Express the work done in Kwh if a motor that does 24000 J of work in two minutes

Answer:

"byte"

A byte usually consists of eight zero and one digits

Honeywell used the byte in its early computers and IBM used a hexadecimal system which consisted of 16 zero and one digits

The force of inertia acting on the box of books on the metal rolling cart was opposed by the friction force between the box and the cart's surface, which prevented the box from sliding or falling off the cart when Karol stopped suddenly.

The friction force between the wheels of the cart and the ground opposes the motion of the cart, causing it to slow down and eventually come to a stop. As the cart slows down, the box of books on top of it also experiences a force in the opposite direction due to its inertia. However, the force of friction between the box and the cart is greater than the force of inertia acting on the box, which causes it to stay in place on the cart.

This is because the friction force depends on the weight of the box and the coefficient of friction between the box and the surface of the cart. If the weight of the box is greater, the friction force will also be greater, making it more difficult for the box to slide or fall off the cart. Similarly, if the surface of the cart is rougher, the friction force will also be greater, providing more resistance to the box's motion.

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Yes, it is possible for a first or second class lever to have a mechanical advantage less than one beacuse a first-class lever has the fulcrum between the effort and the load, while a second-class lever has the load between the effort and the fulcrum.

The mechanical advantage of a lever can be calculated using the formula:
Mechanical Advantage (MA) = Effort Arm Length / Load Arm Length
For a mechanical advantage less than one, the effort arm length must be shorter than the load arm length. This means the fulcrum (or pivot point) must be closer to the effort than the load in a first-class lever, and closer to the load than the effort in a second-class lever. When this occurs, the lever will require a larger effort to move the load, resulting in a mechanical advantage less than one.

So, it is possible for a first or second class lever to have a mechanical advantage less than one.

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The brass instrument that has a movable slide is the trombone. The slide allows the player to vary the length of the tubing, thus changing the pitch of the notes.

By moving the slide in and out, the player can play a wide range of notes and create smooth glissandos between notes. Unlike other brass instruments such as the trumpet or French horn, which use valves to change the length of the tubing, the trombone uses a slide. The slide is made up of two parallel tubes, which are connected by a U-shaped bend, allowing the player to move the slide in and out to adjust the length of the tubing.

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Answer:

The shorter the wavelengths and higher the frequency corresponds with greater energy. So the longer the wavelengths and lower the frequency results in lower energy. The energy equation is E = hν.

Using the mirror and magnification formulas, the image distance and height are calculated to be -71.80 cm and 1.38 cm, respectively.

We can use the mirror formula to find the image height. The mirror formula is

1/f = 1/d_o + 1/d_i

where f is the focal length of the mirror, d_o is the object distance (distance of the object from the mirror), and d_i is the image distance (distance of the image from the mirror).

The distances are considered positive when measured to the right of the mirror (in the direction of the incident light), and negative when measured to the left.

First, we need to find the image distance. Since the object is placed in front of a concave mirror, the image is formed on the same side of the mirror as the object, and is real and inverted. Using the mirror formula and the given values, we get:

1/74 = 1/838 + 1/d_i

Solving for d_i, we get:

d_i = 1 / (1/74 - 1/838) = -71.80 cm

The negative sign indicates that the image is formed on the same side of the mirror as the object.

Next, we can use the magnification formula to find the image height. The magnification formula is

m = -d_i / d_o

where m is the magnification, which is negative for a real and inverted image.

Substituting the values we have found, we get

m = -(-71.80 cm) / 838.00 cm = 0.086

Finally, we can find the image height by multiplying the object height by the magnification

h_i = m * h_o = 0.086 * 16.00 cm = 1.38 cm

Therefore, the image height is 1.38 cm.

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Acceleration is defined as the rate of change of velocity with respect to time. In this case, the person's speed increased by 20 m/s over a time of 20 seconds. To calculate the acceleration, we can use the formula:

Acceleration = (Change in velocity) / (Time)

In this case, the change in velocity is 20 m/s, and the time is 20 seconds. Substituting these values into the formula, we get:

Acceleration = 20 m/s / 20 s

Simplifying the equation:

