10 points
True or False? Charging by friction happens when you rub two objects
together and PROTONS move from one object to the other.

Answers

Answer 1

Answer:

true

Explanation:

yes because I said so


Related Questions

A high-speed flywheel in a motor is spinning at 450 rpm (revolutions per minute) when a power failure suddenly occurs. The flywheel has mass 45.0 kg and diameter 74.0 cm . The power is off for 25.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 180 complete revolutions. A. At what rate is the flywheel spinning when the power comes back on? (rad/s)B. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on? (s)C. How many revolutions would the wheel have made during this time?

Answers

Answer:

a) final angular speed w2 = 414rpm

b) time to stop t = 312.5s

c) angular distance from start to stop s = s = 1171.88 rev

Explanation:

Average angular speed wa = (w1+w2)/2 = Angular distance/time = s/t

Given;

w1 = 450 rpm = 450/60 rev/s = 7.5 rev/s

s = 180 rev

t = 25s

a)

s/t = (w1+w2)/2

w2 = 2s/t - w1

Substituting the values;

w2 = 2×180/25 - 7.5 = 6.9 rev/s

w2 = 6.9×60rpm = 414rpm

b)

Angular deceleration a = (w1-w2)/t = (7.5-6.9)/25

a = 0.024 rev/s^2

time t for w2 = 0

t = (w1-w2)/a = (7.5-0)/0.024

t = 312.5 s

c)

Angular distance from start to stop s;

s = w1t - 0.5at^2

s = 7.5×312.5s - 0.5×0.024×312.5^2

s = 1171.88 rev

A chain lying on the ground is 11 meters long and its mass is 95 kilograms. The chain is threaded through a pulley, which is fixed to the ground, and pulled directly up so that it forms the shape of an L. How much work is required to raise one end of the chain to a height of 7 meters?

Answers

Answer: 296.1 J or 6.98 kJ

Explanation:

Given

Length of chain, l = 11 m

Mass of chain, m = 95 kg

It is worthy of note that chain has two ends. So this depends which of the two ends should end up at

the given height of 7 m.

Scenario 1. If it is the leading end:

Weight of chain raised

W = mg/l

W = 9.8×95/11

W = 84.6 N

Height raised = ½ m (which is the height of the centre of gravity of the raised portion).

Work done = 84.6 × 1/2 * 7 = 296.1 Joules

Scenario 2. If it is the trailing end:

Weight of chain raised = 9.8×95 = 931 N.

Height raised (average) = 12.5 m (5.5 m for the chain midpoint + 7 m off the ground)

Work done = 931 × 7.5 = 6.98 kJoules

Thus, the work required is either of 296.1 J or 6.98 kJ

The work required to lift one end of a 95 kg chain to 7 meters height is 3261.85 J.

To calculate the work required to raise one end of the chain to a height of 7 meters, we consider the work done against the force of gravity. The chain has a mass of 95 kilograms, and we need to move it through a vertical distance of 7 meters.

Work is defined as the product of force and the distance over which the force is applied. In this case, the force is the weight of the chain, which is the product of its mass and the acceleration due to gravity (g = 9.81 m/s²). However, the chain gets lighter as you pull it up, since a smaller length of chain is left hanging at every moment. To solve this, we find the average force, which would be the weight of half the chain, as only half of it will be lifted through the full height on average.

The chain's total weight is mass multiplied by the acceleration due to gravity so we have 95 kg × 9.81 m/s².

The weight of half the chain is therefore

0.5 × 95 kg × 9.81 m/s²

The work done to lift half the mass through the full height is the average force times the height (7 m).

The calculation will then be:

{W} = 0.5 ×95 kg × 9.81 m/s² ×7 m

W= 3261.85 J

So, we multiply these values to find the total work done in lifting one end of the chain to a height of 7 meters.

En un planeta distante,un objeto en caida libre cambia su velocidad de 10m/s a 90m/s en 2 segundos ¿Cual es la aceleración gravitacional de dicho planeta?

Answers

Answer:

a = 40 m / s²

Explanation:

For this exercise we can use the vertical launch equations to which these are valid in any system with inertial

      v = v₀ + a t

 

      a = (v-v₀) / t

We calculate

     a = (90 - 10) / 2

     a = 80/2

     a = 40 m / s²

This is the value of the planet's gravity acceleration

Two skaters, one with mass 65 kg and the other with mass 40 kg, stand on an ice rink holding a pole of length 10 m and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the 40 kg skater move

Answers

Final answer:

The 40 kg skater moves approximately 6.19 meters towards the 65 kg skater when they pull themselves along a 10 m pole on ice, utilizing the conservation of momentum and considering the center of mass of the system.

Explanation:

To solve how far the 40 kg skater moves when two skaters pull themselves along a pole on ice, we use the principle of conservation of momentum. This principle states that in the absence of external forces, the total momentum of a system remains constant. Since both skaters start from rest, their initial total momentum is zero. As they pull towards each other, their momenta are equal in magnitude and opposite in direction, keeping the total momentum of the system at zero.

