Compared with some original speed, how much work must the brakes supply to stop a car that is moving twice as fast

Answers

Answer 1

Stopping a car moving at twice the original speed requires four times as much work compared to the original speed because the kinetic energy is quadrupled when the speed is doubled.

Stopping a car moving at twice the original speed requires a significant increase in the work done by the brakes. The kinetic energy of the car is proportional to the square of its speed, so if the speed is doubled, the kinetic energy is quadrupled.

To see this more clearly, consider the equation for kinetic energy:

KE = 1/2 * m * v²

where KE is the kinetic energy, m is the mass of the car, and v is its velocity. If the speed of the car is doubled, the kinetic energy becomes:

KE' = 1/2 * m * (2v)² = 2 * (1/2 * m * v²) = 2 * KE

This means that the kinetic energy of the car moving at twice the original speed is twice that of the original speed. To bring the car to a stop, this entire amount of kinetic energy must be dissipated by the brakes, which requires four times as much work compared to the original speed.

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Related Questions

What is the average momentum of an avalanche that moves a 40-cm-thick layer of snow over an area of 100 m by 500 m over a distance of 1 km down a hill in 5.5 s

Answers

The average momentum of the avalanche is approximately 1,090,909,090 kg*m/s.  By using momentum formula Momentum = mass x velocity

To calculate the average momentum of an avalanche, we first need to find its mass, then its velocity, and finally, use the momentum formula. Here are the steps:

1. Calculate the volume of the snow layer:
Volume = thickness x length x width = 0.4 m (40 cm) x 100 m x 500 m = 20,000 m³

2. Find the mass of the snow layer, assuming the snow density is 300 kg/m³ (a typical value):
Mass = volume x density = 20,000 m³ x 300 kg/m³ = 6,000,000 kg

3. Calculate the average velocity of the avalanche:
Distance = 1 km = 1,000 m
Time = 5.5 s
Velocity = distance / time = 1,000 m / 5.5 s ≈ 181.82 m/s

4. Compute the average momentum:
Momentum = mass x velocity = 6,000,000 kg x 181.82 m/s ≈ 1,090,909,090 kg*m/s

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A circular saw blade 0.200 m in diameter starts from rest. In 6.00 s it accelerates with constant angular acceleration to an angular ve- locity of 140 rad>s. Find the angular acceleration and the angle through which the blade has turned.

Answers

The angular acceleration of the blade is 23.3 rad/s^2, and the angle turned by the blade is 420 radians.

We can use the equations of rotational kinematics to solve this problem. The initial angular velocity is zero, and the final angular velocity is 140 rad/s. The time taken is 6.00 s, and the diameter of the circular saw blade is 0.200 m.

The equation for angular acceleration is:

α = (ωf - ωi) / t

where α is the angular acceleration, ωi is the initial angular velocity, ωf is the final angular velocity, and t is the time taken.

Plugging in the values given in the problem, we get:

α = (140 rad/s - 0 rad/s) / 6.00 sα = 23.3 rad/s^2

The equation for the angle turned by the blade is:

θ = ωi t + (1/2) α t^2

where θ is the angle turned by the blade, ωi is the initial angular velocity, α is the angular acceleration, and t is the time taken.Plugging in the values given in the problem, we get:

θ = 0 rad + (1/2) x 23.3 rad/s^2 x (6.00 s)^2θ = 420 rad.

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A metal surface is illuminated by light with a wavelength of 450 nmnm . The maximum kinetic energy of the emitted electrons is found to be 1.90 eVeV . Part A What is the maximum electron kinetic energy if the same metal is illuminated by light with a wavelength of 350 nmnm

Answers

Okay, here are the steps to solve this problem in two parts:

Part A (wavelength of 350 nm):
* Shorter wavelength (350 nm) means higher energy photons.
* The energy of a photon is proportional to its frequency (E=hν), and frequency is inversely proportional to wavelength (ν=c/λ).
* So 350 nm wavelength corresponds to higher frequency and more energy than 450 nm.

* To get photoemitted electrons, the photon energy must exceed the metal's work function (φ).
* The maximum KE of emitted electrons depends on the excess energy (hν - φ).
* If the maximum KE is 1.90 eV for 450 nm light, and only 350 nm light is used:
The excess energy (hν - φ) will be greater, so the maximum KE of emitted electrons will also be greater.

* Assume the work function remains the same. Then the increased photon energy at 350 nm will result in a higher maximum KE for emitted electrons.
* Without more information, we can't determine the exact maximum KE, but it would be > 1.90 eV.

* In summary, shorter wavelength (higher energy) photons will produce a higher maximum KE of emitted electrons for the same metal, assuming the work function remains constant. The excess energy leads to faster, more energetic electrons.

