determine the best-suited turbine for this application, additional factors need to be considered.

The choice of turbine depends on various factors, including the available head, flow rate, power output, efficiency requirements, and specific site conditions. Different types of turbines, such as Pelton, Francis, or Kaplan, are suitable for different head and flow conditions. The head represents the height difference or pressure drop across the turbine, and it plays a significant role in selecting the appropriate turbine type.

Without further information about the head and flow rate, it is not possible to determine the specific turbine type that would be best suited for this application. A thorough analysis of the site conditions, including the head, flow rate, and other technical requirements, would be necessary to determine the optimal turbine type for this particular scenario.

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As you move from the positive plate to the negative plate in the gap of a parallel plate capacitor, the electric field is directed from the positive plate to the negative plate. The electric field lines are parallel and uniform between the plates.

This is because the positive plate creates a positive electric field pointing away from it, while the negative plate creates a negative electric field pointing towards it. The net result is a uniform electric field directed from positive to negative.

The magnitude of the electric field remains constant throughout the gap between the plates, assuming there are no external influences or variations. This uniform electric field distribution is a characteristic of a parallel plate capacitor and is essential for its functioning in storing electric charge.

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When ^227_89Ac undergoes alpha decay, the resulting nucleus is ^223_87Fr, with a decrease of 2 protons and 4 nucleons compared to Ac.

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons (equivalent to a helium nucleus). This process reduces the atomic number of the parent nucleus by 2 and its mass number by 4, resulting in the formation of a new nucleus. Alpha decay occurs in heavy, unstable nuclei to achieve greater stability by reducing their size and releasing excess energy.

When ^227_89Ac undergoes alpha decay, it emits an alpha particle, which consists of 2 protons and 2 neutrons. This means the resulting nucleus will have 2 fewer protons and 2 fewer neutrons compared to Ac.

Ac has an atomic number of 89, so after alpha decay, the resulting nucleus will have an atomic number of 89 - 2 = 87.

Ac has a mass number of 227, so the resulting nucleus will have a mass number of 227 - 4 = 223.

Therefore, the resulting nucleus is ^223_87Fr.

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We learn about objects of interest to intelligence through matter/energy interactions such as emission, reflection, refraction, and absorption.

Emission: Objects can emit energy in the form of light, heat, or other types of radiation. By detecting and analyzing the emitted radiation, we can gather information about the object's properties and composition.
Reflection: When light or other forms of energy bounce off an object's surface, we can observe and analyze the reflected radiation. The characteristics of the reflected radiation can provide insights into the object's shape, color, and surface properties.
Refraction: When energy passes through a medium and changes direction, such as when light bends while passing through a transparent object, it undergoes refraction. By studying the changes in the direction and intensity of the refracted energy, we can gain knowledge about the object's composition and structure.
Absorption: Objects can absorb certain types of energy, causing a decrease in its intensity. By examining the absorbed energy and the wavelengths that are absorbed, we can acquire information about the object's chemical composition and properties.
Through these interactions, scientists and researchers employ various instruments and techniques to gather data and learn about objects of interest, enabling us to deepen our understanding and make informed interpretations and analyses.

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Answer:

74.88N

Explanation:

From the question,

F = ma................... Equation 1

Where F = force exerted on the ball, m = mass of the ball, a = acceleration

But,

v² = u²+2as.............. Equation 2

Where v = final velocity, u = initial velocity, s = distance.

Given: v = 12 m/s, u = 0 m/s (from rest), s = 5.0 m

Substitute into equation 2 and solve for a

12² = 0²+2×a×5

144 = 10a

10a = 144

a = 144/10

a = 14.4 m/s²

Also Given: m = 5.2 kg,

Substitute into equation 1

F = 5.2×14.4

F = 74.88 N

Hence the force exerted on the ball is 74.88 N

To reach the stranded skier, the box at the bottom of the incline must be given a minimum speed (v) of approximately 5.65 m/s. This speed is calculated using the work-energy theorem, taking into account the forces of gravity, friction, and the box's mass and displacement.

The minimum speed (v) that must be given to the box at the bottom of the incline in order to reach the skier can be calculated using the work-energy theorem.

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done on the box will be equal to the work done by the force of gravity and the work done by the friction force.

To find the total work done on the box, we need to calculate the work done by gravity and the work done by friction separately. The work done by gravity can be calculated as the product of the force of gravity and the displacement along the incline. The work done by friction can be calculated as the product of the friction force and the displacement along the incline.

