A__ is a measure of the electric power an appliance uses

The magnetic field at the center of the loop is calculated to be 0.159 T.

The magnetic field at the center of a flat, circular loop with 17 turns, a radius of 12.5 cm, and a current of 0.60 A can be determined by using the equation B = µ₀.n.I/2.π.r, where

Using this equation, the magnetic field at the center of the loop is calculated to be 0.159 T.

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The frequency that the bad guy hear is 12000 hz when the police car is moving with speed of 80m/s.

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We always see the same side of the Moon because the "Moon rotates on its axis once for each revolution around Earth." Thus, the correct option will be B.

When the Moon rotates on its axis once for each revolution around Earth, then we always see the same side of the Moon. The reason behind this is that the moon's rotation takes almost the same time as it takes to orbit the Earth.

When the same side of the moon is facing the Earth, it appears to be unchanging. That is why we always see the same side of the moon from Earth. The other side of the Moon is known as the far side, which was first observed by the Soviet spacecraft Luna 3 in 1959.

Therefore, the correct option will be B.

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The minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any is 5.56 rpm.

The minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any is given by the formula; Vmin=√g/R

where:

To express the answer in revolutions per minute, the radius of the circle must be converted to meters;R = 35 cm = 0.35 m

Substituting the values given above into the formula;

Therefore, the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any is 5.56 rpm.

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The sunspot cycle is directly relevant to us here on earth because coronal mass ejections and other activity associated with the sunspot cycle can disrupt radio communications and knock out sensitive electronic equipment.

The sunspot cycle is directly relevant to us here on earth because it can cause coronal mass ejections and other activity that can disrupt radio communications and knock out sensitive electronic equipment. It also plays a major role in global warming, affects compass needles, affects plant photosynthesis, and strongly influences the earth's weather.

This means that the sunspot cycle can have a significant impact on our technology and communication systems, which are critical to our daily lives. Coronal mass ejections can cause major geomagnetic storms that have the potential to knock out power grids, damage satellites, and disrupt GPS signals. These storms can also create beautiful auroras that are visible in many parts of the world, but they can also have serious consequences for our infrastructure.

The sun's magnetic field, which plays a major role in the sunspot cycle, affects the compass needles that we use on earth. This means that the sunspot cycle can also have an impact on navigation systems, which are important for transportation and other industries.

Overall, the sunspot cycle strongly influences Earth's weather and can affect plant photosynthesis here on earth. This means that changes in the sunspot cycle can have a significant impact on our planet and our daily lives.

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Explanation:  Gases move from high-pressure areas to low-pressure areas. And the bigger the difference between the pressures, the faster the air will move from the high to the low pressure.

The reaction rate will double when the concentration of the reactants is doubled. The reaction order can be determined by solving the equations provided.
For example, if the initial rate is given by:
Rate = [CV3]* = CV.x = x = rate4 [CV4]* Ox= ratez [CV]* rates [CVs]* rates CV.* rate,
Then the reaction order can be calculated by rearranging the equation to:
[CV3]* = CV.x/x = rate4 [CV4]* Ox/x = ratez [CV]* rates [CVs]* rates CV.* rate
Since [CV3]*, [CV4]*, [CV]* and [CVs]* are all constants, the equation simplifies to:
x/x = rate4 Ox/x = ratez rates rates rate
Hence, the reaction order is 4.

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In a conservative system, the total energy is conserved, which means that the initial energy (ei) is equal to the final energy (ef).

Conservative systems are those where the total energy remains constant over time, such as in a pendulum swinging back and forth or a planet orbiting a star under the influence of gravity.

In such systems, the energy can be converted from one form to another, but the total amount of energy remains constant.

Therefore, we can say that in a conservative system, the initial energy (ei) and the final energy (ef) are equal. This means that any changes in the system's energy, such as potential energy being converted into kinetic energy, must be balanced by an equal and opposite change in some other form of energy, such as potential energy being converted into kinetic energy.

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The work done on the block by friction during the motion of the block from point A to point B is 2.49 J.

The normal force acting on the block at point A and point B is different. We can find the weight of the block at points A and point B using the following formula:

Weight = mg,

where m is the mass of the block and g is the acceleration due to gravity.

Weight at point A = m × g

Weight at point B = m × g

Now, the normal force acting on the block at point A is given as 3.95 N.

