Igneous rock can be changed into which type of rock through the application of heat and pressure?


metamorphic rock


clastic sedimentary rock


non-foliated igneous rock


foliated igneous rock

According to Newton's third law of motion, the reaction force exerted by the air on the balloon is equal in magnitude but opposite in direction to the action force exerted by the balloon on the air.

Given that the magnitude of the action force (F_balloon) is -3 N, the magnitude of the reaction force (F_air) will also be 3 N. The negative sign indicates that the forces are in opposite directions, but when considering magnitudes, we ignore the negative sign.

Therefore, the magnitude of the reaction force of the air pushing on the balloon is 3 N.

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Explanation:

Assuming that the person is the only force applying work to the box, the total work done by the person is 993.4 J. This is calculated by multiplying the force of gravity (98.1 N) with the vertical distance (1.40 m) to obtain 137.4 J, and then adding the work done in carrying the box horizontally (8.00 m × 98.1 N = 756.8 J). Therefore, the total work done is 137.4 J + 756.8 J = 993.4 J.

Answer:

T is less than or equal to 19 N

Explanation:

Charge-coupled devices (CCDs) are electronic imaging sensors that are widely used in digital cameras, telescopes, and other imaging devices. Some of the major advantages of CCDs over other imaging techniques include:

High sensitivity: CCDs are highly sensitive to light, allowing them to capture clear and detailed images even in low-light conditions. This makes them ideal for applications such as astrophotography and microscopy, where light levels are often very low.

Low noise: CCDs produce very little electronic noise, which means that images captured using CCDs tend to be very clean and free of unwanted artifacts. This makes CCDs particularly well-suited for scientific imaging applications, where accuracy and precision are critical.

High resolution: CCDs can capture images with very high spatial resolution, allowing them to resolve fine details in images. This makes them ideal for applications such as digital microscopy, where the ability to capture and analyze fine details is essential.

Overall, CCDs offer a powerful combination of high sensitivity, low noise, and high resolution that makes them well-suited for a wide range of imaging applications, from scientific research to artistic photography.

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Answer:

The spring force is called a restoring force because the force exerted by the spring is always in the opposite direction to the displacement. This is why there is a negative sign in the Hooke's law equation. Pulling down on a spring stretches the spring downward, which results in the spring exerting an upward force.

Answer:

A

Explanation:

Answer:

Can't see the pic

Explanation:

The continuous motion of air and water distributes the sun's energy.

The continuous motion of air and water helps to distribute the sun's energy throughout the earth, and thus maintain fairly constant average temperatures.

The equatorial regions of earth receive more heat than any regions of higher latitudes. By means of the convection air currents, whereby hotter and the lighter air from the equatorial regions is circulated for colder and denser air from the all regions of higher latitudes, heat is distributed between these regions.

Also, the heat of the surface waters at the equatorial regions is circulated in the form of convection currents as seas and oceans move around, to other known regions of the earth in order to maintain a fairly constant average temperature.

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If you are at zero degrees latitude and zero degrees longitude, you are in the Atlantic Ocean.

This location is often referred to as the "Prime Meridian" and is the line of longitude that separates the Eastern Hemisphere from the Western Hemisphere. It is also the starting point for measuring longitude, with longitudes to the east and west being measured in degrees relative to this line. The Prime Meridian intersects the Equator at zero degrees latitude, which is the line of latitude that circles the Earth at 0 degrees North and South and separates the Northern Hemisphere from the Southern Hemisphere.

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(A) The pressure transients at the mid-point of the pipeline are approximately 1,208,277 Pa.
(B) Friction in the pipeline affects the calculated pressure transients by increasing the overall resistance to flow

(a) The pressure transients at the mid-point of the pipeline can be calculated using the water hammer equation. Water hammer refers to the sudden changes in pressure and flow rate that occur when there are rapid variations in fluid flow. The equation is given by:

ΔP = (ρ × ΔV × c) / A

Where:

ΔP = Pressure change

ρ = Density of water

ΔV = Change in velocity

c = Wave speed

A = Cross-sectional area of the pipe

First, let's calculate the change in velocity:

ΔV = Q / A

Q = Flow rate = 0.12 m/s

A = π × ((d/2)^2 - ((d-2t)/2)^2)

d = Internal diameter of the pipe = 300 mm = 0.3 m

t = Pipe wall thickness = 5 mm = 0.005 m

Substituting the values:

A = π × ((0.3/2)^2 - ((0.3-2(0.005))/2)^2

A = π × (0.15^2 - 0.1495^2) = 0.0707 m^2

ΔV = 0.12 / 0.0707 = 1.696 m/s

Next, let's calculate the wave speed:

c = √(E / ρ)

E = Young's modulus of steel = 210x10^9 Pa

ρ = Density of water = 1000 kg/m^3

c = √(210x10^9 / 1000) = 4585.9 m/s

Finally, substituting the values into the water hammer equation:

ΔP = (1000 × 1.696 × 4585.9) / 0.0707 = 1,208,277 Pa

Therefore, the pressure transients at the mid-point of the pipeline are approximately 1,208,277 Pa.