Does this help explain the reasoning behind why the maximum electron KE would be greater for 350 nm light compared to 450 nm light, keeping the metal and its work function the same? Let me know if any part is unclear or if you have any other questions!

Part B (maximum electron kinetic energy):
* For 350 nm light illuminating the same metal, the maximum electron KE would be greater than 1.90 eV, as explained.
* The only information given is that 1.90 eV is the maximum KE for 450 nm light.
* Without more details on the metal's work function or specific photon energies for 350 nm, an exact maximum KE cannot be calculated.
* But in general, for the same metal, higher energy photons (shorter wavelength, 350 nm) will produce faster, more energetic electrons, resulting in a higher maximum KE than lower energy photons (longer wavelength, 450 nm).

* The excess energy from photon absorption goes into electron kinetic energy after photoemission. More photon energy means more kinetic energy for emitted electrons.
* In summary, without further metal/photon specifics, the best that can be said is that the maximum electron KE would be greater than 1.90 eV if the metal were illuminated by 350 nm light versus 450 nm light. Does this match your understanding? Let me know if any part of the explanation needs more clarification.

I can also re-explain or provide another example if needed. The core ideas are:
1) Higher photon energy means higher maximum KE of emitted electrons for the same metal.
2) Excess energy from photon absorption determines electron KE after photoemission. More excess energy leads to faster, more energetic electrons.
3) At 350 nm, maximum electron KE would be > 1.90 eV based on it being higher than at 450 nm, assuming a constant work function.

Please let me know if this helps summarize or if you have any other questions!

What factor will control whether or not the universe keeps expanding or eventually starts to contract

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To answer what factor will control whether or not the universe keeps expanding or eventually starts to contract.

The factor that will control whether or not the universe keeps expanding or eventually starts to contract is the amount of matter and energy in the universe. If there is enough matter and energy, the gravitational pull will eventually cause the expansion to slow down and stop, and the universe will begin to contract. However, if there is not enough matter and energy, the expansion will continue indefinitely. Scientists are still studying the composition of the universe to determine whether or not there is enough matter and energy to cause a contraction.

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Calculate the angle for the third-order minimum of 595-nm wavelength yellow light falling on double slits separated by 0.185 mm.

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The angle for the third-order minimum of 595-nm wavelength yellow light falling on double slits separated by 0.185 mm is approximately 0.552°.

To calculate the angle for the third-order minimum of 595-nm wavelength yellow light falling on double slits separated by 0.185 mm, we can use the formula for destructive interference in a double-slit experiment:

d * sin(θ) = m * λ

where:
- d is the distance between the slits (0.185 mm or 0.000185 m)
- θ is the angle we want to find
- m is the order of the minimum (m = 3 for the third-order minimum)
- λ is the wavelength of the light (595 nm or 5.95 * 10^-7 m)

Rearrange the formula to solve for the angle θ:

sin(θ) = (m * λ) / d

Substitute the values:

sin(θ) = (3 * 5.95 * 10^-7 m) / 0.000185 m

sin(θ) ≈ 0.00963

Now, find the angle using the inverse sine function:

θ ≈ arcsin(0.00963)

θ ≈ 0.552°

So, the angle for the third-order minimum of 595-nm wavelength yellow light falling on double slits separated by 0.185 mm is approximately 0.552°.

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While reading the strength of magnetic field at each position from the magnet (or coil) using the magnetometer application in your phone, you noticed the reading was fluctuating randomly. Which is the best way to estimate the uncertainty in the strength of magnetic field measurement

Answers

The correct option is E,  Make the repeated measurements at equal distances and using STDEV/(Sqrt of range of samples) is the best way to estimate the uncertainty in the power of magnetic field measurement in this scenario.

A magnetic field is a physical phenomenon that results from the motion of electric charges. It is a force field that surrounds a magnet or an electrically charged particle and exerts a force on other magnets or charged particles in the vicinity.

The magnetic field is characterized by its direction and strength, which can be visualized using field lines that represent the path a hypothetical small magnetic north pole would take if it were placed in the field. The direction of the field lines indicates the direction of the force that a north pole would experience if it were placed in the field. Magnetic fields are generated by moving charges, such as the flow of current in a wire or the motion of electrons within an atom.

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Complete Question:

at the same time as reading the power of magnetic discipline at every role from the magnet (or coil) the usage of the magnetometer utility in your phone, you noticed the analyzing was fluctuating randomly. that's the best way to estimate the uncertainty in the energy of magnetic area measurement?