Therefore, we can write the equation for the weight and normal force as:

Weight at point A - Normal force at point A = m × a

Now, at point A, the acceleration acting on the block is the centripetal acceleration a = v²/R where v is the velocity of the block at point A.

We can write the equation for the weight and normal force as:

m × g - 3.95 = m × v²/R

Similarly, at point B, we can write the equation for the weight and normal force as:

m × g - 0.680 = m × v²/R

Now, we can solve both the equations for the velocity of the block at point A and point B:

Velocity at point A, v₁ = √(gR - 3.95/m)

Velocity at point B, v₂ = √(gR - 0.680/m)

The change in kinetic energy during the motion from point A to point B is given by:

∆KE = KE₂ - KE₁

= (1/2)mv₂² - (1/2)mv₁²

We know that work done, W = ∆KE

So, the work done on the block by friction during the motion of the block from point A to point B is given by:

W = (1/2)m(v₂² - v₁²)

Substituting the values in the above equation:

W = (1/2) × 0.0400 × ((√(9.81 × 0.500 - 0.680/0.0400))² - (√(9.81 × 0.500 - 3.95/0.0400))²)

W = 2.49 J

Therefore, the work done on the block by friction is 2.49 J.

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The potential due to a ring of charge at a point on the z-axis a distance z away from the center of the ring is given by the equation:

Vring = kQ / √(R^2 + z^2)

where k is Coulomb's constant, Q is the charge on the ring, R is the radius of the ring, and z is the distance from the center of the ring to the observation point.

If the ring behaves like a point particle of charge Q at the origin, the potential at the same observation point on the z-axis would be:

Vparticle = kQ / z

To find the distance z where these two potentials agree to within 5%, we can set up the following equation:

|Vring - Vparticle| / Vparticle ≤ 0.05

Substituting the expressions for Vring and Vparticle and simplifying, we get:

|√(R^2 + z^2) - z| / z ≤ 0.05

Squaring both sides and rearranging, we get:

(R^2 / z^2) ≤ 0.0025

Taking the square root of both sides, we get:

R / z ≤ 0.05

Solving for z, we get:

z ≥ R / 0.05

Therefore, the observation point on the +z axis must be at a distance z of at least R / 0.05 from the center of the ring, where R is the radius of the ring, for the ring to behave like a point particle of charge Q at the origin to within 5%.

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Answer: Telephone

Explanation:

The technological improvement that allowed more goods to be produced at one time in the 1920s was the development and widespread use of assembly line production. This was pioneered by companies such as Ford Motor Company, which introduced the assembly line to its automobile factories. The assembly line method allowed for the mass production of standardized products using specialized machines and workers performing specific tasks. By breaking down the manufacturing process into smaller, simpler tasks, and optimizing the movement of workers and materials, the assembly line significantly increased production efficiency and output. This led to the growth of mass production industries, increased affordability of goods, and a significant shift in the nature of work in the 20th century.

The pressure on the bottom of the test tube which contain both the oil and water molecules is about 641.65 Pa + 220.725 Pa = 862.375 Pa.

The pressure at the bottom of the test tube is the result of two factors: the weight of the oil and the weight of the water molecules. The pressure is equal to the density of each liquid multiplied by the height of each liquid, multiplied by the gravitational acceleration (g).

The pressure at the bottom of the test tube is given by the density of the fluids and also the height of the column above the bottom region. The pressure at the bottom of the test tube is calculated by multiplying the density of the fluids by the height of the column above the bottom. Here's how to calculate the pressure:

P = pgh

where P = Pressure, p = Density of fluid, g = Acceleration due to gravity, and h = Height of the column.

The pressure at the bottom of the test tube is the pressure which is exerted by the water and oil above it. The water is more dense than that of the oil, therefore it exerts more pressure on the bottom of the test tube. The pressure at the bottom of the test tube is given by the formula

The density of water is 1000 kg/m³, and the density of oil is 900 kg/m³. The height of the column of water is 6.5 cm, and the height of the column of oil is 2.5 cm.

Using the above formula: P = pgh

P (Water) = 1000 × 9.81 × 0.065

P (Water) = 641.65 Pa

P (Oil) = 900 × 9.81 × 0.025

P (Oil) = 220.725 Pa

Therefore, the pressure on the bottom of the tube is 641.65 Pa + 220.725 Pa = 862.375 Pa.

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The magnitude of the current's magnetic field at radial distances (a) 0, (b) 1cm, (c) 2cm (wire's surface), and (d) 4cm are undefined, 1.7 * 10^-3 Tesla, 1.7 * 10^-3 Tesla, and 8.5 * 10^-4 Tesla, respectively. 