(b) Friction in the pipeline affects the calculated pressure transients by increasing the overall resistance to flow. As water moves through the pipe, it encounters frictional forces between the water and the pipe wall. This friction causes a pressure drop along the length of the pipeline.

The presence of friction results in a higher effective wave speed, which affects the calculation of pressure transients. The actual wave speed in the presence of friction can be higher than the wave speed calculated using the Young's modulus of steel alone. This higher effective wave speed leads to a reduced pressure rise during the transient event.

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load=effort÷load arm

load=50N÷100

load=0.5

it can carry load of 0.5

Answer:Argon

Neon

Helium

Krypton

Xenon

Answer:

The thermal efficiency, \(\eta _{reheat}\), of the Rankine cycle with reheat is 36.81%

Explanation:

p₁ = 8 MPa = 80 Bars

T₁ = 450°C = 723.15 K

From steam tables, we have;

v₁ = 0.0381970 m³/kg

h₁ = 3273.23 kJ/kg

s₁ = 6.5577 kJ/(kg·K) = s₂

The p₂ = 0.8 MPa

T₂ = Saturation temperature at 0.8 MPa = 170.414°C = 443.564 K

h₂ = 2768.30 kJ/kg

\(T_{2'}\) = 400°C = 673.15 K

\(h_{2'}\) = at 400°C and 0.8 MPa = 3480.6 kJ/kg

p₃ = 10 kPa = 0.1 Bar

T₃ = Saturation temperature at 10 kPa = 45.805 °C = 318.955 K

h₃ = 2583.89 kJ/kg

h₄ = \(h_{3f}\) = 191.812 kJ/kg

The thermal efficiency, \(\eta _{reheat}\), of a Rankine cycle with reheat is given as follows;

\(\eta _{reheat} = \dfrac{\left (h_{1}-h_{2} \right )+\left (h_{2'}-h_{3} \right )-W_{p}}{h_{1}-\left (h_{4}+W_{p} \right )+\left (h_{2'}-h_{2} \right )}\)

Therefore, we have;

\(\eta _{reheat} = \dfrac{(3273.23 -2768.30 ) + (3480.6 -2583.89 ) - 8.04)}{(3273.23 -(191.812 + 8.04) + (3480.6 -2768.30 ) } = 0.3681\)

Which in percentage is 36.81%.

Answer:

A This is a nuclear reaction, because the atoms nucleus changed.

Answer:

Workdone = 17482.36 Joules

Explanation:

Given the following data;

Force, F = 120 N

Angle, d = 28.0°

Distance, x = 165 m

To find the work done, we would use the following formula;

Workdone = FxCosd

Substituting into the formula, we have;

Workdone = 120*165*Cos(28)

Workdone = 19800 * 0.8830

Workdone = 17482.36 Joules

Answer:

D: The ability to do work or cause change.

Explanation:

Energy can be defined as the ability to do work. We apply energy in basically everything we do in life, energy is needed to walk or move from one position to another, it is required to eat and do other basic things of life. The body system converts the food that is being consumed into energy, this energy can be switched from one form to another.

There are various forms of energy which include solar energy, wind energy, electrical energy, mechanical energy, kinetic energy, potential energy, etc. This form of energy can be converted from one form to another but cannot be destroyed.

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Answer:

   I = 2 A

Explanation:

Electric current is defined as the amount of face per unit of time

         I = Q / t

where Q is the charge and t is the time

let's reduce the time to the SI system

         t = 1 min (60s / 1 min) = 60 s

let's calculate

         I = 120/60

         I = 2 A

Answer:

If Earth's diameter doubled but density was similar to the old Earth, the planet's mass would go up and gravity would be twice as strong. That would instantly make tides twice as much as it was before.

Explanation:

brainliest please

Answer:

E=2.58155844×10^-29

Explanation:

we use this formula

E= hc/ lambda

E= [(6.626×10^-34)( 3×10^8)]/(7.7×10^-3)

E=2.58155844×10^-29

To test the circuit breaker's tripping behavior, you can perform an overload test.

The following are the procedures for testing the circuit breaker:

Inspection:

Examine the circuit breaker for any visual harm or defects.

Ensure that all screws and terminals are secure and free of debris.

Continuity:

Use a multimeter to check the circuit breaker's continuity.

Ensure that there is continuity between the input and output wires of the circuit breaker and that the multimeter's resistance reading is zero when the breaker is closed.

Load:

Examine the circuit breaker's load-carrying capacity.

To do so, use a power source that is within the circuit breaker's current range and attach the circuit breaker to it.

As a result, the power source's current would flow through the circuit breaker.