Let's denote the distance the 65 kg skater moves as d65 and the distance the 40 kg skater moves as d40. Because the pole has a length of 10 m, these two distances must add up to 10 m, so d65 + d40 = 10. To find out how far each skater moves, we look at their center of mass. The system's center of mass doesn't change because there are no external forces acting on the system.

The position of the center of mass xCM can be found using the formula xCM = (m1 * x1 + m2 * x2) / (m1 + m2), where m1 and m2 are the masses of the skaters, and x1 and x2 are their respective positions. Considering the system initially (before movement), the center of mass is at the midpoint of the pole as the skaters are at the ends. Therefore, xCM = 5 m.

Because the total mass of the system is 105 kg and the center of mass is 5 m from the starting point of either skater, we find that the 40 kg skater moves closer to the center of mass than the 65 kg skater to maintain the center of mass position. By directly calculating it, d40 = 10 * (65 / 105) = 6.19 m. Thus, the 40 kg skater moves approximately 6.19 meters towards the 65 kg skater.

1) what is a calorie?

Answers

i read it’s a unit of energy

. Conservation along the horizontal using a bicycle wheel: Stand on the platform holding a bicycle wheel with its axis horizontal with your arms fully extended. While holding the platform stationary, let a colleague spin the wheel. Release the platform then turn the wheel’s spin axis "up." Observe the platform turn with a ‘down’ spin vector, ω. Return the wheel to the starting position and the platform should stop. Turn the wheel’s spin axis "down." Return the wheel to the starting position. Describe the platform’s reaction in terms of conservation of vertical angular momentum: Use L1 as the angular momentum of the person and the platform, and L2 as the wheel’s angular momentum. Use vector diagrams to show how the vertical angular momentum of the system, Ltotal = L1 + L2, starts at zero and remains zero throughout the exercise. That is, show that L1i + L2i = L1f + L2f.

Answers

Answer:

w = I₂ / (I₁ -I₂) w₀ ,    L₂ = 2 L₁

Explanation:

This is an angular momentum exercise,

      L = I w

where bold indicates vectors

We must define the system as formed by the bicycle wheel, the platform, we create a reference system with the positive sign up

At the initial moment the wheel is turning and the platform is without rotation

The initial angle moment is

                 Lo = L₂ = I₂ w₀

L₁ is the angular momentum of the platform and L₂ is the angular momentum of the wheel.

In the Final moment, when the wheel was turned,

                [tex]L_{f}[/tex] = L₁ - L₂

                L_{f} = (I₁ - I₂) w

the negative sign of the angular momentum of the wheel is because it is going downwards since the two go with the same angular velocity

as all the force are internal, and there is no friction the angular momentum is conserved,

             L₀ =L_{f}

             I₂ w₀ = (I₁ -I₂) w

             w = I₂ / (I₁ -I₂) w₀

we can see that the system will complete more slowly

 we can also equalize the angular cognition equations

                 L₀ = Lf

                 L₁ = L₂-L₁

                 L₂ = 2 L₁

In this part we can see that the change in the angular momentum of the platform is twice the change in the angular momentum of the wheel.

 

This experiment with a rotating bicycle wheel demonstrates the conservation of angular momentum. Rotating the wheel changes the platform's rotational direction to maintain a balanced angular momentum. Returning the wheel to its initial position stops the platform's rotation, preserving the system's angular momentum at zero.

In this experiment, we examine the principles of conservation of angular momentum using a rotating bicycle wheel. When you and a platform are stationary, and a colleague spins the bicycle wheel, the system's initial angular momentum is zero.

Here’s a step-by-step explanation:

Initially, both you (holding the wheel) and the platform have zero angular momentum:[tex]L_1_i + L_2_i = 0 + 0 = 0.[/tex]Your colleague spins the wheel, creating an angular momentum [tex]L_2.[/tex]If you turn the wheel's spin axis up, it creates a torque, causing you and the platform to rotate in the opposite direction. This rotation ensures that the sum of angular momenta remains zero, thus: [tex]L_1 + L_2 = 0.[/tex]Returning the wheel to the starting horizontal position stops the platform’s spin, returning the total angular momentum of the system back to zero.When you turn the wheel’s spin axis down, the platform reacts oppositely, maintaining the conservation of vertical angular momentum.After returning the wheel to horizontal, the platform stops again, demonstrating that the vertical angular momentum of the system remains balanced: [tex]L_1_f + L_2_f = 0.[/tex]

This experiment vividly shows how changing the direction of angular momentum through torque affects rotational motion while conserving the overall angular momentum of the system.

Each of the following would cause an increase in blood pressure except __________.

Answers

Answer:

an inhibitor of angiotensin II

Explanation:

Angiotensin, specifically angiotensin II binds to many receptors in the body to affect several systems. It can normally increase blood pressure by constricting the blood vessels but with the introduction of an inhibitor, it wouldn't bring about an increase in blood pressure.

Which element is a metal? A.lithium B.Nitrogen C.sulfur D.chlorine

Answers

Answer:

the answer is b

Explanation:

hope it helps ...