A). Make the repeated measurements at several distances and use STDEV/(Sqrt of variety of samples)

B). Use the decision of the electricity readings in the utility

C). Use the resolution of the meter scale

D). Use the quadrature rule

E). Make the repeated measurements at equal distance and use STDEV/(Sqrt of range of samples)

The K-T extinction (also known as the dinosaur killer event) occurred about 66 million years ago. What date is this (approximately) on the cosmic calendar?

Answers

A cosmic calendar is a visualization tool used to represent the history of the universe on a calendar year, where January 1 represents the Big Bang and December 31 represents the present day.

The universe is the vast expanse of space and all matter and energy within it. It includes everything from the smallest subatomic particles to the largest galaxies and beyond. The universe is estimated to be approximately 13.8 billion years old, having originated in the Big Bang, a colossal explosion that occurred nearly 14 billion years ago. The universe is constantly expanding, with galaxies moving away from each other at ever-increasing speeds.

The universe is composed of different types of matter, including dark matter and ordinary matter. The latter includes atoms, which are the building blocks of all physical matter. The universe is also filled with energy in various forms, including light and electromagnetic radiation.

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An electrolytic cell consists of two inert Pt- electrodes and 0.766 M NaOH (aq) solution under standard conditions.

(a) What are the half-reactions for the anode and cathode?

(b) What is the standard cell potential for this cell?

(c) What external potential must be applied to this cell so that it will function as an electrolytic cell?

(d) How many electrons flow through the cell if the cell is driven by 2.0A current for 1.5 hours?

Answers

The half-reactions are as follows Anode (oxidation): 2H2O(l) → O2(g) + 4H+(aq) + 4e- Cathode (reduction): 2H2O l + 2e- → H2(g) + 2OH-aq. The find the standard cell potential (Excel), we first need to find the standard reduction potentials (E°) for each half-reaction.

The standard reduction potential table

E°(O2/H2O) = +1.23 V

(For the anode reaction, reverse the sign as it is an oxidation reaction) E°(H2O/H2) = -0.83 V Now, we can calculate the

Excel = Cathode - Encode = (-0.83) - (-1.23) = +0.40 V

As an electrolytic cell requires an external potential to drive the non-spontaneous reaction, the applied external potential must be greater than the standard cell potential External potential > +0.40 V To find the number of electrons that flow through the cell, we can use the formula Number of electrons.

= (Current × Time) / (Faraday's constant)

First, we need to convert the time to seconds:

1.5 hours × 3600 s/hour = 5400 s

Then, we can calculate the number of electrons Number of electrons = (2.0 A × 5400 s) / (96,485 C/mol) ≈ 0.111 mol

of electrons in summary, the anode and cathode half-reactions involve the production of O2 and H2 gas, respectively, and the standard cell potential is +0.40 V. An external potential greater than +0.40 V must be applied for the cell to function as an electrolytic cell. Finally, 0.111 mol of electrons flow through the cell when driven by a 2.0 A current for 1.5 hours.

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(a) What is the gravitational potential energy of a two-particle system with masses 9.3 kg and 7.3 kg, if they are separated by 2.1 m

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the gravitational potential energy of the two-particle system is approximately 2.046 × 10^(-10) N*m.by using formula of

U = (G * m1 * m2) / r where gravitational constant is 6.674 × 10^(-11) N(m/kg)^2

Gravitational potential energy is the energy an object has due to its position in a gravitational field, specifically the energy it would take to separate two objects against the force of gravity.

To calculate the gravitational potential energy (U) of a two-particle system with masses m1 (9.3 kg) and m2 (7.3 kg) separated by a distance r (2.1 m), you can use the following formula:

U = (G * m1 * m2) / r

where G is the gravitational constant, approximately 6.674 × 10^(-11) N(m/kg)^2.

Now, we can plug in the values into the formula:

U = (6.674 × 10^(-11) N(m/kg)^2 * 9.3 kg * 7.3 kg) / 2.1 m

Next, we perform the calculations:

U ≈ (4.297 × 10^(-10) N*m^2/kg^2) / 2.1 m

U ≈ 2.046 × 10^(-10) N*m

So, the gravitational potential energy of the two-particle system is approximately 2.046 × 10^(-10) N*m.

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5. Two friends were studying kung fu and wanted to know which would give them the most kinetic energy that
could be transferred by their kicks: working on becoming faster (speed) or working on building more muscle
(mass). Using the graph below, explain which option would be best. (2 points)
Kinetic Energy increase
per velocity (speed) and mass increase
Z
1 2 3 4 5
67
Unitary increase
8
9
10
Velocity increase
Mass increase

Answers

We can see here that the option that would be best is: Kinetic Energy increase.