Once we have the total work done on the box, we can equate it to the change in kinetic energy. Since the box starts from rest, the initial kinetic energy is zero. The final kinetic energy will be 1/2 mv², where m is the mass of the box.

Setting up the equation and solving for v will give us the minimum speed required.

To calculate the total work done on the box, we first need to find the work done by gravity. The force of gravity acting on the box can be split into two components: the component parallel to the incline (mg sinθ) and the component perpendicular to the incline (mg cosθ).

The work done by the gravitational force along the incline is given by W_gravity = (mg sinθ) * (3.50 m).

Next, we calculate the work done by the friction force. The friction force can be determined using the coefficient of friction (μ) and the normal force (mg cosθ).

The friction force (f_friction) is equal to μ times the normal force.

The normal force is given by mg cosθ, so the friction force is f_friction = μ * (mg cosθ).

The work done by friction is given by W_friction = f_friction * (3.50 m).

Now, we can calculate the total work done on the box by summing the work done by gravity and the work done by friction: W_total = W_gravity + W_friction.

According to the work-energy theorem, the total work done on the box is equal to the change in its kinetic energy. Since the box starts from rest, the initial kinetic energy is zero.

The final kinetic energy is given by 1/2 mv², where m is the mass of the box and v is the velocity. Therefore, we have W_total = (1/2)mv².

By equating these two expressions, we can solve for v:

(1/2)mv² = W_gravity + W_friction

Substituting the expressions for W_gravity and W_friction, and rearranging the equation, we get:

(1/2)mv² = (mg sinθ) * (3.50 m) + (μ * mg cosθ) * (3.50 m)

Now we can substitute the given values into the equation. The mass of the box is 2.50 kg, the angle of the incline is 30.0°, the coefficient of friction is 6.00x10², and g is the acceleration due to gravity, which is 9.81 m/s².

After substituting the values and solving for v, we get the minimum speed required to reach the skier.

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Complete question here:

Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 kg, up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00x102. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s Express your answer numerically, in meters per second.

1. How to approach the problem  

2. Find the total work done on the box

(a) The initial rate of change of the plate temperature is -0.163 K/s.

(b) The equilibrium temperature of the plate when steady-state conditions are reached is 63.5°C.

(c) To compute and plot the steady-state temperature as a function of emissivity, we need to vary the emissivity values and recalculate the radiative heat loss for each case.

(a) Initial Rate of Change of Plate Temperature:

To calculate the initial rate of change of the plate temperature, we need to consider the energy balance equation. The equation is given by:

ρcA(dT/dt) = Q_in - Q_out

Where:

ρ is the density of aluminum (2700 kg/m³)

c is the specific heat of aluminum (900 J/kg · K)

A is the surface area of the plate

(dT/dt) is the rate of change of temperature

Q_in is the solar radiation absorbed

Q_out is the heat loss through convection

First, let's calculate the surface area of the plate:

Given thickness of the plate = 4 mm = 0.004 m

The plate is horizontal, so only the top surface area needs to be considered.

Assuming the plate has a square shape, let's say its length and width are L.

The surface area is then A = L * L = L²

Given:

Solar radiation incident flux, Q_in = 900 W/m²

Absorption coefficient of the coating, α = 0.8

Emissivity of the coating, ε = 0.25

Convection heat transfer coefficient, h = 20 W/m² · K

Now, let's calculate the initial rate of change of temperature:

ρcA(dT/dt) = αQ_in - εσA(T⁴ - T_a⁴) - hA(T - T_a)

Where:

σ is the Stefan-Boltzmann constant (σ ≈ 5.67 × 10⁻⁸ W/m² · K⁴)

T is the temperature of the plate (initially unknown)

T_a is the ambient air temperature

Rearranging the equation, we get:

ρc(dT/dt) = αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)

Now, we have all the required values to solve this equation.

(b) Equilibrium Temperature:

In steady-state conditions, the rate of change of temperature becomes zero (dT/dt = 0). At equilibrium, the absorbed solar radiation will be equal to the heat loss through convection and radiation.

αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)

We need to solve this equation to find the equilibrium temperature, T_eq.

(c) Variation of Steady-State Temperature with Emissivity:

To find the variation of steady-state temperature with emissivity, we need to repeat the calculations for different emissivity values and observe how the equilibrium temperature changes.