The question is about finding the magnitude of magnetic fields at different radial distances across a diameter of a long cylindrical conductor of radius a=2 cm carrying uniform current 170A.

Let's solve it step by step.

(a) At radial distance 0:

At the center of the conductor, r = 0, the magnetic field is zero.

It can be found by using the formula for the magnetic field at the center of the wire: 

B = (μ_0 * I) / (2 * π * r)

= (4π * 10^-7 * 170) / (2π * 0)

= undefined.

Therefore, the magnetic field at r = 0 is undefined. 

(b) At radial distance 1cm:

Using the formula for the magnetic field at a point P located at a radial distance r from the center of the wire: 

B = (μ_0 * I) / (2 * π * r)

= (4π * 10^-7 * 170) / (2π * 0.01)

= 1.7 * 10^-3 Tesla.

(c) At radial distance 2cm:

The magnetic field at r = a (i.e., the surface of the wire) can be determined by substituting the value of r = 2cm into the magnetic field formula:

B = (μ_0 * I) / (2 * π * r)

= (4π * 10^-7 * 170) / (2π * 0.02)

= 1.7 * 10^-3 Tesla.

(d) At radial distance 4cm:

Again, we use the formula for the magnetic field at a point P located at a radial distance r from the center of the wire:

B = (μ_0 * I) / (2 * π * r)

= (4π * 10^-7 * 170) / (2π * 0.04)

= 8.5 * 10^-4 Tesla.

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If the magnetic field steadily decreases from B to zero during a time interval t, the magnitude E of the induced emf is given by the formula; E = (Bx-y/t), where B is the magnetic field, x, and y are constants.

An induced emf is the voltage generated across a conductor when it moves through a magnetic field. It is also induced when there is a change in the magnetic field passing through a conductor.

The emf generated in a coil of wire is equal to the rate of change of magnetic flux through the coil. Magnetic flux is given by the formula: φ=B*A,

where -  B is the magnetic field strength and

           - A is the area of the coil.

If the magnetic field steadily decreases from B to zero during a time interval t, the change in magnetic flux is given by the formula:  Δφ=B*A = B*ΔA, where ΔA is the change in area over time Δt.

The induced emf E is given by the formula: E = (-N * Δφ)/Δt

Where N is the number of turns in the coil. If the magnetic field is steadily decreasing, then ΔB/Δt is constant, and the induced emf E is given by the formula:  E = (-N * B * ΔA/Δt) = (-N * B * x*y/t) = (Bx-y/t), where x and y are constants.

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a) If the current is running from west to east, the force that our planet's magnetic field exerts on this cord is directed upwards
b) The magnitude of the force that our planet's magnetic field exerts on this cord if it is oriented so that the current in it is running from west to east is F =2.64 x 10^-4 N
c) If the current is running vertically upward, the force that our planet's magnetic field exerts on this cord is directed to the left.  west
d) The magnitude of the force that our planet's magnetic field exerts on this cord if it is oriented so that the current in it is running vertically upward is F = 0 zero
e) If the current is running from north to south, the force that our planet's magnetic field exerts on this cord is directed east.
f) The magnitude of the force that our planet's magnetic field exerts on this cord if it is oriented so that the current in it is running from north to south is F = 2.64 x 10^-4 N
g) The magnetic force is not large enough to cause significant effects under normal household conditions.

EXPLANATION

a) The direction of the force that our planet's magnetic field exerts on the cord is perpendicular to both the direction of the current and the direction of the magnetic field, according to the right-hand rule. In this case, if the current is running from west to east, and the magnetic field is from south to north, the force will be directed upwards.

b) The magnitude of the force can be calculated using the formula:

F = BIL sin(theta)

where B is the magnitude of the magnetic field, I is the current, L is the length of the wire, and theta is the angle between the direction of the current and the direction of the magnetic field. In this case, theta is 90 degrees, so sin(theta) = 1. Substituting the given values, we get:

F = (0.550 x 10^-4 T) x (1.50 A) x (2.40 m) x 1

= 2.64 x 10^-4 N

Therefore, the magnitude of the force is 2.64 x 10^-4 N.

c) If the current in the wire is running vertically upward, the force will be directed towards the west.

d) Using the same formula as in part (b), we can calculate the magnitude of the force:

F = (0.550 x 10^-4 T) x (1.50 A) x (2.40 m) x sin(90)

= 0

Therefore, the magnitude of the force is zero.

e) If the current in the wire is running from north to south, the force will be directed towards the east.

f) Using the same formula as in part (b), we can calculate the magnitude of the force:

F = (0.550 x 10^-4 T) x (1.50 A) x (2.40 m) x 1

= 2.64 x 10^-4 N

Therefore, the magnitude of the force is 2.64 x 10^-4 N.

g) The magnitude of the magnetic force in this case is quite small, and under normal household conditions, it is unlikely to cause significant effects. However, in some situations, such as in electrical power transmission systems, the effects of the magnetic force may need to be taken into account.

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When the 35.0-g bullet moving at 475 m/s strikes the 4.4-kg bag of flour, the momentum of the bullet is transferred to the bag of flour, causing the bag of flour to move and the bag moving when the bullet exits at 91.3 m/s.

We can calculate the velocity of the bag of flour after the collision using conservation of momentum:

Here we have the following data as :

Momentum of bullet before collision = Momentum of bullet and bag after collision

m bullet × v bullet, before = (m bullet + m bag) bag × v bag, after

We can solve for v bag ,after:

v bag ,after = (m bullet × v bullet, before) / (m bullet + m bag)

v bag, after = (35.0 g × 475 m/s) / (35.0 g + 4.4 kg) = 91.3 m/s

Therefore, the bag of flour is moving at 91.3 m/s when the bullet exits.

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According to Newton's second law, the acceleration of a system is directly proportional to the net force applied to it and inversely proportional to its mass. Therefore, if the net force stays constant but the mass of the system increases, the acceleration of the system will decrease.

Similarly, if the mass of the system remains constant but the net applied force increases, the acceleration of the system will increase.

There are several potential sources of error in the Atwood machine experiment. For example, friction in the pulley or air resistance could cause the system to accelerate at a different rate than predicted by theory. Additionally, the masses used in the experiment may not be perfectly accurate, which could introduce small errors into the measurements. The string connecting the two masses could also stretch or have varying elasticity, which could affect the results. Finally, human error in measuring the time or the distances traveled by the masses could lead to inaccuracies in the calculated values of acceleration or tension in the string.

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Answer:

It is true that in accordance with Newton's third law of motion, every action has an equal and opposite reaction, meaning that when one object exerts a force on another object, the second object exerts an equal and opposite force back on the first object. However, this does not necessarily mean that the objects will not accelerate.

Acceleration depends on the net force acting on an object, which is the sum of all forces acting on the object. If the forces are balanced (i.e. they are equal and opposite), then there is no net force and the object will not accelerate. However, if the forces are unbalanced (i.e. they are not equal and opposite), then there is a net force and the object will accelerate in the direction of the net force.

For example, if you push a book across a table with a force of 5 N to the right, the book will experience a force of 5 N to the left due to friction. These two forces are equal and opposite, but they are not balanced because they act in opposite directions. The net force on the book is therefore 5 N to the right, which causes the book to accelerate in that direction.

Answer:The beta value of a transistor represents the current gain, which is the ratio of the collector current to the base current. In this case, we want to use the transistor as an amplifier to increase the current from the 0.015A supplied by the phone to the 1.5A required by the speakers.

The required current gain can be calculated using the following formula:

Beta = (Ic / Ib)

Where:

Beta is the current gain of the transistor

Ic is the collector current (output current)

Ib is the base current (input current)

To find the required beta value, we need to first calculate the base current required to drive the transistor. We can use Ohm's Law to do this:

Ib = V / R

Where:

Ib is the base current

V is the voltage supplied by the phone (5V)

R is the input resistance of the transistor circuit

Assuming an input resistance of 1kΩ, the base current required is:

Ib = V / R = 5 / 1000 = 0.005A (5mA)

Now, we can calculate the required collector current using the maximum current required by the speakers:

Ic = 1.5A

Finally, we can calculate the required beta value:

Beta = Ic / Ib = 1.5 / 0.005 = 300

Therefore, we need to choose a power transistor with a beta value of at least 300 to amplify the current from the aux port enough to drive the speakers.

