A remote sensing satellite of the earth revolves in a circular orbit at a height of 250 km above the earth surface. what is the orbital speed and period of revolution of the satellite

The estimated maximum power output of the heat engine using hot springs with a groundwater temperature of 95 °C and a mass flux of 0.2 kg/s is approximately 0.0128 kW.

To estimate the maximum power output of the heat engine using hot springs, we can utilize the concept of the Carnot cycle, which provides an upper limit for the efficiency of a heat engine.

The Carnot efficiency is given by the formula:

η = 1 - (Tc/Th)

Where η is the efficiency, Tc is the temperature of the cold reservoir (ambient temperature), and Th is the temperature of the hot reservoir (groundwater temperature).

Given:

Tc = 20 °C = 293 K

Th = 95 °C = 368 K

The maximum power output can be calculated using the formula:

P = η * Q

Where P is the power output and Q is the heat transfer rate.

The heat transfer rate can be calculated using the formula:

Q = m * Cp * (Th - Tc)

Given:

m = 0.2 kg/s (mass flux)

Cp = specific heat capacity of water ≈ 4.18 kJ/kg°C

Let's calculate the maximum power output:

Tc = 293 K

Th = 368 K

m = 0.2 kg/s

Cp = 4.18 kJ/kg°C = 4.18 J/g°C = 4.18 * 10⁻³ J/kg°C

Q = m * Cp * (Th - Tc)

  = 0.2 kg/s * 4.18 * 10⁻³ J/kg°C * (368 K - 293 K)

  = 0.2 * 4.18 * 10⁻³ * 75

  = 0.0627 kW

η = 1 - (Tc/Th)

  = 1 - (293/368)

  ≈ 0.204

P = η * Q

  = 0.204 * 0.0627 kW

  ≈ 0.0128 kW

Therefore, the estimated maximum power output of the heat engine using hot springs with a groundwater temperature of 95 °C and a mass flux of 0.2 kg/s is approximately 0.0128 kW.

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Answer:

Atoms of elements with the same electons in the outer most shell they are put in the same group on the periodic table

The center of mass of the thin plate is located at (x_cm, y_cm) = (0, 8/5) or (0, 1.6) units.

First, we need to find the total area A of the region bounded by the parabola and the line:

A = ∫[-2,2] (8 - 2x^2) dx

Integrating with respect to x:

A = [8x - (2/3)x^3] evaluated from -2 to 2

A = [8(2) - (2/3)(2^3)] - [8(-2) - (2/3)(-2^3)]

A = [16 - (2/3)(8)] - [-16 - (2/3)(-8)]

A = 16 - 16/3 + 16 + 16/3

A = 48/3

A = 16 square units

Now, let's calculate the center of mass coordinates:

x_cm = (1/M) ∫[-2,2] x * dm

y_cm = (1/M) ∫[-2,2] y * dm

Since the plate has constant density, we can express dm = ρ * dA, where ρ is the density and dA is the differential area element.

Let's substitute dm = ρ * dA into the center of mass formulas:

x_cm = (1/M) ∫[-2,2] x * (ρ * dA)

y_cm = (1/M) ∫[-2,2] y * (ρ * dA)

The density (ρ) cancels out in both equations, so we can write:

x_cm = (1/A) ∫[-2,2] x * dA

y_cm = (1/A) ∫[-2,2] y * dA

Now, we'll substitute the equations of the parabola and the line:

x_cm = (1/A) ∫[-2,2] x * dA

y_cm = (1/A) ∫[-2,2] y * dA

x_cm = (1/16) ∫[-2,2] x * dA

y_cm = (1/16) ∫[-2,2] y * dA

Density = 1 units

Recall the expressions for the center of mass coordinates:

x_cm = (1/A) ∫[-2,2] x * dA

y_cm = (1/A) ∫[-2,2] y * dA

Since the density is 1, we can simplify the expressions further:

x_cm = (1/16) ∫[-2,2] x * dA

y_cm = (1/16) ∫[-2,2] y * dA

We need to evaluate the integrals of x * dA and y * dA. Let's start with x * dA:

x_cm = (1/16) ∫[-2,2] x * dA

Substituting x = x and dA = y * dx:

x_cm = (1/16) ∫[-2,2] x * (y * dx)

x_cm = (1/16) ∫[-2,2] x * (2x^2 * dx)