What is kinetic energy?

The energy that an object has as a result of motion is known as kinetic energy. It is calculated by multiplying an object's mass by the square of its velocity, divided by half.

Joules (J) are the metric unit for kinetic energy. Kinetic energy is calculated using the equation KE = 1/2mv2, where m is the object's mass and v is its speed.

Compared to the increase in kinetic energy per unit increase in mass, the increase in kinetic energy per unit increase in velocity is significantly greater.

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Levi keeps the wires so they are as far apart as possible and sets the current in wire 2 to 5 A. Now he varies the current in wire 1. Mira pays attention to the force per unit length on wire 1. What does she observe as a result

Answers

Answer:

The two wires exert the same force on each other.

Explanation:

Mira will observe that varying the current in wire 1 affects the force per unit length on wire 1.

When the wires are kept as far apart as possible and the current in wire 2 is set to a constant value of 5 A, varying the current in wire 1 will affect the force per unit length on wire 1.

According to Ampere's law, the magnetic field created by a current-carrying wire is directly proportional to the current passing through the wire. When the current in wire 1 is varied, it will create a magnetic field around wire 1.

If the current in wire 1 is increased, the magnetic field around wire 1 will also increase. As a result, the force per unit length on wire 1 will increase. Mira will observe a stronger force acting on wire 1 as the current in wire 1 is increased.

On the other hand, if the current in wire 1 is decreased, the magnetic field around wire 1 will weaken, leading to a decrease in the force per unit length on wire 1. Mira will observe a weaker force acting on wire 1 as the current in wire 1 is decreased.

Therefore, Mira will observe that varying the current in wire 1 affects the force per unit length on wire 1, with an increase in current leading to a stronger force and a decrease in current resulting in a weaker force.

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When Herschel counted stars in the sky along the Milky Way, he concluded that the Sun was close to the center of the Milky Way. Why was he wrong

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Herschel's assumption of uniform star distribution in the Milky Way was incorrect. Technological advancements allowed us to map the galaxy's structure and determine the Sun's position in an outer arm.

Herschel's conclusion that the Sun was close to the center of the Milky Way was based on the assumption that the stars in the Milky Way were uniformly distributed. However, this assumption turned out to be incorrect. Later studies, such as those by Harlow Shapley, demonstrated that the Milky Way is a barred spiral galaxy and that the Sun is actually located in one of its outer arms, known as the Orion Arm.

Additionally, Herschel's counting of stars was limited by the technology of his time, which did not allow him to see through the dust and gas that make up the Milky Way's disk. Today, with modern telescopes, we can observe stars and other objects in different wavelengths, allowing us to peer deeper into the galaxy and map its structure.

Herschel's conclusion about the location of the Sun in the Milky Way was based on limited information and a flawed assumption. Subsequent observations and technological advancements have since allowed us to better understand the structure of our galaxy, revealing that the Sun is located much farther away from the center than previously thought.

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The mass of one small ball is 1.50 g, and the mass of another is 870.0 g. If the center-to-center distance between these two balls is 10.0 cm, find the magnitude of the gravitational force that each exerts on the other.

Answers

Answer:

Approximately [tex]8.70 \times 10^{-14}\; {\rm N}[/tex], assuming that both balls are of uniform density.

Explanation:

The gravitational attraction between two spheres of uniform density is:

[tex]\begin{aligned}F &= \frac{G\, M\, m}{r^{2}}\end{aligned}[/tex],

Where:

[tex]G \approx 6.67 \times 10^{-11}\; {\rm m^{3} \cdot kg^{-1} \cdot s^{-2}}}[/tex] is the gravitational constant,[tex]M[/tex] and [tex]m[/tex] are the mass of the two spheres, and[tex]r[/tex] is the distance between the center of the two spheres.

Apply unit conversion and ensure that mass and distance are both measured in standard units:

[tex]\displaystyle m = 1.50\; {\rm g} \times \frac{1\; {\rm kg}}{10^{3}\; {\rm g}} = 1.50 \times 10^{-3}\; {\rm kg}[/tex].

[tex]\displaystyle M = 870.0\; {\rm g} \times \frac{1\; {\rm kg}}{10^{3}\; {\rm g}} = 0.8700\; {\rm kg}[/tex].

[tex]\displaystyle r = 10.0\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}}= 0.100\; {\rm m}[/tex].