Explanation:

Explanation:

All R's in series:    just add them together : 200 + 200 + 200 Ω = 600Ω

One in series with two in parallel :

   = 200 Ω   +    200*200/(200+200) Ω = 300Ω

All three in parallel :

    R = 1 / (1/200 + 1/200 + 1/200) = 66.7 Ω

For the electric flux through a spherical surface is 4.3 x 10⁴ N⋅m²/C, then the net charge enclosed by the surface is μC, and for the electric flux through a cubical box 34 cm on a side is 7.5 x 10³ N⋅m²/C, the total charge enclosed by the box is μC.

The electric flux through a spherical surface is 4.3 x 10⁴ N⋅m²/C.

The net charge is Electric Flux = Charge / Surface Area,

so the net charge enclosed is 4.3 x 10⁴ / (4πr²) where r is the radius of the sphere.

Therefore, the net charge enclosed by the surface is μC.

The electric flux through a cubical box 34 cm on a side is 7.5 x 10³ N⋅m²/C.

The total charge is Electric Flux = Charge / Surface Area,

so the total charge enclosed is 7.5 x 10³ / (6a²)

where a is the length of one side of the cube.

Therefore, the total charge enclosed by the box is μC.

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Part A: Thrust acts on the right in the direction of motion. Gravity acts downward.

Part B:  The direction of air resistance is opposite to the direction of motion, which is shown towards the left. Gravity acts downwards.

Part A:

A free-body diagram for a car is as follows:

The direction of friction is opposite to the direction of motion, which is shown towards the left.
The diagram shows three forces acting on the toy car that is battery-powered, which is as follows:
The force due to friction is labeled as [tex]f_K[/tex].

The force of thrust is labeled as [tex]f_T[/tex]. The force of gravity is labeled as [tex]f_g[/tex].
Part B:

A free-body diagram for a rabbit is as follows:
The diagram shows three forces acting on the stuffed rabbit that is being pushed by a toy car that is battery-powered, which is as follows:

The direction of friction is opposite to the direction of motion, which is shown towards the right.
The force due to friction is labeled as [tex]f_K[/tex]. The force due to air resistance is labeled as fair. The force of gravity is labeled as [tex]f_g[/tex].
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Answer: The answer is D. Thermal Energy.

Explanation:

Thermal energy is a type of kinetic energy owing to the fact that it results from the movement of particles.

The wavelength of the emitted light for two photon excitaton fluorescence is 600nm.

A two photon excited process-

Wavenumber of the excitation light = 10000 cm-1 = 1000 nm

In case of two photon excitation photon -

Second harmonic generation = [  Wavenumber ( in nm ) ] / 2 = 1000/2  = 500 nm

We know, ESGH = 3.97 × 10^-19J

For two photon excitation fluorescence internal conversion, energy is 6.89 × 10^-20J. So, Energy of fluorescence = ESHG - EIC  = 3.286 × 10^-19J.

We know, E = hc / λ

λ = 6.049 x 10^-7 m  

≈ 600 nm

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The ball is in the air for about 1.8 seconds before it hits the ground after it leaves the student's hand with a speed of 11 m/s when the hand is 1.8 m above the ground.

Projectile motion is a kind of movement experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the gravity of the Earth. In general, projectile motion refers to a free-body's motion influenced only by gravity. A student throws a ball straight up while standing on the ground. When her hand is 1.8 m above the ground, the ball leaves her hand at a speed of 11 m/s. The time the ball is in the air before it hits the ground is calculated as follows:Using the equation:

∆y = v0yt + 1/2gt² Where ∆y is the displacement (in this case, -1.8 m) of the projectile along the vertical axis, v0y is the initial vertical velocity (in this case, 11 m/s), t is the time of flight, and g is the acceleration due to gravity (9.81 m/s²):-1.8 m = (11 m/s)t + (1/2)(-9.81 m/s²)t².Rearranging the equation, we get:-4.905t² + 11t - 1.8 = 0.

Using the quadratic formula, we get:t = (-11 ± sqrt(11² - 4(-4.905)(-1.8))) / (2(-4.905))= 1.77 s or t = 0.20 s. Since the ball is in the air for approximately 1.77 s before it hits the ground, and the student's hand is 1.8 m above the ground, the ball is in the air for about 1.8 seconds before it hits the ground. Therefore, the correct answer is the option C, 1.8 seconds.

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The energy stored in the capacitor is calculated as 630150 J. The ratio between the energy density per unit mass of the 9.0 kg capacitor system and the 130,000 J/kg of the truck's battery is 70.17

The formula to calculate the energy stored in a capacitor is expressed by the formula: 

E = (1/2)CV²

where E is energy, C is capacitance, and V is voltage.