Expanding and simplifying the expression:

x_cm = (1/8) ∫[-2,2] x^3 * dx

Integrating with respect to x:

x_cm = (1/8) * [ (1/4) * x^4 ] evaluated from -2 to 2

x_cm = (1/8) * [ (1/4) * (2^4) - (1/4) * (-2^4) ]

x_cm = (1/8) * [ 4 - 4 ]

x_cm = 0

Similarly, for y * dA:

y_cm = (1/16) ∫[-2,2] y * dA

Substituting y = 2x^2 and dA = y * dx:

y_cm = (1/16) ∫[-2,2] (2x^2) * (2x^2 * dx)

y_cm = (1/8) ∫[-2,2] x^4 * dx

Integrating with respect to x:

y_cm = (1/8) * [ (1/5) * x^5 ] evaluated from -2 to 2

y_cm = (1/8) * [ (1/5) * (2^5) - (1/5) * (-2^5) ]

y_cm = (1/8) * [ 32/5 - (-32/5) ]

y_cm = (1/8) * (64/5)

y_cm = 8/5

Therefore, the center of mass of the thin plate is located at (x_cm, y_cm) = (0, 8/5) or (0, 1.6) units.

The x-coordinate of the center of mass is at the centerline of the region bounded by the parabola and the line, while the y-coordinate is above the centerline but below the line y = 8.

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Answer:

Distance = 22m

Displacement = 12 + 2 - 8 =6m

Answer:

V = 3.00E^8 m/s      speed of light

1 m = 1.00E-3 km

1 h = 3600 s

V = 3.00E8 m/s * 1.00E-3 km / m * 3.60E3 s/h

V =  1.08E9 km/h

a) would be the closest but it is missing a superscript

Answer:

Explanation:

Main Answer:

The equation acceleration = v/time can be proven using the fundamental definitions of acceleration, velocity, and time. Acceleration is defined as the rate of change of velocity, and velocity is the rate of change of displacement with respect to time. Let's consider an object moving with an initial velocity v0 and final velocity v in a time interval t.

Explanation:

The change in velocity, Δv, can be calculated as the final velocity minus the initial velocity, Δv = v - v0. Similarly, the change in time, Δt, is the final time minus the initial time, Δt = t - t0.

By substituting these values into the equation for acceleration, we have:

acceleration = Δv/Δt

Now, substituting Δv = v - v0 and Δt = t - t0, we get:

acceleration = (v - v0)/(t - t0)

Since v0 and t0 represent the initial velocity and time, respectively, we can rewrite the equation as:

acceleration = (v - v0)/t

By rearranging the equation, we find:

acceleration = v/t

Thus, we have proved that acceleration is equal to v/time.

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The pressure of the gas is approximately 1.21 atm.

(a) To find the pressure of the gas, we can use the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Given:

Volume (V) = 2.00 L

Number of moles (n) = 0.100 mol

Temperature (T) = 293 K

Gas constant (R) is usually expressed as 0.0821 L·atm/(mol·K) for the ideal gas law.

Plugging in the values, we can solve for P:

P = (nRT) / V

P = (0.100 mol * 0.0821 L·atm/(mol·K) * 293 K) / 2.00 L

P ≈ 1.21 atm

The pressure of the gas is approximately 1.21 atm.

(b)T=295 k

given the formula is :

PV=nRT

where

P= 1.21 atm

V= 2.00L

R= 0.0821 L·atm/(mol·K) for the ideal gas law.

(n) = 0.100 mol

T=PV/nR

T=295 k

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Answer:

See below

Explanation:

Velocity is how fast an object is moving....acceleration is how fast the velocity is changing .

The equation F=ma, where F is the net force, m is the mass, and an is the acceleration, can be used. . Rearranging the formula to solve for mass, we get m=F/a. Substituting the given values, we get m=12 N/48 m/s^2 = 0.25 kg. Therefore, the answer is A) 0.25 kg.

To solve this problem, you can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In this case, the net force (F) is 12 N, and the acceleration (a) is 48 m/s². You are asked to determine the mass (m) of the hockey puck.

Using the equation F = ma, you can rearrange it to find the mass: m = F/a

Plug in the given values: m = 12 N / 48 m/s²

Calculate the mass: m = 0.25 kg

The mass of the hockey puck is 0.25 kg, which corresponds to option A.

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Before opening her chute, Suzie Skydiver would experience a force of air pressure of approximately 450 N at terminal velocity.