Acceleration = 1 m/s²

Therefore, the person's acceleration is 1 m/s².

~~~Harsha~~~

The bulk modulus is a proportionality constant that relates the pressure acting on an object to the fractional force change in volume.

It is defined as the ratio of the change in pressure applied to the material to the resulting fractional change in volume. The bulk modulus is typically denoted by the symbol K and has units of pressure. It is an important property of materials, as it can be used to predict how they will behave under different conditions, such as changes in pressure or temperature.

The fractional change in length, Young's modulus, and spring constant - are all related to different properties of materials. Shear refers to the deformation of a material when a force is applied parallel to its surface, while fractional change in length refers to the elongation or contraction of a material when a force is applied perpendicular to its surface. Young's modulus is a measure of a material's stiffness, or resistance to deformation, when a force is applied perpendicular to its surface.
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The flow meter ball is designed to measure the rate of gas flow through a system. It is a simple mechanical device that relies on the buoyancy of the gas to float at a certain height in the flow tube. The position of the ball indicates the rate of flow of the gas.

However, the buoyancy of the gas is affected by a number of factors, including the type of gas being used, the temperature and pressure of the gas, and the presence of other gases or contaminants in the system. In the case of shielding gases, different gases have different densities and viscosities, which can affect the buoyancy of the flow meter ball.

In summary, while the shielding gases may be flowing at the same rate, the position of the flow meter ball can be affected by a number of factors, including the type of gas being used, the temperature and pressure of the gas, and the presence of other gases or contaminants in the system.

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The relative density of paraffin is 1.8, which means that it is 1.8 times denser than water.

The relative density of a substance is defined as the ratio of its density to the density of a reference substance. In this case, we can use water as the reference substance, which has a density of 1 g/cm³ at room temperature.

To calculate the relative density of paraffin, we first need to determine the volume of the bottle. We can do this by subtracting the weight of the empty bottle (25g) from the weight of the full bottle when filled with water (50g), which gives us a volume of 25 cm³.

Next, we can use the mass of the full bottle when filled with paraffin (45g) and the volume of the bottle (25 cm³) to calculate the density of paraffin. Density is defined as mass per unit volume, so we can use the formula:

density = mass / volume

density of paraffin = 45g / 25 cm³

density of paraffin = 1.8 g/cm³

Finally, we can calculate the relative density of paraffin by dividing its density by the density of water:

relative density of paraffin = density of paraffin / density of water

relative density of paraffin = 1.8 g/cm³ / 1 g/cm³

relative density of paraffin = 1.8

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The wavelength of electromagnetic radiation emitted are a. 121.6 nanometers. b. 656.3 nanometers. c.656.3 nanometers.

a. The wavelength of electromagnetic radiation emitted when the photon makes a transition from n = 2 to n = 1 is 121.6 nanometers.

This is known as the Lyman series in the hydrogen atom.

b. The wavelength of electromagnetic radiation emitted when the photon makes a transition from n = 3 to n = 2 is 656.3 nanometers.

This is known as the Balmer series in the hydrogen atom.

c. The wavelength of electromagnetic radiation emitted when the photon makes a transition from n = 3 to n = 1 is 656.3 nanometers.

This is also part of the Balmer series in the hydrogen atom.

The wavelength of electromagnetic radiation emitted during a transition in a hydrogen atom can be calculated using the Rydberg formula:

[tex]1/λ = R(1/n1^2 - 1/n2^2)[/tex]

where λ is the wavelength,

R is the Rydberg constant

[tex](1.097 \times 10^7 m^-1)[/tex]  

and n1 and n2 are the initial and final quantum numbers, respectively.

The transitions from higher energy levels to lower energy levels release energy in the form of photons with characteristic wavelengths. These wavelengths correspond to different series named after their discoverer, such as the Lyman series, Balmer series, etc. The energy released during these transitions is quantized, which means that only certain discrete wavelengths can be emitted or absorbed.

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The gravitational force between two objects can be calculated using the formula:

F = G * (m1 * m2) / r^2

Where F is the gravitational force, G is the gravitational constant (6.67 x 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.

Substituting the given values into the formula, we get:

F = (6.67 x 10^-11 N*m^2/kg^2) * (5.98 x 10^24 kg) * (7.36 x 10^22 kg) / (3.85 x 10^8 m)^2

F = 1.99 x 10^20 N

Therefore, the gravitational force between the earth and the moon is approximately 1.99 x 10^20 N.

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The energy density in the magnetic field 25 cm from the long straight wire carrying a current of 12 A is approximately 3.6 × 10^-4 J/m^3, which is option (C).

The energy density in the magnetic field around a long straight wire carrying a current I at a distance r from the wire can be calculated using the formula:

u = (B^2) / (2*μ0)

where B is the magnetic field strength and μ0 is the permeability of free space.

For a long straight wire, the magnetic field strength at a distance r from the wire is given by:

B = (μ0*I) / (2*π*r)

Substituting the given values, we get:

B = (4π × 10^-7 T·m/A) × (12 A) / (2π × 0.25 m) = 1.52 × 10^-4 T

Now we can use the formula to find the energy density:

u = (B^2) / (2*μ0) = [(1.52 × 10^-4 T)^2] / (2 × 4π × 10^-7 T·m/A) ≈ 3.6 × 10^-4 J/m^3

Therefore, the energy density in the magnetic field 25 cm from the long straight wire carrying a current of 12 A is approximately 3.6 × 10^-4 J/m^3, which is option (C).

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The scattered photon energy is 8.4 MeV, and the recoiling proton energy is 1.6 MeV.


When a photon scatters off a proton, some of its energy is transferred to the proton, resulting in the proton recoiling. The scattered photon's energy can be calculated using the Compton scattering formula, which relates the initial and final energies of the photon and the scattering angle. In this case, the scattered photon's energy is 8.4 MeV.

The energy transferred to the proton can be calculated using energy conservation principles. Since the photon's initial energy was 10 MeV, and 8.4 MeV was scattered, the energy transferred to the proton is 1.6 MeV. This energy is imparted to the proton as kinetic energy, causing it to recoil. The final energy of the proton can be calculated using conservation of momentum, but without further information on the system, it cannot be determined in this scenario.

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To increase the electric potential between two points, we need to increase the potential difference between them. This can be achieved by either increasing the electric potential energy of the charged particle at one point or decreasing it at another point.

When we talk about electric potential, we are referring to the amount of electric potential energy per unit charge. Electric potential energy is the energy possessed by a charged particle due to its position in an electric field. It is directly proportional to the amount of charge present and the electric field strength. When we do work to move more charge against an electric field, we increase the electric potential energy of the charged particle.
However, electric potential is not directly related to the amount of charge moved against the electric field. Instead, it is a measure of the potential difference between two points in an electric field. Electric potential is defined as the amount of electric potential energy per unit charge required to move a charge from one point to another.
In summary, doing more work to move more charge against an electric field will increase the electric potential energy of the charged particle, but it does not necessarily increase the electric potential between two points. To increase the electric potential between two points, we need to increase the potential difference between them by altering the electric potential energy at one or both points.

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The planet that had the Great Dark Spot in 1989 but lost it by 1995 was Neptune. The Great Dark Spot was a massive storm in the atmosphere of Neptune, similar to the Great Red Spot on Jupiter.

It was discovered by the Voyager 2 spacecraft in 1989 and was observed to be approximately the size of Earth.

However, when the Hubble Space Telescope observed Neptune in 1995, the Great Dark Spot had disappeared. This could be due to the dynamic nature of Neptune's atmosphere, which is constantly changing and evolving.

Neptune is the eighth planet from the Sun and is known for its vibrant blue color and strong winds.

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The waves created by the generator are traveling at a speed of 60 m/second.

The speed of a wave can be calculated by multiplying its frequency by its wavelength. In this case, the frequency of the wave is given by the number of vibrations per unit time, and the wavelength is given as 0.5 m.

The frequency of the wave generator can be calculated as 600 vibrations per 5 seconds, or 120 vibrations per second.

Therefore, the speed of the waves created by the generator can be calculated as:

Speed = Frequency x Wavelength

Speed = 120 vibrations/second x 0.5 m/vibration

Speed = 60 m/second

Therefore, the waves created by the generator are traveling at a speed of 60 m/second.

It is worth noting that the speed of a wave is determined by the properties of the medium through which it is traveling.

In this case, the speed of the wave is determined by the tension and density of the rope. If these properties were to change, the speed of the wave would also change.

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The tension in the string is 4.86 N. With the wave velocity, v, we can find the tension, T, using the formula T = µ * v^2. Finally, we get the tension in the string by plugging in the values and solving the equation.

We can use the formula for the frequency of a string:
f = (n/2L) * sqrt(T/μ)
where:
- f is the frequency
- n is the harmonic number (in this case, the third harmonic means n = 3)
- L is the length of the string
- T is the tension in the string
- μ is the mass per unit length of the string

We are given the length and mass per unit length of the string, as well as the frequency of the third harmonic. We can rearrange the formula to solve for T:
T = (μ * (2L * f/n)^2)
Plugging in the values we know:
T = (0.020 kg/m * (2 * 0.25 m * 180 Hz/3)^2)
T = 4.86 N
To find the tension in a string with given length, mass per unit length, and frequency of the third harmonic, we can use the formula T = (μ * (2L * f/n)^2) and plug in the values to get our answer. including an introduction, explanation of the formula and calculation process, and conclusion.

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the magnification of the image is -1.25, which means that the image is 1.25 times larger than the object, but inverted.

To find the magnification of the image, we can use the formula:
magnification = image height / object height