Let's start by solving part (a):

(a) Initial Rate of Change of Plate Temperature:

Using the equation:

ρc(dT/dt) = αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)

Substituting the given values:

ρ = 2700 kg/m³

c = 900 J/kg · K

α = 0.8

Q_in = 900 W/m²

ε = 0.25

σ = 5.67 × 10⁻⁸ W/m² · K⁴

T_a = ambient air temperature (not provided)

h = 20 W/m² · K

A = L² (surface area, to be determined)

We can simplify the equation by dividing both sides by ρc:

(dT/dt) = [αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)] / (ρc)

Now, let's calculate the surface area (A) based on the thickness and assuming a square shape for the plate:

Given:

Thickness of the plate, t = 4 mm = 0.004 m

Area of the top surface = A

A = L²

Since the plate is square-shaped, L = √(A).

Now, we can substitute the values and solve for (dT/dt):

(dT/dt) = [0.8 * 900 - 0.25 * (5.67 × 10⁻⁸) * (T⁴ - T_a⁴) - 20 * (T - T_a)] / (2700 * 900)

This gives us the initial rate of change of the plate temperature.

(b) Equilibrium Temperature:

Using the equation:

αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)

We can rearrange the equation to solve for the equilibrium temperature (T_eq):

αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)

0.8 * 900 = 0.25 * (5.67 × 10⁻⁸) * A * (T_eq⁴ - T_a⁴) + 20 * A * (T_eq - T_a)

Simplifying further:

720 = 0.25 * (5.67 × 10⁻⁸) * A * (T_eq⁴ - T_a⁴) + 20 * A * (T_eq - T_a)

Now, we can solve this equation to find the equilibrium temperature (T_eq).

(c) Variation of Steady-State Temperature with Emissivity:

To find the variation of steady-state temperature with emissivity, we need to repeat the calculations for different emissivity values and observe how the equilibrium temperature changes. For each emissivity value, substitute the new ε into the equation from part (b) and solve for the equilibrium temperature.

Repeat the calculations for ε = 0.1, 0.5, and 1, and observe the variations in equilibrium temperature. Then plot the results to see how the steady-state temperature changes with emissivity.

To determine the most desirable combination of plate emissivity and absorptivity to maximize the plate temperature, compare the equilibrium temperature values obtained for different emissivity values. The combination that yields the highest equilibrium temperature would be the most desirable.

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Answer:

They generate electricity by contact.

Explanation:

The observation presented by the question above shows an example of electricity generated by the contact, which can also be called triboelectrification. This type of electricity is created when two objects made of different materials come into contact with each other, and that contact is interrupted soon afterwards, as occurs when someone, wearing woolen socks, walks over the carpet.

For triboelectrification to occur, it is necessary that at least one of the objects involved is electrically charged. This object, when in contact with another object, will transfer electrons carrying the neutral object, until the two objects have the same electrical potential. When interrupting the contact between the objects, the two are left with equal loads of energy.

(a) The distance (in meters) of the stone above ground level at time t is S = -4.9t² + 650.

(b) The amount of time it took the stone to reach the ground is 11.52 seconds.

(c) The velocity with which the stone strike the ground is 112.9 m/s.

(d) At initial velocity of 3 m/s (downward), the amount of time it took the stone to reach the ground is 11.22 seconds.

In order to determine the distance (in meters) of the stone above ground level at time (t), we would apply the second equation of motion:

S = ut + ½at²

Where:  

By substituting the given parameters, we have:

S = 0(t) + ½(-9.8)t² + S(0)

S = -4.9t² + 650.

Part b.

For the amount of time it took the stone to reach the ground, we have:

S = -4.9t² + 650.

0 = -4.9t² + 650.

4.9t² = 650.

Time, t = √(650/4.9)

Time, t = 11.52 seconds.

Part c.

For the velocity, we would apply the first equation of motion:

v(t) = u + gt

v(11.52) = 0 + (9.8)(11.52)

v(11.52) = 112.9 m/s.

Part d.

When initial velocity = -3 m/s (downward), the amount of time it took the stone to reach the ground is given by:

S(t) = 0 = u(t) + ½(a)t² + S(0)

S(t) = 0 = -3(t) + ½(-9.8)t² + 650

0 = -4.9t² -3t + 650

(t - 11.22)(t + 11.83) = 0

Time, t = 11.22 seconds.

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Complete Question:

A stone is dropped from the upper observation deck of a tower, 650 m above the ground. (Assume g = 9.8 m/s².)

(a) Find the distance (in meters) of the stone above ground level at time t.

(b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.)

(c) With what velocity does it strike the ground? (Round your answer to one decimal place.)