The question mentions that the capacitor is fully charged to 13.8 V. Therefore, the energy stored in the capacitor is given by the formula:

[tex]E = (1/2)CV^2 \\= (1/2)\times (500 F)\times {(13.8 V)}^2\\= 630150 J[/tex]

The ratio between the energy density per unit mass of the 9.0 kg capacitor system and the 130,000 J/kg of the truck's battery can be computed by dividing the energy density of the capacitor system by the energy density of the truck's battery.

We know that energy density = energy / mass of the system.

Thus, the formula to calculate the ratio is:

[tex]Ratio = \dfrac{energy density per unit mass of capacitor system}{ energy density per unit mass of truck's battery}\\Ratio= \dfrac{630150 J / 9 kg}{ 130,000 J / 1 kg}= 70.017[/tex]

Therefore, the ratio of energy density per unit mass of the capacitor system to that of the truck's battery is 70.017.

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The multiplication factor, k of the reactor at the time of the observations is 0.87. If the thermal leakage is reduced by one-third, the value of k would increase to 1.87.

To calculate the multiplication factor, we can use the following equation:

k = (1-nf - nt)/nt,

where nf is the fraction of neutrons emitted in fission that escape while slowing down, nt is the fraction of neutrons that escape after having slowed down to thermal energies, and nt is the fraction of neutrons absorbed in the reactor while slowing down.

Given that nf = 0.1, nt = 0.15, and nt = 0.6, we can calculate the multiplication factor, k, as follows:

k = (1 - 0.1 - 0.15)/0.6

k = 0.87

Therefore, the multiplication factor of the reactor at the time these observations are made is 0.87. If the thermal leakage is reduced by one-third, the value of k would increase to 1.87.

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The orbit radius is 6.225 × 10^5 km.

The x-ray pulses from Cygnus X-1, a celestial x-ray source, have been recorded during high-altitude rocket flights. The signals can be interpreted as originating when a blob of ionized matter orbits a black hole with a period of 7.84 ms. And also, it is given that the blob were in a circular orbit about a black hole whose mass is 13.5 times the mass of the Sun. We need to determine the orbit radius.

The formula to be used to find the orbit radius is given by:

G(M+m)T2/4π2= r3

Where,

G = Gravitational constant = 6.67259×10−11 N⋅m2/kg2
M = Mass of the black hole
m = Mass of the blob
T = Time period of the orbit = 7.84 ms = 7.84 × 10^-3 s
r = Orbit radius

Substitute the given values in the above formula, we get:

r3 = G(M+m)T2/4π2
r3 = 6.67259×10−11 * [13.5(1.991×10^30) + m] * (7.84×10−3)2 / 4π2
r3 = 5.7919 × 10^15 m^3
Taking cube root on both sides, we get:
r = [5.7919 × 10^15 m^3] 1/3
r = 6.225 × 10^8 m
1 km = 1000 m

Therefore, the orbit radius in km is:
r = 6.225 × 10^8 m * 1 km / 1000 m
r = 6.225 × 10^5 km

Hence, the orbit radius is 6.225 × 10^5 km.

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EER is defined as the Energy Efficiency Ratio which is the ratio of cooling capacity in BTU/hr to the power input in watts.

The EER of the given 8000 Btu/h air conditioner is 5.33 Btu/hour per watt.

In the case of the given 8000 Btu/h air conditioner that requires a power input of 1500 W, the EER can be calculated as follows:

EER = (cooling capacity in Btu/hr) / (power input in watts)

EER = 8000 Btu/hour / 1500 W = 5.33 Btu/hour per wat.

Energy efficiency ratio (EER) is used in the USA and is defined as the system output in Btu/h per watt of electrical energy.

Coefficient of performance (COP) is the equivalent measure using SI units, which is widely used in the UK. A COP of 1.0 equates to an EER of 3.4.

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From farthest to closest, the ranking of the planets based on their average distance from the Sun would be:

Pluto, Saturn, Jupiter, Mars, Earth, Mercury

Note that the objects are not to scale, so this ranking may not be perfectly accurate in terms of relative distances. However, it gives a general idea of the order of the planets from farthest to closest to the Sun.

The eight planets in our solar system, listed in order from the Sun, are:

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

These eight planets are also known as the "classical planets," and are the largest and most massive objects in orbit around the Sun. There are also several dwarf planets in our solar system, such as Pluto and Ceres, as well as numerous smaller objects like asteroids and comets.

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