However, since we don't know the actual heights of the object and image, we need to use another formula that relates the distance of the object and image from the lens:
1/f = 1/d_o + 1/d_i
where f is the focal length of the lens, d_o is the distance of the object from the lens, and d_i is the distance of the image from the lens.
We know that the object is 18 cm from the lens, and the image is 22.5 cm from the lens. We can rearrange the formula to solve for the focal length:
1/f = 1/18 + 1/22.5
1/f = 0.0556
f = 18 cm
Now that we know the focal length of the lens, we can use the magnification formula:
magnification = -d_i / d_o
where the negative sign indicates that the image is inverted. Substituting the distances we know, we get:
magnification = -22.5 / 18
magnification = -1.25
Therefore, the magnification of the image is -1.25, which means that the image is 1.25 times larger than the object, but inverted.

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When a charged positive rod is held near, but does not touch a neutral electroscope, the charge on the knob becomes positive. This is because the positive charge on the rod induces a separation of charges in the electroscope, causing the electrons in the knob to move away from the rod and towards the leaves, leaving the knob positively charged. This is known as electrostatic induction.

It is important to note that the electroscope does not become positively charged, as the charges induced are only temporary and there is no transfer of charge from the rod to the electroscope. The electroscope remains neutral overall, but the separation of charges allows for the detection of the presence of the charged rod.

In summary, holding a charged positive rod near a neutral electroscope induces a temporary separation of charges, causing the knob to become positively charged.

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The statement "A push system means providing the next station with exactly what is needed when it is needed" is (a) false

A push system is actually the opposite of providing exactly what is needed when it is needed. In a push system, goods are produced and pushed onto the next station or customer regardless of their immediate need or demand.

This can lead to excess inventory and waste if the products are not sold or used in a timely manner. A pull system, on the other hand, responds to customer demand and only produces what is needed when it is needed.

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The weight of an object is the force exerted on it by the gravitational field of a massive object, such as the Earth. The strength of this force depends on the mass of the object and the gravitational field strength at its location.

The gravitational field strength on the surface of the Earth varies with distance from the Earth's center and is directly proportional to the mass of the Earth. As the Earth spins faster, the centrifugal force caused by the rotation of the Earth reduces the effective gravitational force felt by a person on the surface. This is because the centrifugal force acts in the opposite direction to gravity and reduces the net force acting on the person. Thus, the faster the Earth spins, the less a person weighs.

On the other hand, the weight of a person on a rotating space station is affected by two forces: the gravitational force due to the mass of the Earth and the centrifugal force due to the rotation of the space station. The centrifugal force is proportional to the square of the rotation rate and the distance from the center of rotation. As the space station spins faster, the centrifugal force becomes stronger, and thus the net force acting on the person also increases. This results in an increase in the person's weight.

Therefore, the faster the Earth spins, the less a person weighs due to the reduction in the effective gravitational force, whereas the faster a space station spins, the more a person weighs due to the increase in the centrifugal force.

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The height and image distance from the convex lens is 30 cm and

-22.5 cm respectively.

Focal length of the convex lens, f = 45 cm

Height of the object, h₀ = 20 cm

Distance of the object from the convex lens, u = -15 cm

According to the lens formula,

1/v - 1/u = 1/f

1/v = 1/f + 1/u

1/v = (1/45) + (1/-15)

1/v = -2/45

Therefore, the image distance from the convex lens,

v = -45/2

v = -22.5 cm

According to the magnification formula of the convex lens,

m = v/u = hi/h₀

Therefore, the height of the image,

hi = h₀v/u

hi = 20 x (-22.5/-15)

hi = 30 cm

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Your question was incomplete, but most probably your question would be:

Convex lens of focal length 45 cm has an object kept at distance 15cm from it. If height of object is 20 cm, determine position and height of image.

The electric potential at a point does not necessarily indicate the magnitude of the electric field at that point. The two quantities are related, but they measure different properties of the electric field. A counterexample to this relationship is provided by the conducting spherical shell example, where the electric potential is large but the electric field is zero.

When the electric potential at a point is large, it does not necessarily mean that the electric field at that point is also large. The electric potential is a scalar quantity that measures the work done per unit charge in moving a charge from a reference point to the point in question. On the other hand, the electric field is a vector quantity that measures the force per unit charge experienced by a charge at the point in question.
Consider a point charge that is surrounded by a conducting spherical shell. The electric field inside the shell is zero due to the shielding effect of the charges on the shell. However, the electric potential at any point inside the shell is proportional to the charge of the point charge divided by the distance from the point charge. Therefore, the electric potential at any point inside the shell is large due to the close proximity of the point charge. However, the electric field at any point inside the shell is zero.

