Calculate the mass in kg of a ball at a height of 3m above the ground with a potential energy of 120J.

Answers

Answer 1

The mass of the ball at a height of 3m above the ground with a potential energy of 120J can be calculated using the equation:

Mass = Potential Energy/Gravity * Height

Mass = 120J/(9.81m/s² * 3m)

Mass = 4.1 kg

Answer 2

Answer:

4 kg

Explanation:

Using,

Energy/ Work done = Force x Distance (Height)

E = F • s

But recall, that F = mg

Therefore,

E = m • g • s

Making mass (m), the subject of the formula

m = E / (g • s)

m = 120 / (10 • 3)

m = 120 / 30

m = 4 kg

But if g = 9.8 ms-¹

Then,

m = 120 / (9.8 • 3)

m = 120 / 29.4

m = 4.08 kg

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Related Questions

3. Large amplitude vibrations produced when the of receiver of the applied forced vibration matches the

Answers

An object's amplitude dramatically increases when the frequency of the applied forced vibrations matches the object's natural frequency. Resonance describes this behavior.

Theory A wave's amplitude directly relates to the quantity of energy it can carry. A wave with a high amplitude carries a lot of energy, whereas one with a low amplitude carries only a little. A wave's strength is determined by the typical energy that moves through a given area in a certain amount of time and in a particular direction.The sound wave's amplitude grows in proportion to its strength. We perceive louder noises to be of higher intensity. Comparative sound intensities are frequently expressed using decibels (dB)

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A compact car can climb a hill in 10 s. The top of the hill is 30 m higher than the bottom, and the car’s mass is 1,000 kg What is the power output of the car?

Answers

Answer:

the power output of the car is 29.43 kW (rounded to two decimal places).

Explanation:

To find the power output of the car, we need to use the formula:

power = work / time

where work is the change in potential energy of the car as it climbs the hill, which can be calculated using the formula:

work = force x distance

where force is the force required to lift the car against gravity, which is given by:

force = mass x gravity

where mass is the mass of the car, and gravity is the acceleration due to gravity (9.81 m/s^2).

So, the force required to lift the car against gravity is:

force = 1000 kg x 9.81 m/s^2 = 9810 N

The distance the car travels up the hill is 30 m.

Therefore, the work done by the car is:

work = force x distance = 9810 N x 30 m = 294300 J

The time taken by the car to climb the hill is 10 s.

Therefore, the power output of the car is:

power = work / time = 294300 J / 10 s = 29430 W

Estimat the number and wattage of lamps. which would be required to illuminate a workshop space 60x1.5 meteres by means of lamps mounted 5 metres above the working Plane The average illumination required is about 100 wt. coefficient of utilisation = 0.4 luminous efficiency 16 lumens per watt. Assume a space-height ratio of unity and a cundle Power depreciation of 20%​

Answers

The number and wattage of lamps required to illuminate the workshop would be approximately 8 lamps and 70 watts respectively.

Wattage calculation

To estimate the number and wattage of lamps required to illuminate a workshop space of 60x1.5 meters, we can follow these steps:

Calculate the area of the workshop:

Area = length x widthArea = 60m x 1.5mArea = 90 square meters

Determine the total lumens required:

Lumens = area x average illuminationLumens = 90 sq m x 100 luxLumens = 9000 lumens

Adjust for the coefficient of utilization and luminous efficiency:

Effective lumens = lumens / (coefficient of utilization x luminous efficiency)Effective lumens = 9000 / (0.4 x 16)Effective lumens = 1406.25 lumens

Adjust for space-height ratio and candle power depreciation:

Effective lumens per lamp = effective lumens x space-height ratio x (1 - depreciation)Effective lumens per lamp = 1406.25 x 1 x (1 - 0.2)Effective lumens per lamp = 1125 lumens

Determine the number of lamps required:

Number of lamps = total lumens required / effective lumens per lampNumber of lamps = 9000 / 1125Number of lamps = 8 lamps (rounded up)

Determine the wattage of each lamp:

Wattage per lamp = effective lumens per lamp / luminous efficiencyWattage per lamp = 1125 / 16Wattage per lamp = 70.3 watts (rounded up)

Therefore, approximately 8 lamps with a wattage of 70 watts each would be required to illuminate the workshop space.

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Select the correct answer. In a given chemical reaction, the energy of the products is greater than the energy of the reactants. Which statement is true for this reaction? A. Energy is absorbed in the reaction. B. Energy is released in the reaction. C. No energy is transferred in the reaction. D. Energy is created in the reaction. E. Energy is lost in the reaction. Reset Next

Answers

A - energy is absorbed during the reaction

an electron is moving parallel to an electric field (from higher to lower voltage). its potential energy is

Answers

The potential energy of an electron moving parallel to an electric field decreases as it moves from higher voltage to lower voltage. The work done by the electric field on the electron is equal to the decrease in potential energy. The potential energy of the electron is proportional to its charge and the voltage difference between the two points.