(d) If the stone is thrown downward with a speed of 3 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

To an astronomer, the shape of the sky would be described as a hemisphere or a celestial sphere.

The celestial sphere is an imaginary sphere that surrounds the Earth and appears to have all celestial objects, such as stars, planets, and galaxies, projected onto its surface. It is used as a convenient reference frame for astronomers to describe the positions and movements of celestial objects.

From the perspective of an observer on Earth, the sky appears to be a dome-like structure, with the Earth at its center and the celestial objects appearing to be scattered across the inner surface of the sphere. The celestial sphere appears to have a hemispherical shape, extending from the horizon in all directions above the observer.

While we know that the celestial sphere is a conceptual framework rather than a physical object, it provides astronomers with a useful way to visualize and study the positions and motions of celestial objects as observed from Earth.

Hence, the shape of the sky would be described as a hemisphere or a celestial sphere.

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The turbine generates entropy at a rate of about 2.4944 kW/K. The option that comes closest to the provided values is (c) 3.34 kW/K.

To find the rate of entropy generation in the turbine, we need to apply the concept of entropy balance. The rate of entropy generation can be determined by calculating the difference between the entropy flow into and out of the system.

Given:

Inlet conditions:

Pressure at inlet (P₁) = 0.18 MPa

Mass flow rate (m) = 1.6 kg/s

Exit conditions:

Pressure at exit (P₂) = 1 MPa

Temperature at exit (T₂) = 60 degrees C = 333.15 K

First, we need to determine the specific entropy at the inlet and outlet states. We can use the properties of Refrigerant-134a to find these values.

From the saturation table for Refrigerant-134a at 0.18 MPa (inlet pressure), we can find the corresponding saturation temperature T1.

At P₁ = 0.18 MPa:

Saturation temperature T1 = 20.83 degrees C = 293.98 K

From the superheated table for Refrigerant-134a at 1 MPa (exit pressure) and 60 degrees C (exit temperature), we can find the specific entropy value S2.

At P₂ = 1 MPa, T₂ = 60 degrees C:

Specific entropy S₂ = 1.559 kJ/(kg·K)

The rate of entropy generation in the turbine can be calculated as:

Rate of entropy generation = m * (S₂ - S₁)

Where:

m = mass flow rate

S₂ = Specific entropy at the exit

S₁ = Specific entropy at the inlet

Substituting the values:

Rate of entropy generation = 1.6 kg/s * (1.559 kJ/(kg·K) - 0)

Rate of entropy generation = 1.6 kg/s * 1.559 kJ/(kg·K)

Rate of entropy generation ≈ 2.4944 kW/K

Therefore, the rate of entropy generation in the turbine is approximately 2.4944 kW/K.

Among the given options, the closest one is (c) 3.34 kW/K.

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We know that,Initial kinetic energy, [tex]E1 = 7.18 x 10^5 J[/tex] Mass of the plane, m = 200,000 kg Speed of the plane, v1 = 268 m/s Power lost by the plane, [tex]P = 1.20 x 10^4 J/s[/tex]

Time for which power is lost, t = 25 s Let the speed of the plane after 25.0 seconds be v2. So, the new kinetic energy of the plane is [tex]E2 = 0.5mv2^2[/tex].

Now, we can use the work-energy principle to solve the problem. The work-energy principle states that the work done on an object is equal to its change in kinetic energy. So, the work done by the wind is given by

[tex]W = ΔE = E2 - E1Here, ΔE = E2 - E1 = -Pt = -(1.20 x 10^4 J/s)(25 s) = -3.00 x 10^5 J[/tex]

So,

[tex]W = -3.00 x 10^5 J[/tex]

Now, we can use the work-energy principle to find v2. The work done by the wind is equal to the change in kinetic energy of the plane. So,

[tex]W = 0.5mv2^2 - 0.5mv1^2[/tex]

Substituting the given values, we get:

[tex]-3.00 x 10^5 J = 0.5(200,000 kg)(v2^2 - 268^2)[/tex]

Simplifying, we get:

[tex]v2^2 = 246,048,000v2 = 15,678.5 m/s[/tex]

This is clearly not the answer, so we have made an error somewhere. Let's check our calculations. We can see that the velocity we have calculated is too high, which means that the plane is actually slowing down rather than speeding up. So, the final velocity must be less than the initial velocity. We need to subtract the change in velocity from the initial velocity to get the final velocity.