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The distance from the central axis of the wire at which the magnetic field due to the wire is equal to the Earth's magnetic field is approximately 3.98 cm. The answer is closest to option (D) 4.0 cm.

We can use the formula for the magnetic field due to a long straight wire:

B = (μ0 / 2π) * (I / r)

where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance from the wire.

We want to find the distance at which the magnetic field due to the wire is equal to the Earth's magnetic field, so we can set the two fields equal to each other and solve for r:

(μ0 / 2π) * (5.0 A / r) = 5.0 × 10-5 T

Solving for r, we get:

r = (μ0 / 2π) * (5.0 A / 5.0 × 10-5 T) = 3.98 cm

So the distance from the central axis of the wire at which the magnetic field due to the wire is equal to the Earth's magnetic field is approximately 3.98 cm. The answer is closest to option (D) 4.0 cm.

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Machines provide various advantages, including force multiplication, force reduction, distance multiplication, and direction change.

A machine can make work easier in various ways, including the following:

a. A machine makes work easier by multiplying the force you exert. This means that when you apply a smaller force to the machine, it can amplify that force to accomplish tasks that require greater force.

b. A machine makes work easier by reducing the amount of force needed to do the job. By utilizing mechanical advantage, machines allow us to accomplish tasks with less effort. For example, using a lever or a pulley can reduce the force needed to lift a heavy object.

c. A machine makes work easier by multiplying the distance over which you exert force. This involves trading off  the force for distance, so you might need to apply a smaller force over a longer distance to achieve the same work. An example of this would be using a ramp to push an object up to a higher level.

d. A machine makes work easier by changing the direction in which you exert force. Some machines, such as pulleys and gears, help to change the direction of the applied force, making it more convenient or efficient to perform a task.

Machines provide various advantages, including force multiplication, force reduction, distance multiplication, and direction change.  By employing these principles, machines enable us to perform tasks more efficiently and with less effort

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The power output is 2 Watts. Power is a measure of the rate at which work is done or energy is transferred.

In this scenario, you exerted 100 J (joules) of energy to lift a box in 50 s (seconds). Power (P) is calculated by dividing the amount of work done (W) by the time taken (t): P = W/t.

In this case, P = 100 J / 50 s = 2 J/s, which is equal to 2 Watts (W). Therefore, your power output is 2 Watts, indicating that you are transferring energy or performing work at a rate of 2 Joules per second. This measurement quantifies how quickly you are able to lift the box.

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We can use the transformer equation to solve for the number of turns in the primary coil:

Vp/Vs = Np/Ns

where Vp and Vs are the voltages in the primary and secondary coils, respectively, and Np and Ns are the number of turns in the primary and secondary coils, respectively.

Substituting the given values:

110/24 = Np/100

Np = 458.3

Rounding off, the number of turns required in the primary coil is approximately 458.

Therefore, the answer is (A) 458.

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The ball will roll backwards relative to the wagon and then eventually come to a stop. Relative to the ground, the ball will roll forward until it also comes to a stop.

When the wagon abruptly moves forward, the ball's initial inertia causes it to remain at rest while the wagon moves forward, causing it to roll backwards relative to the wagon. As the wagon continues to move forward, the ball will eventually overcome its inertia and start rolling forward, eventually coming to a stop as the wagon continues to move.  

Relative to the ground, the ball will roll forward as the wagon moves, due to the wagon's forward motion. However, as the wagon comes to a stop, so too will the ball. It's important to note that the motion of the ball relative to the ground is affected by both the wagon's forward motion and the ball's inertia, while the motion of the ball relative to the wagon is affected solely by the ball's inertia.

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The ball's initial speed decreases as it moves upward from the vertical spring launcher. At its maximum height, the ball has zero speed and begins to fall back down. Since the block is sliding at a constant speed with negligible friction, the position of the launcher relative to the ground remains unchanged.

Since the block is sliding at a constant speed across a horizontal surface with negligible friction, it means the block is moving at a steady pace without being slowed down by the surface. Meanwhile, the ball is launched vertically upward from the launcher.
When the ball reaches its maximum height, it will momentarily have zero vertical speed before starting to descend due to gravity. However, during its upward and downward journey, the ball maintains the same horizontal speed as the block, since there is negligible friction between the block and the surface.
As a result, the position of the ball relative to the launcher will remain the same horizontally when it reaches its maximum height. The ball will be vertically above the launcher at the maximum height, maintaining the same horizontal position throughout its motion.

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