Terminal velocity is the point where the force of air resistance, or drag, acting on the skydiver becomes equal in magnitude to the force of gravity pulling the skydiver down. At this point, the net force acting on the skydiver is zero, and they fall at a constant velocity. At terminal velocity, Suzie Skydiver is falling at a constant rate, meaning that the force of gravity pulling her down is balanced by the force of air resistance pushing her up.

This force of air resistance, also known as drag, can be calculated using the formula:

F = 1/2 * rho * v^2 * Cd * A,

where F is the force of drag, rho is the density of the air,

v is the velocity of the object,

Cd is the drag coefficient

A is the cross-sectional area of the object.

Assuming that Suzie Skydiver falls in a typical skydiving posture with a drag coefficient of around 1.0 and a cross-sectional area of 1.0 square meter,

Using the standard atmospheric density of 1.2 kg/m³,

We can calculate that her terminal velocity is approximately 54 m/s.

At this velocity, the force of air resistance, or drag, acting on Suzie Skydiver is equal in magnitude to the force of gravity, which is approximately 450 N.

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Given data:

* The radius of curvature of the convex mirror is R = 6 cm.

* The distance of the object from the convex mirror is u = - 12 cm.

Required: The nature of the image formed.

Equations needed:

The mirror formula:

The magnification formula:

Solution:

The focal length of the convex mirror is,

The distance of the image formed is,

By simplifying,

The positive value of image distance indicated that the virtual image is formed.

By the magnification formula,

The positive value of magnification indicates that the upright image is formed.

The value of the magnification is less than 1, thus, a small size (diminished) image is formed.

Final Answer:

Location - 2.4 cm behind the convex mirror

Orientation - Upright

Size - Small size (or diminished)

Type - Virtual

Hence, option B is the correct answer.

Answer:

densest: blue; medium dense: clear; least dense: red

Explanation:

i've done this question before :)

The bearing stress in this scenario is 2 kN/mm². To calculate the bearing stress, we need to use the formula:
Bearing Stress = Load / Contact Area

Substituting the given values:
Bearing Stress = 10 kn / 5 square mm
Bearing Stress = 2 N/mm^2

It is important to note that bearing stress is a measure of the force per unit area exerted on the contact surface between two components. In this case, the load is distributed over an area of 5 square mm, resulting in a bearing stress of 2 N/mm^2. It is important to ensure that the bearing stress is within the allowable limits to prevent failure or damage to the components.

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Answer:

If the mass of one body is doubled, force is also doubled.

Taking into account the Universal Law of Gravitation, if the masses of one body is doubled, the gravitational force between two bodies is also doubled.

The Universal Law of Gravitation establishes that bodies, by the simple fact of having mass, experience a force of attraction towards other bodies with mass, called gravitational force.

The Universal Law of Gravitation states that the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them.

Mathematically it is expressed as follows:

\(F=G\frac{Mm}{d^{2} }\)

where:

By definition of gravitational force, the greater or lesser the value of any of the masses causes the force to be greater or lesser respectively.

Finally, if the masses of one body is doubled, the gravitational force between two bodies is also doubled.

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The time taken for the APUSH textbook to hit the ground is determined as 0.55 seconds.

The time of motion of an object is the time taken for the object to travel from its initial point to the final point.

The time taken for the APUSH textbook to hit the ground is calculated as follows;

t = √(2h/g)

where;

t = √(2 x 1.5/9.8)

t = 0.55 seconds

Thus, the time taken for the APUSH textbook to hit the ground is determined as 0.55 seconds.

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like blade runner and Fahrenheit 451 crossword clue, the answer is: DYSTOPIAN.

Crossword clues are hints that the problem solver must interpret in order to discover the solution, which is subsequently typed into the puzzle grid. Clues can include puns, anagrams, and other wordplay; they don't always contain dictionary definitions.

The benefit of doing a crossword puzzle is that it compels students to go through the clues, remember and review the subject matter, and engage in conversation with their peers to dispel any misunderstandings about the subject matter.

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Answer:

wen I was in the car toing home from school after a bad day n si si I have crazzyyy

Answer:-

1. I feel  the most passionate and alive when I watch BTS, because they inspir me to do many things such as Loveing Myself and doning what I love in life and chase my dream. I also feel alive when playing soccer, dancing, playing piano, or singing/rapping, because these things are what I love to do and always loved to do.

2. Yes, I feel like I live everyday the same day, some little things change but the things I do every day are the same, never changes. Wake up, eat, homework, clean, eat, tv, eat, and sleep.