When an electron moves parallel to an electric field, its potential energy is conserved. The potential energy of an electron is proportional to its charge and the voltage through which it moves. As the electron moves from higher voltage to a lower voltage, its potential energy decreases. The work done by the electric field on the electron is equal to the decrease in potential energy. When the electron is at rest, it has a certain potential energy due to its position in the electric field. If the electron is allowed to move freely, it will accelerate towards the lower voltage region, gaining kinetic energy. As it moves, the electric field continues to do work on the electron, converting its potential energy into kinetic energy. If the electric field is uniform, the potential energy of the electron will be given by the equation U = -qV, where q is the charge of the electron and V is the voltage difference between the two points. The negative sign indicates that the potential energy decreases as the voltage difference decreases.

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Find the equivalent resistance of the combination shown in Figure 4, assuming that
R5 = 17 Ω and R6 = 26 Ω.

Answers

Answer:

Explanation:

R/^5*r^6 Ok so  then this is simple once u get the answer u need to use the given formula in order to plug in the numbres sorry .

So basically

12 x r^6(u must fill in the number s ) and then u need to do `13x14xr the answer and use the rest of the numbers in order to figure out the quantities of each side for the shape . Then ur answer would be the r^x + x = ???

So yeah hope this helped

I think

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imagine that the blue light and orange light from the source were blocked. what color would how be present in the spectrum of light observed

Answers

Everything but blue & orange would now be present in the spectrum of light observed.

Spectrum refers to a range of different wavelengths of electromagnetic radiation. Electromagnetic radiation is a form of energy that travels through space and includes different types such as radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Each type of electromagnetic radiation has a different wavelength and frequency, and together they make up the electromagnetic spectrum.

The concept of spectrum is used in a variety of fields, including physics, astronomy, and telecommunications. The spectrum of electromagnetic radiation is essential for many technologies, such as radios and televisions, cell phones, and medical imaging devices, as they all rely on the transmission and reception of specific wavelengths of electromagnetic radiation.

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Complete Question: -

Imagine that the blue light and orange light from the source were blocked. What color(s) would now be present in the spectrum of light observed?

A metal wire, fixed at one end, has length l and cross-sectional area A. The wire extends a distance e which mass m is hung from the other end of the wire.What is an expression for the Young Modulus E of the metal?

Answers

The expression for the Young Modulus E of the metal is E = mgl / Ae. The Young Modulus E of the metal is calculated using the equation  E = (F l) / (A e2 m), where F is the force applied to the wire.


To find the expression for the Young modulus E of a metal wire with length l, cross-sectional area A, and mass m hung from the other end of the wire, we need to use the following formula:Stress (σ) = Load (F) / Area (A)Strain (ε) = Extension (Δl) / Original length (l)Young Modulus (E) = Stress (σ) / Strain (ε)We know that the metal wire is fixed at one end and the wire extends a distance e when a mass m is hung from the other end of the wire. Therefore, the extension Δl is equal to e.

Let's assume that g is the acceleration due to gravity. Therefore, the load F is equal to m * g.Substituting the values of F, A, and Δl in the above formula, we get:Stress (σ) = F / A = (m * g) / AStrain (ε) = Δl / l = e / lYoung Modulus (E) = Stress (σ) / Strain (ε)= (m * g) / (A * e / l) = mgl / AeTherefore, an expression for the Young Modulus E of the metal is E = mgl / Ae.

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125cm³ of a gas was collected at 15 °C and 755 mm of mercury pressure. Calculate the volume of the gas that will be collected at standard temperature and pressure

Answers

Answer:

119,2 см³

Explanation:

по формуле Клопейрона (P1×V1):T1=(P2×V2):T2

если из этой формулы найти V2, ответ будет равен примерно на 119,2 см³

clock a remains in place and clock b is carried around the earth ( 40,000 km). by how many seconds will is clock b slower if carried on

Answers

Clock a remains in place and clock b is carried around the earth (40,000 km). According to Einstein's theory of relativity, The clock b is slower by approximately 44.6 seconds.

According to Einstein's theory of relativity, time dilation takes place when an object moves at a velocity close to the speed of light. The closer the velocity is to the speed of light, the more time slows down. This is why time on Earth is slower at high altitudes than it is on the ground.

According to the theory, the same effect happens when objects are moving at a high speed, which is why clocks that are taken on an airplane, for example, appear to be ticking more slowly.

1. The following equation is used to determine the time dilation:

t = t0 / √(1 – v²/c²),

where t is the time elapsed, t0 is the time at rest, v is the velocity, and c is the speed of light. When the earth rotates on its axis, every point on the planet's surface moves at a different velocity, with the highest velocity at the equator, and the velocity decreases as we move towards the poles. The earth's circumference at the equator is roughly 40,000 kilometers (24,901 miles).
As a result, a person standing on the equator would be traveling at a speed of around 1,674 kilometers per hour (1,040 miles per hour) because the earth spins once every 24 hours. We must first determine the velocity of a point on the earth's surface at the equator before we can use the equation to calculate time dilation.