Observe the tripping behavior:

After the power source has been connected, increase the current gradually to see if the circuit breaker trips at the appropriate value.

To do this, use a variable resistor or a load bank device to regulate the current flowing through the circuit breaker.

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Answer:

40.18m

Explanation:

9.8 m/s/s is gravity.

9.8*4.1=40.18m

The force the replacement aorta can exert with 1.70 cm stretch distance is unknown.

Without knowing the properties of the replacement aorta, such as its elasticity and diameter, it is impossible to accurately determine the greatest force it can exert with a maximum stretch distance of 1.70 cm.

Additionally, the force exerted on the aorta is influenced by various factors, such as blood pressure and heart rate.

Thus, the force capability of the replacement aorta should be evaluated using specific tests that can determine its strength and elasticity under different physiological conditions.

This information can help healthcare professionals determine the suitability of the replacement aorta for a particular patient and reduce the risk of complications.

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The heat that is required to freeze the water is 1250 kJ. Option B

The freezing point of water is the temperature at which liquid water turns to solid. We know that heat is withdrawn from water when it is frozen. This heat is added when the water melts.

The latent heat of fusion = mL

m =  3.75 kg

L = 333kJ/Kg

Hence;

H =  3.75 kg * 333kJ/Kg

H = 1250 kJ

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Answer:

A thumb only has one joint and two phalanges. The thumb is referred to as “the big finger".

Brainlist pls!

Answer:

a. T₂ = 60 N

b. W = 25.98 N

Explanation:

a.

Taking right side as positive x direction, the equation for sum of forces in x-direction can be written as follows:

\(T_{2}Sin\ 30^o - T_{1} = 0\\T_{2}Sin\ 30^o = T_{1}\\T_{1} = (T_{2})(Sin\ 30^o)\\where,\\T_{2} = 30\ N\\Therefore,\\T_{1} = (30\ N)(Sin\ 30^o)\\\)

T₁ = 15 N

b.

Taking upward direction as positive y direction, the equation for sum of forces in y-direction can be written as follows:

\(T_{2}Cos\ 30^o - W = 0\\T_{2}Cos\ 30^o = W\\W = (30\ N)Cos\ 30^o\\\)

W = 25.98 N

The distance between the [210] family of lattice planes in a simple cubic lattice is equal to the length of one side of the cube divided by the square root of 10.

The general formula for calculating the distance between two parallel planes in a lattice is given by: \(d(hkl) = a / sqrt(h^2 + k^2 + l^2)\) where d(hkl) is the distance between the planes, a is the lattice parameter (the length of one side of the unit cell), and h, k, and l are the Miller indices of the plane.

The distance between the [210] family of lattice planes can be calculated as: \(d(210) = a / sqrt(2^2 + 1^2 + 0^2) = a / sqrt(5)\)

Therefore, the distance between adjacent lattice points along a face diagonal is equal to the length of one side of the cube (a), multiplied by the square root of 2. Therefore:\(d(210) = a / sqrt(5)= (side length of the cube) / sqrt(10)\)

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The electric field at (3,3,-2) m due to 9.4 nC at(2, -2, 3) and -9.8 nC at (-3, 3, 1) is 0.28 N/C.

The electric field at the given position is determined from the summation of the field due to the two charges.

The electric field at (3,3,-2) m due to 9.4 nC at(2, -2, 3) is calculated as follows;

\(|r_1_2| = \sqrt{(2 -3)^2 + (-2-3)^2 + (3 --2)^2} = \sqrt{(1)^2 + (-5)^2 + (5)^2} \\\\|r_1_2|= 7.14 \ m\)

\(E_{12} = \frac{kQ}{r^2} (\frac{3i + 3j - 2k}{|r|} )\\\\E_{12} = \frac{9\times 10^9 \times 9.4\times 10^{-9}}{7.14^2}(\frac{3i + 3j - 2k}{7.14} )\\\\E_1_2 = 0.23(3i + 3j - 2k)\\\\E_1_2 = 0.69i + 0.69j - 0.46k\)

The electric field at (3,3,-2) m due to -9.4 nC at(-3,3,1) is calculated as follows;

\(|r_2_3| = \sqrt{(-3-3)^2 + (3-3)^2 + (1--2)^2 } \\\\|r_2_3| = 6.71 \ m\)

\(E_{23} = \frac{kQ}{r^2} (\frac{3i + 3j - 2k}{|r|} )\\\\E_{23} = \frac{9\times 10^9 \times (-9.8\times 10^{-9})}{6.71^2}(\frac{3i + 3j - 2k}{6.71} )\\\\E_2_3 = -0.29(3i + 3j - 2k)\\\\E_2_3 = -0.87i - 0.87j + 0.58k\)

E(12) + E(23) = -0.18i - 0.18j + 0.12k

\(|E| = \sqrt{(0.18)^2 + (0.18)^2 +(0.12)^2} \\\\|E| = 0.28 \ N/m\)

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