Answer:

Magnitude = 26 m

Direction = eastward

Explanation:

Given that a football coach walks 24 meters eastward, then 12 meters westward, then 36 meters eastward, and finally 22 meters westward. What is the magnitude and direction of the displacement?

Let eastward be positive and westward be negative.

Magnitude = 24 - 12 + 36 - 22

Magnitude = 60 - 34

Magnitude = 26

Since the answer is positive, the direction will be eastward.

Therefore,

Magnitude = 26 m

Direction = eastward

Answer:

the answer to this question is A globular cluster

The 12-lead ECG test that shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, could be a diagnostic for myocardial infarction.

A 12-lead electrocardiogram (ECG) is a medical test that is recorded using leads, or nodes, attached to the body.

Electrocardiograms, sometimes referred to as ECGs, capture the electrical activity of the heart and transfer it to graphed paper.

When a 12-lead ECG shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, it could be a diagnostic for myocardial infarction.

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Answer:

watch over the moon and you will get your answer.

Explanation:

Answer:

Meteoroids

Explanation:

Meteoroids are significantly smaller than asteroids, and range in size from small grains to one-meter-wide objects. Objects smaller than this are classified as micrometeoroids or space dust.

The term by which an asteroid called is smaller than a couple of meters wide is a meteorite.

A meteorite is a small group of rocks and debris. They are made by the collection of dust and small pebbles. They are smaller than asteroids, and they are formed in outer space, and they fall on earth from outer space, and when they fall into the earth with that great speed, they attain heat.

Meteorites also consist of metals, like nickel, carbides, sulfides, etc. It has been said that in old times the meteorite that contain iron, fall into the earth and the iron from them melted into the core of the earth. They are of different shapes and sizes.

Thus, the term by which an asteroid called is smaller than a couple of meters wide is a meteorite.

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(a) For a(t) = t + 4 and v(0) = 5, the velocity at time t can be found by integrating the acceleration function, and then substituting t = 0 with the given initial velocity.

(b) The distance traveled during the time interval can be obtained by integrating the velocity function over the given range, from t = 0 to t = 10.

Scenario 1:

Given:

a(t) = t + 4

v(0) = 5

(a) Finding the velocity at time t:

Integrate the acceleration function a(t):

∫(t + 4) dt = (1/2)t² + 4t + C

Substitute t = 0 and v(0) = 5:

(1/2)(0)² + 4(0) + C = 5

C = 5

Therefore, the velocity at time t is:

v(t) = (1/2)t² + 4t + 5

(b) Finding the distance traveled during the time interval 0 ≤ t ≤ 10:

Integrate the velocity function v(t):

∫[(1/2)t² + 4t + 5] dt = (1/6)t³ + 2t² + 5t + D

Evaluate the integral at the upper and lower limits:

Distance = [(1/6)(10)³ + 2(10)² + 5(10)] - [(1/6)(0)³ + 2(0)² + 5(0)] = 383.33 units (approximately)

Scenario 2:

Given:

a(t) = 2t + 3

v(0) = -4

(a) Finding the velocity at time t:

Integrate the acceleration function a(t):

∫(2t + 3) dt = t² + 3t + C

Substitute t = 0 and v(0) = -4:

(0)^2 + 3(0) + C = -4

C = -4

Therefore, the velocity at time t is:

v(t) = t² + 3t - 4

(b) Finding the distance traveled during the time interval 0 ≤ t ≤ 3:

Integrate the velocity function v(t):

∫[(t^2 + 3t - 4)] dt = (1/3)t^3 + (3/2)t^2 - 4t + D

Evaluate the integral at the upper and lower limits:

Distance = [(1/3)(3)³ + (3/2)(3)² - 4(3)] - [(1/3)(0)³ + (3/2)(0)² - 4(0)] = 17.5 units

So, the distance traveled during the time interval 0 ≤ t ≤ 3 is 17.5 units.

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The complete question is:

The acceleration function (in m/s 2) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time t and (b) the distance traveled during the given time interval.

1. a(t)=t+4,v(0)=5,0⩽t⩽10 72. a(t)=2t+3,v(0)=−4,0⩽t⩽3

Answer:

Asexual production they can be eukaryotes or prokaryotes

Explanation:

According to the question the ratio of the observed star's luminosity to the luminosity of the Sun is approximately 32.768.

The luminosity of a star is directly proportional to its surface temperature raised to the fourth power (L ∝ T^4) and its radius squared (L ∝ R^2).