2. We use the formula

v = 2πr / T,

where v is velocity, r is the radius of the earth, and T is the time it takes the earth to complete one rotation. The formula is as follows:

v = 2πr / Tv

= 2 x 3.14 x 6,378 km / 24 hv

= 1,674 km/h

3. Substituting these values into the equation, we get:

t = t0 / √(1 – v²/c²)t = t0 / √(1 – (1,674 m/s)² / (299,792,458 m/s)²)t = t0 / √(1 – 2.8 x 10^-8)t = t0 / 0.9999999714

This means that the clock on the equator will tick slightly slower than it would at rest. The difference in time can be calculated by subtracting the two values:

t – t0 = t0 / 0.9999999714 – t0t – t0 = t0 (1 – 0.9999999714)t – t0 = 0.0000000286 t0

4. We must first calculate the amount of time elapsed on the equator if a clock b is carried 40,000 km around the earth. It is easy to calculate the distance and speed, but we must also consider that the earth is rotating as well. As a result, we must determine the combined speed of the earth's rotation and the motion of clock b relative to the earth's surface.

5. To calculate this combined velocity, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. If we imagine the velocity of the earth's rotation as the base of the triangle and the velocity of clock b as the height of the triangle, we can use this theorem to calculate the combined velocity as follows:

combined velocity = √(1,674² + vclock²)

where v clock is the velocity of clock b. Since clock b is being transported at the equator, it has the same velocity as the earth's rotation. As a result, we can substitute 1,674 km/h for v clock:

combined velocity = √(1,674² + 1,674²)

combined velocity = √(2 x 1,674²)

combined velocity = 2,367 km/h

6. Substituting the combined velocity into the equation for time dilation, we obtain:

t – t0 = t0 (1 – √(1 – v²/c²))t – t0 = t0 (1 – √(1 – (2,367 km/h)² / (299,792,458 m/s)²))t – t0

= t0 (1 – √(1 – 1.579 x 10^-11))t – t0

= t0 (1 – 0.999999999920215)t – t0

= 0.000000000079785 t0

Converting this value to seconds, we get:

0.000000000079785 t0 = 79.785 ns

Now we can combine the time dilation for the earth's rotation and the motion of clock b to obtain the total time dilation:

t – t0 = 0.0000000286 t0 + 0.000000000079785 t0t – t0 = 0.000000028679785 t0

Substituting the value of t0 (one second) into the equation, we get:

t – 1 = 0.000000028679785 seconds

Therefore, clock b will be approximately 44.6 seconds slower than clock a after being carried 40,000 km around the earth.

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A beam consisting of five types of ions labeled A, B, C, D, and E enters a region that contains a uniform magnetic field as shown in the figure below. The field is perpendicular to the plane of the paper, but its precise direction is not given. All ions in the beam travel with the same speed. The table below gives the masses and charges of the ions. Note: 1 mass unit = 1.67 x 10â€"27 kg and e = 1.6 x 10â€"19 C
Which ion falls at position 2?

Answers

At position 2, ion B falls. It is less deflected because it has a lesser mass than ions C, D, and E and the same charge as ion A.

A force perpendicular to the charged particle's velocity and the magnetic field's direction is applied when it reaches the magnetic field. The right-hand rule asserts that the palm will face the direction of the force if the thumb of the right hand points in the direction of the particle's velocity and the fingers point in the direction of the magnetic field. The particle's charge, velocity, and magnetic field intensity all affect how much force is generated.

Since all ions are moving at the same speed in this scenario, the force exerted on each ion is proportional to its charge to mass ratio. Ion B has the smallest mass of all the ions, so the least force and is least deflected of the ions, falling at position 2.

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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration? What is the Coefficient of friction?

Answers

Answer:

Explanation:

The centripetal acceleration of the race car is given by the formula:

a = v^2 / r

where v is the speed of the race car and r is the radius of the turn.

Substituting the given values, we get:

a = (27.5 m/s)^2 / 57.0 m = 13.3 m/s^2

So the centripetal acceleration of the race car is 13.3 m/s^2.

To find the coefficient of friction, we need to use the formula:

f = μN

where f is the force of friction, μ is the coefficient of friction, and N is the normal force.

The normal force is equal to the weight of the car, which we can calculate as:

N = mg

where m is the mass of the car and g is the acceleration due to gravity (9.81 m/s^2).

Assuming the mass of the car is 1500 kg, we get:

N = 1500 kg × 9.81 m/s^2 = 14,715 N

The force of friction is equal to the centripetal force required to keep the car moving in a circle:

f = ma = (1500 kg)(13.3 m/s^2) = 19,950 N

Substituting the values of N and f into the formula for friction, we get:

19,950 N = μ(14,715 N)

Solving for μ, we get:

μ = 1.35

So the coefficient of friction is 1.35.

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