Substitute these value into the equation and evaluate:

[tex]\begin{aligned}F &= \frac{G\, M\, m}{r^{2}} \\ &= \frac{(6.67 \times 10^{-11}\; {\rm m^{3}\cdot s^{-1}\cdot kg^{-2}})\, (0.8700\; {\rm kg})\, (1.50\times 10^{-3}\; {\rm kg})}{(0.100\; {\rm m})^{2}} \\ &= \frac{(6.67 \times 10^{-11})\, (0.8700)\, (1.50\times 10^{-3})}{(0.100)^{2}}\; {\rm kg\cdot m\cdot s^{-2}} \\ &= \frac{(6.67 \times 10^{-11})\, (0.8700)\, (1.50\times 10^{-3})}{(0.100)^{2}}\; {\rm N} \\ &\approx 8.70 \times 10^{-14}\; {\rm N}\end{aligned}[/tex].

(a) Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the electric field is 900 V/m. The room temperature mobility of electrons is 0.38 m2/V-s.

Answers

If the room temperature mobility of electrons is 0.38 m2/V-s,  the drift velocity of electrons in germanium at room temperature when the magnitude of the electric field is 900 V/m is 342 m/s.

The formula to calculate drift velocity (v_d) is:

v_d = μ * E

where μ is the mobility of electrons in the material and E is the magnitude of the electric field.

Given that the mobility of electrons in germanium at room temperature is 0.38 m²/V-s and the magnitude of the electric field is 900 V/m, we can calculate the drift velocity as:

v_d = 0.38 * 900 = 342 m/s

Therefore, the drift velocity of electrons in germanium at room temperature when the magnitude of the electric field is 900 V/m is 342 m/s.

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After a piece of copper wire from a hardware store is heated and returned to room temperature, it becomes softer. This is because:

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When a piece of copper wire from a hardware store is heated and then returned to room temperature, it becomes softer due to a process known as annealing. Annealing is a heat treatment that alters the physical and sometimes chemical properties of a material, making it more ductile and less hard.

During the heating process, the copper atoms gain energy, which allows them to move more freely within the material. This increased mobility leads to a redistribution of dislocations and a reorganization of the crystal lattice structure. When the wire is cooled down to room temperature, the atoms slowly return to their original positions, but with a more uniform and less stressed arrangement. This new arrangement results in a material with improved ductility and reduced hardness, making the copper wire softer.

In summary, heating a copper wire and allowing it to cool down to room temperature results in a process called annealing. This process redistributes dislocations and reorganizes the crystal lattice structure, ultimately making the material more ductile and less hard.

Consequently, the copper wire becomes softer, which can be useful for applications that require increased flexibility and reduced brittleness.

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What types of events have scientists so far been able to detect with gravitational wave observatories, such as LIGO

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Gravitational wave observatories, such as LIGO, have allowed scientists to detect a range of cosmic events that produce gravitational waves. These events include the collision of black holes, the merger of neutron stars, and even the vibrations produced by the formation of the universe shortly after the Big Bang.

Scientists have been able to detect several types of events with gravitational wave observatories, such as LIGO. By detecting these gravitational waves, scientists are able to gain a better understanding of the universe and the fundamental laws of physics that govern it. These events include:
1. Binary black hole mergers: When two black holes orbit each other and eventually merge, they produce gravitational waves. LIGO has detected multiple instances of these mergers.
2. Binary neutron star mergers: Similar to black hole mergers, when two neutron stars orbit each other and merge, they emit gravitational waves. LIGO and Virgo observatories detected a neutron star merger in 2017.
These detections have provided valuable insights into the astrophysics of black holes and neutron stars, as well as improved our understanding of the fundamental physics of gravitational waves.

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29. A flight to be conducted in VFR on top conditions at 12,500 ft MSL. What is the in-flight visibility and distance from clouds required for operation in Class E airspace during daylight hours

Answers

It is necessary to maintain a visibility of 5 statute miles and a distance of 1,000 feet below, 1,000 feet above, and 2,000 feet horizontally from clouds.

To answer your question regarding the in-flight visibility and distance from clouds required for operation in Class E airspace during daylight hours at 12,500 ft MSL under VFR on top conditions:

The in-flight visibility required for operating in Class E airspace at an altitude of 12,500 ft MSL during daylight hours is 5 statute miles. The distance from clouds required for this operation is to maintain at least 1,000 feet below, 1,000 feet above, and 2,000 feet horizontally from clouds.

In summary, when conducting a flight in VFR on top conditions at 12,500 ft MSL in Class E airspace during daylight hours, you must maintain a visibility of 5 statute miles and a distance of 1,000 feet below, 1,000 feet above, and 2,000 feet horizontally from clouds.

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Planet Tehar has a radius of 10,000 km. An object dropped near Tehar's surface falls with an acceleration of 36 m/s2. What is the strength of Tehar's gravitational field at a height of 50,000 km above its surface

Answers

The strength of Tehar's gravitational field at a height of 50,000 km above its surface is approximately 1.44 m/s^2.