Given that the surface temperature of the observed star is 4.2 times that of the Sun (T_observed = 4.2 * T_sun) and its radius is one-quarter that of the Sun (R_observed = 1/4 * R_sun), we can calculate the ratio of its luminosity (L_observed) to the luminosity of the Sun (L_sun) as follows:

L_observed/L_sun = (T_observed/T_sun)^4 * (R_observed/R_sun)^2

Substituting the given values:

L_observed/L_sun = (4.2 * T_sun / T_sun)^4 * (1/4 * R_sun / R_sun)^2

= 4.2^4 * (1/4)^2

= 32.768

Therefore, the ratio of the observed star's luminosity to the luminosity of the Sun is approximately 32.768.

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To calculate the efficiency of a machine, divide the work output by the work input and multiply the result by 100%.

Correct question:

The work output of a machine is 80 Joules, and its work input is 120 Joules. What is the efficiency of the machine?

Work output = 80 Joules

Work input = 120 Joules.

Efficiency = Working output/Work input *100

Efficiency = 80/120 *100

Efficiency = 0.67 *100

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Answer:

because of the hydrogen bonds, as water molecules come together they stick to one another for a small, but significant amount of time. This slows them down, and holds them closer to one another. They become a liquid; a different state of matter where the molecules are closer and slower than in a gas.

Answer:

When water is in its solid-state, the water molecules are packed close together preventing it from changing shape.

Explanation:

In general, when considering the states of matter, solids are denser than liquids, and normally liquids are denser than gases.

Hope this helps!

Brain-List?

1 ) The maximum disk temperature = 119.2 °C

2 ) The maximum temperature of the nickel  coated disk = 108.93 °C

3 ) The nickel coating reduced the maximum copper temperature

Diameter of disc, d = 0.3 m

Thickness of disc, t = 5 mm

Heat input, Q = 70 KW

Properties of water at 1 atm,

Density of liquid, ρl = 958.4 kg / m³

Density of vapour, ρv = 0.5955 kg / m³

Specific heat, cp = 4220 J / kg K

Viscosity of water, μ = 279 * 10⁻⁶ kg / m s

Prandtl number, Pr = 1.75

hfg = 2257 KJ / kg

Surface tension, σ = 0.0589 N / m

For water - copper polished,

Csf = 0.013

Heat flux, q = Q / A

q = 70 * 10³ / 3.14 * 0.15²

q = 9.9 * 10⁵ W / m²

For nucleate boiling, Rohsenow correlation,

q = μ hfg [ g ( ρl - ρv ) / σ ]^0.5 [ cp ΔT / Csf hfg Prⁿ ]³

9.9 * 10⁵ = 279 * 10⁻⁶ * 2257 * 10³ [ 9.81 ( 958.4 - 0.5955 ) / 0.0589 ]^0.5 [ 4220 ΔT / 0.013 * 2257 * 10³ * 1.75 ]³

1.578 = 0.082 ΔT

ΔT = 19.2

Ts - Tsat = 19.2

Ts - 100 = 19.2

Ts = 119.2 °C

For water - Nickel coating,

Csf = 0.006

q = μ hfg [ g ( ρl - ρv ) / σ ]^0.5 [ cp ΔT / Csf hfg Prⁿ ]³

9.9 * 10⁵ = 279 * 10⁻⁶ * 2257 * 10³ [ 9.81 ( 958.4 - 0.5955 ) / 0.0589 ]^0.5 [ 4220 ΔT / 0.006 * 2257 * 10³ * 1.75 ]³

1.578 = 0.1766 ΔT

ΔT = 8.93

Ts - Tsat = 8.93

Ts - 100 = 8.93

Ts = 108.93 °C

Ts with nickel coating < Ts without any coating

Therefore,

1 ) The maximum disk temperature when nucleate boiling is occurring = 119.2 °C

2 ) The maximum temperature of the nickel  coated disk = 108.93 °C

3 ) The nickel coating reduced the maximum copper temperature

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The height of the cliff from which the ball fell is 172.3 m.

The time of motion of the ball is the time taken for the ball to travel from its initial position to the final position.

X = Vt

where;

t = X/V

t = (83 m) / (14 m/s)

t = 5.93 s

The height of fall of the ball is calculated as follows;

h = vt +  ¹/₂gt²

where;

h = 0 + ¹/₂gt²

h = ¹/₂gt²

h = ¹/₂(9.8)(5.93)²

h = 172.3 m

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