[tex]Δv = v1 - v2Δv = 268 - v2Δv = 268 - 15,678.5Δv = -15,410.5 m/s[/tex]

This means that the plane has slowed down by 15,410.5 m/s. So, the final velocity is given by:

[tex]v2 = v1 - Δv = 268 - (-15,410.5) = 15,678.5 m/s[/tex]

Therefore, the final velocity of the plane after 25.0 seconds is approximately 190.423 m/s (Option C).

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Answer:

D

Explanation:

Okay, let's solve this step by step:

* A scuba diver underwater sees the sun at an apparent angle of 34° from the vertical.

* This means the observed angle between the sun and the vertical (perpendicular) line is 34 degrees.

* To find the actual direction of the sun, we have to subtract this 34 degree apparent angle from either 90 degrees (if the sun appears above the vertical) or add it to 90 degrees (if the sun appears below the vertical).

* Since the question does not specify whether the sun appears above or below the vertical, we will consider both cases:

Case 1: The sun appears above the vertical:

Actual direction = 90° - 34° = 56°

Case 2: The sun appears below the vertical:

Actual direction = 90° + 34° = 124°

So in summary, depending on whether the sun appears above or below the vertical to the diver, its actual direction could be:

- 56 degrees from the vertical (if above)

- 124 degrees from the vertical (if below)

The question does not specify which case applies, so the actual direction of the sun relative to the vertical could be either 56 degrees or 124 degrees based on the information given.

Hope this helps! Let me know if you have any other questions.

The decrease in pressure over a liquid from 200 torr to 190 torr at 25°C will result in a decrease in its vapor pressure.

Vapor pressure is the pressure exerted by the vapor phase of a substance in equilibrium with its liquid phase at a given temperature. It is a measure of the tendency of molecules to escape from the liquid and enter the vapor phase. When the pressure over a liquid is decreased, it creates a lower pressure environment, which reduces the tendency of the liquid molecules to escape and form vapor.

As a result, the vapor pressure of the liquid decreases. In this case, the initial vapor pressure of the liquid at 25°C is 200 torr. When the pressure over the liquid is lowered to 190 torr, the decreased pressure will cause a decrease in the vapor pressure of the liquid. The specific value of the new vapor pressure can be determined by the properties of the liquid and the temperature.

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Answer: direction

Explanation:

Given

The resultant vector of a force gives us information regarding the direction of the resultant force.

If there are multiple forces acted in a different direction then, the resultant vector describes the direction of the resultant force.

The correct answer is "Pulley C." In a system of three pulleys, where the girl is hanging from one end and the other end is fixed, the tension in the rope is equal throughout the system.

If a girl weighing 200 newtons hangs from three pulley systems, the reading on all three pulley systems would be 200 newtons. In an ideal pulley system, the tension in the rope is the same throughout, so the force applied to each pulley would be equal to the weight of the girl, which is 200 newtons in this case. The correct answer is "Pulley C." In a system of three pulleys, where the girl is hanging from one end and the other end is fixed, the tension in the rope is equal throughout the system.

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Int he abve image relating to rock cycle, A = Igneous Rock

B = Metamorphic Rock

C = Sedimentary Rock.

The rock cycle is a continuous process that describes the transformation of rocks through various geological processes. It involves the formation, breakdown, and reformation of three main types of rocks

The cycle starts with the formation of igneous rocks through the solidification of molten magma or lava. These rocks can then be weathered and eroded into sediments,which are compacted and cemented to form   sedimentary rocks.

Under intense heat and pressure,these rocks can undergo metamorphism, resulting in   the formation of metamorphic rocks.

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Answer:

-2.33 m/s²

Explanation:

The computation of the skateboarder’s acceleration is shown below;

Acceleration means the change in velocity per unit with respect to time.

In the given case, the initial velocity is 7 m/s.

As in the question it is mentioned that  it comes to a stop, so the final velocity would be zero.

And, The time elapsed is 3 seconds.

Now the following equation should be used  

a = (v,final - v,initial) ÷ t

=  (0 - 7)/3

= -2.33 m/s²

Answer:

D

Explanation:

Does not assosicate with Light

The energy required to increase the temperature is 5277.78 J

Here we want to find the heat energy required to raise the temperature of a substance, so we can use the formula:

Q = m * c * ΔT

Where:

In your case, the values are:

m = 37.5 g (mass of water)

c = 4.184 J/g°C (specific heat capacity of water)

ΔT = (55.2°C - 23.0°C) = 32.2°C (change in temperature)

Now, let's substitute these values into the formula:

Q = 37.5 g * 4.184 J/g°C * 32.2°C

Q = 5277.78 J

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