To understand why, we need to use the formula for gravitational field strength: g = GM/r^2, where g is the gravitational field strength, M is the mass of the planet, and r is the distance from the center of the planet.

Since the planet has a radius of 10,000 km, its diameter is 20,000 km. Therefore, its total distance from the center to a point 50,000 km above the surface is 60,000 km.

Using the formula, we can calculate the gravitational field strength as follows:

g = GM/r^2

g = (G * M) / (60,000 km)^2

g = (6.67 x 10^-11 Nm^2/kg^2) * (5.97 x 10^24 kg) / (60,000,000 m)^2

g ≈ 1.44 m/s^2

Therefore, the strength of Tehar's gravitational field at a height of 50,000 km above its surface is approximately 1.44 m/s^2.

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A refrigerator has a coefficient of performance of 2.25, runs on an input of 95.0 W of electrical power, and keeps its inside compartment at 5.00 degrees (Celsius).

If you put a dozen 1.00 L plastic bottles of water at 31.0 degrees (Celsius) this refrigerator, how long will it take for them to be cooled down to 5.00 degrees (Celsius)? (Ignore any heat that leaves the plastic.)

Answers

The bottles of water will drop to a temperature of 5.00 degrees Celsius in around 5.5 hours.

The following formula must be used to determine how long it will take to cool the water bottles:

Q = mcΔT

Where T is the temperature difference between the initial and final temperatures, m is the mass of the water, Q is the quantity of heat transmitted, and c is the heat capacity of water.

The labour performed by the refrigerator, which is determined by: The amount of heat transported is equal to this amount.

W = QH - QC

Where QC is the heat emitted to the room and QH is the heat the refrigerator releases from its internal compartment.

We can write: Using the coefficient of performance.

QH / COP = W

Because we are aware that the electrical input is 95.0 W, we may write:

W = 95.0 W x t

t is the time expressed in hours.

The result of adding these equations is:

COP = 95.0 W x t for QH

Rearranging gives us:

QH=95.0 WxtxCOP

Now, we can apply this equation to determine how much heat is transmitted from the water bottles to the interior compartment:

QH = mcΔT

To solve for t, we obtain:

t = (95.0 W x COP) / (mcT)

When we enter the values, we obtain:

(12 kg times 4184 J/kg°C over a range of 31.0 and 5.00°C)) / (95.0 W x 2.25)

5.5 hours is t.

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The flutes on a twist drill serve which one of the following functions: (a) adds rigidity to the drill, (b) improves hole size accuracy, (c) lubricates the cutting edges, (d) provides passageways for extraction of chips, or (e) strengthens the drill

Answers

The flutes on a twist drill serve multiple functions, including (b) improving hole size accuracy by helping to maintain a consistent diameter throughout the drilling process, (d) providing passageways for extraction of chips to prevent clogging and overheating, and (c) to some extent, lubricating the cutting edges to reduce friction and heat buildup.

However, they do not add rigidity or strengthen the drill.


The flutes on a twist drill serve the function of (d) providing passageways for extraction of chips. This improves the drilling process by efficiently removing debris and allowing for smooth drilling operation.

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an elementary student of mass m=34 kg is swinging on a swing. the length from the top of the swing set to the seat is L=4.7 m. the child is attempting to swing all the way around in a full circle.

-what is the minimum speed in meters per second the child must be moving with at the top of the path in order to make a full circle?

-assuming the child is traveling at the speed found in part a what is their apparent weight in newtons at the top of their path? (at the top, the child is upside-down)

-if the velocity at the very top is the same velocity from part a what is the childs apparent weight in newtons at the very bottoms of the path?

Answers

An elementary student of mass m=34 kg is swinging on a swing. the length from the top of the swing set to the seat is L=4.7 m.

a) The minimum speed the child must be moving at the top of the path in order to make a full circle is  9.14 m/s.

b) The apparent weight of the child at the top of the path is 1005.52 N.

c) The apparent weight of the child at the bottom of the path is 333.54 N.

We can solve this problem using the conservation of energy and the centripetal force equation.

(a) At the top of the swing, the child is momentarily at rest, so all of the kinetic energy has been converted to potential energy. All of the potential energy has been transformed into kinetic energy at the swing's bottom.

The minimum speed required at the top of the path to make a full circle is the speed at which the centripetal force required to keep the child moving in a circle is equal to the gravitational force pulling the child downward.

Setting the centripetal force and gravitational force equal, we have:

[tex]mv^2 / L[/tex]= mg

where m is the mass of the child, v is the speed of the child at the top of the path, L is the length of the swing, and g is the acceleration due to gravity.

Solving for v, we get:

v = [tex]\sqrt{(gL) }[/tex]= [tex]\sqrt{(9.81 m/s^2 * 4.7 m) }[/tex]≈ 9.14 m/s

Therefore, the minimum speed the child must be moving at the top of the path in order to make a full circle is approximately 9.14 m/s.

(b) At the top of the path, the child is momentarily upside-down, so the apparent weight is the sum of the gravitational force and the centripetal force required to keep the child moving in a circle.

The gravitational force on the child is:

[tex]mg = 34 kg * 9.81 m/s^2 = 333.54 N[/tex]

To keep the kid moving in a circle, you need to apply the following centripetal force:

[tex]mv^2 / L = 34 kg * (9.14 m/s)^2 / 4.7 m[/tex] ≈ [tex]671.98 N[/tex]

Therefore, the apparent weight of the child at the top of the path is approximately 1005.52 N (333.54 N + 671.98 N).

(c) At the bottom of the path, the child is moving at the same speed as at the top, so the centripetal force required to keep the child moving in a circle is the same. However, at the bottom of the path, the gravitational force is the only force acting on the child.

The gravitational force on the child is the same as in part (b):

mg = [tex]34 kg * 9.81 m/s^2 = 333.54 N[/tex]

The centripetal force required to keep the child moving in a circle is:

[tex]mv^2 / L = 34 kg * (9.14 m/s)^2 / 4.7 m[/tex] ≈ [tex]671.98 N[/tex]

Therefore, the apparent weight of the child at the bottom of the path is approximately 333.54 N (equal to the gravitational force).

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When operated on a household 110.0 V line, typical hair dryers draw about 1450 W of power. The current can be modeled as a long, straight wire in the handle. During use, the current is about 2.65 cm from the user's hand. What is the current in the hair dryer?

Answers

The current in the hair dryer is approximately 13.18 A.

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume.

To find the current in the hair dryer when it operates on a 110.0 V line and draws 1450 W of power, we can use the formula:

Power (P) = Voltage (V) × Current (I)

We are given the power (1450 W) and the voltage (110.0 V), so we can solve for the current (I) as follows:

1. Rearrange the formula to solve for I:

I = P / V
2. Substitute the given values:

I = 1450 W / 110.0 V
3. Calculate the result:

I ≈ 13.18 A

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What happens to the number of waves when you change the color from green to
violet?

A) increase
B) decrease
C) remain the same
D) there are not any waves present

Answers

When changing the color from green to violet, the number of waves will remain the same.

The color of light is determined by its wavelength, which is the distance between two consecutive peaks or troughs of the wave. Red light has the longest wavelength, while violet light has the shortest wavelength. However, regardless of the color of the light, all electromagnetic waves travel at the speed of light and have the same frequency.

Frequency refers to the number of waves that pass a given point in a second, and is measured in hertz (Hz). Therefore, when the color of light is changed from green to violet, the wavelength will become shorter, but the frequency (and thus the number of waves) will remain the same.

So, the answer is C) remain the same.

If a rod of metal is pulled through a tapered hole smaller than the diameter of the rod, the strength of the metal in the rod increases. This is because:

Answers

When a metal rod is pulled through a tapered hole smaller than its diameter, the strength of the metal increases due to work hardening which is also known as strain hardening.


As the metal rod is forced through the tapered hole, it undergoes plastic deformation. This means that the metal's shape changes permanently without breaking. During this plastic deformation, the metal's crystal structure becomes more disordered, causing an increase in dislocation density which is the number of dislocations per unit volume.

The increase in dislocation density hinders the movement of dislocations in the metal, making it more resistant to further deformation. This increased resistance to deformation leads to an increase in the strength of the metal in the rod, a phenomenon known as work hardening or strain hardening.

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If a boat and its riders have a mass of 800 kgkg and the boat drifts in at 1.6 m/sm/s how much work does Sam do to stop it

Answers

Sam doesn't do any work to stop the boat. This is because the force needed to stop the boat is 0 N, as the boat is not accelerating.


To calculate the work that Sam does to stop the boat, we need to use the formula:
work = force x distance

Using Newton's Second Law of Motion, which states that force equals mass times acceleration (F = ma). Since the boat is drifting in at a constant speed, its acceleration is 0 m/s^2. Therefore, the force needed to stop the boat is also 0 N.
Next, we need to find the distance over which Sam stops the boat. We don't have this information in the question, so we'll assume that Sam stops the boat over a distance of 10 meters. This distance is just an estimate and may not be accurate.

Using the formula for work, we can now calculate the amount of work that Sam does to stop the boat:
work = force x distance
work = 0 N x 10 m
work = 0 J

The answer to the question is 0 J, which means that Sam doesn't do any work to stop the boat. This is because the force needed to stop the boat is 0 N, as the boat is not accelerating. Therefore, Sam doesn't need to exert any force to stop the boat, and hence doesn't do any work.

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Determine the average emissivity of the surface and the rate of radiation emission from the surface, in W/m2.

Answers

The average emissivity of the surface and rate of radiation emission in W/m2 can be calculated using relevant formulas.

The emissivity of a surface is a measure of its ability to emit thermal radiation.

To determine the average emissivity of a surface, the ratio of the actual radiation emitted by the surface to that emitted by a blackbody at the same temperature must be calculated.

The rate of radiation emission from the surface can be determined by multiplying the Stefan-Boltzmann constant by the emissivity of the surface and the fourth power of its temperature.

This will give the rate of energy emitted per unit area of the surface. The resulting value is expressed in watts per square meter (W/m2).

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what initial speed should a particle be given if it is to have a final speed when it is very far from the earth

Answers

To determine the initial speed of a particle required to have a final speed when it is far from the earth, several factors come into play. The gravitational force exerted by the earth on the particle will determine its acceleration. The initial speed must be such that the particle can overcome the gravitational pull of the earth and reach the desired final speed.

The formula that can be used to calculate the initial speed required is:

v1 = √(2GM/R + v2^2)

Where v1 is the initial speed, v2 is the final speed, G is the gravitational constant, M is the mass of the earth, and R is the distance of the particle from the earth.

So, the initial speed required would depend on the final speed and the distance of the particle from the earth. For instance, if the particle is to have a final speed of 10 km/s and is very far from the earth, say 10,000 km away, the initial speed required would be approximately 11.18 km/s.

In conclusion, the initial speed required for a particle to have a final speed when it is far from the earth depends on various factors such as the final speed, the distance of the particle from the earth, and the gravitational force of the earth on the particle.

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Assume that a group of explorers traveled to the Orion Nebula, a star-forming cloud at a distance of 1,300 light-years, using revolutionary technology that allowed them to travel at a speed very close to the speed of light. Observers back on Earth would say it took them __________ to get there, but the travelers would say it took them __________ to get there.

Answers

Observers back on Earth would say it took them 1,300 years to get there, but the travelers would say it took them much less time due to time dilation caused by their high speed.

What is time dilation?

Time dilation is a difference in the elapsed time measured by two observers, caused by a relative velocity between them or a difference in gravitational potential. It is a prediction of the theory of relativity.

What is speed?

Speed is the measure of how fast an object is moving, calculated as the distance traveled per unit of time, without regard to direction or displacement. It is measured in meters per second (m/s).

According to the given information:

Observers back on Earth would say it took them 1,300 years to get there, but the travelers would say it took them much less time due to time dilation caused by their high speed. Time dilation means that time passes slower for objects in motion than for stationary objects. As a result, the travelers would experience time differently and their journey would seem much shorter to them than it would to observers on Earth. However, the exact amount of time the travelers experience would depend on the speed they were traveling at.

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A metal block has a density of 2500 kg per cubic meter and a volume of 2 cubic meters. What is the block's mass

Answers

A metal block has a density of 2500 kg per cubic meter and a volume of 2 cubic meters. The block's mass is 5000 kg.

The mass of the metal block can be calculated using the formula:

mass = density x volume

where density is measured in kilograms per cubic meter (kg/m³) and volume is measured in cubic meters (m³).

In this case, the density of the metal block is 2500 kg/m³ and the volume is 2 m³. Substituting these values into the formula, we get:

mass = density x volume

mass = 2500 kg/m³ x 2 m³

mass = 5000 kg

Therefore, the mass of the metal block is 5000 kg.

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Experiments allow physicists today to reproduce (on very small scales) energy and temperature conditions thought to have prevailed in the early universe as far back in time as about __________.

Answers

Experiments allow physicists today to reproduce (on very small scales) energy and temperature conditions thought to have prevailed in the early universe as far back in time as about one trillionth of a second after the Big Bang.

The study of the early universe is known as cosmology, and physicists use a variety of tools to probe the conditions that existed during its formation. One of the most important of these tools is the Large Hadron Collider (LHC) at CERN, which is capable of producing particle collisions at energies that were last seen in the universe just after the Big Bang. By studying the behavior of particles in these collisions, physicists hope to gain insights into the fundamental forces and particles that govern the universe at its most basic level. Through these experiments, physicists can test theories about the early universe and better understand the nature of the cosmos.

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