Lithium (Li) is a metal located in Group 1 of the periodic table and is known as an alkali metal. It stands out for its low density and capacity to conduct electricity, along with exhibiting the typical properties of metals.

The element lithium (Li) is a metal, and notably, it is the first metal in the periodic table located in Group 1. It has the lowest atomic number (3) among the metals. The defining characteristics of metals include their ability to conduct electricity, possess malleability, and often exhibit a shiny luster. Lithium, being part of the alkali metals category, shares these properties. Furthermore, lithium is the least dense metal and has a unique electronic structure where the third electron occupies an outer shell, differentiating it from nonmetals.

Lithium is used in various applications including batteries and the aerospace industry due to its low density and high reactivity. The chemically similar elements to lithium include other alkali metals such as sodium and potassium, which also share similar chemical properties.

In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how the incident particles scatter. The electron in the ground state of a hydrogen atom is found to be momentarily at a distance a0/2 from the nucleus in 1 300 of the observations. In this set of trials, how many times is the atomic electron observed at a distance 2a0 from the nucleus?

Answers

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

[tex]p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }[/tex]

[tex]p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }[/tex]

[tex]p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }[/tex]

[tex]N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}[/tex]

[tex]N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}[/tex]

[tex]N = 1300(16) e^{-3}[/tex]

N = 1035.57

N = 1036 times

Why is a shadow formed

Answers

Answer:

Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.

Explanation:

Answer:

shadows are made by blocking light

Which of the following is NOT a result of supernova explosions? The neutron core is completely destroyed. Any planets within a few dozen light-years receive a life-threatening dose of radiation. Many of the elements the star fused during its life are blasted out into space. Many new elements, including some heavier than iron, are fused during the supernova explosion.

Answers

Answer:

 The neutron core is completely destroyed

Explanation:

 A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.

Final answer:

The incorrect statement is that a supernova explosion results in the complete destruction of the neutron core; often, the core becomes a neutron star. Supernovae can be life-threatening to nearby planets, but the actual danger zone is closer than a few dozen light-years. They are also responsible for creating new elements heavier than iron.

Explanation:

A supernova explosion is a cataclysmic event at the end of a massive star's life cycle. During a supernova, a number of phenomena occur, including the ejection of elements the star fused during its lifetime, and the creation of new elements heavier than iron. One of the statements provided as a result of supernova explosions is incorrect: 'The neutron core is completely destroyed'. In many cases, the core does not get completely destroyed; instead, it often collapses into a densely packed neutron star. Another result is a potential threat to life on planets within a few dozen light-years due to a life-threatening dose of radiation. It is important to note, though, that planets would need to be quite close to the supernova, much closer than a few dozen light-years, to receive such a dose. Finally, supernovae also contribute to the formation of new elements, with the violent energies enabling the fusion of atoms into elements heavier than iron.

List down different layers of the sun. Rank these layers based on their distance from the sun’s center

Answers

Answer:

Core

Radiative zone

Convective zone

Photosphere

Chromosphere

Transient region

Corona

Ranks of layers based on their distance from the sun’s center

1st-corona

2nd-Transient region

3rd-chromosphere

4th-Photosphere

5th-convective zone

6th-radiative zone

7th-core

Use the following information for determining sound intensity. The number of decibels β of a sound with an intensity of I watts per square meter is given by β = 10 log(I/I0), where I0 is an intensity of 10−12 watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. Find the number of decibels β of the sound.

Answers

Answer:

120 decibels

Explanation:

we know sound in decibels is given by.

[tex]\beta =10log(\frac{I_{} }{I_{0} } )[/tex]

Where I is intensity of the sound and [tex]I_{0}=10^-12W/m^{2 }[/tex] is reference intensity.

substituting All of this in our decibel formula with I =1W/[tex]m^2[/tex]

gives us 120 decibels.

Final answer:

The sound intensity level of a sound is measured in decibels (dB) using a logarithmic scale that uses a reference intensity of 10^-12 W/m², which is the threshold of human hearing. The decibel level of a sound is determined by the ratio of its intensity to this reference intensity, with each increment of 10 dB corresponding to a tenfold increase in intensity.

Explanation:

The sound intensity level ß, measured in decibels (dB), provides a logarithmic measure of sound intensity I in watts per meter squared (W/m²), with the formula

ß = 10 log(I/I0)

where I0 = 10⁻¹² W/m² is the reference intensity, corresponding to the threshold of human hearing. The decibel level of a sound is thus derived from the ratio of its intensity to this reference intensity. A sound with the same intensity as the reference intensity (I = I0) has a decibel level of 0 dB, because log 10(1) = 0.

Interpreting decibels involves understanding that because the formula uses a logarithm, each increment of 10 dB corresponds to a tenfold increase in intensity. Therefore, a sound with an intensity of 10^-11 W/m² would be 10 dB, a sound with an intensity of 10⁻¹⁰ W/m² would be 20 dB, and so on. The decibel scale helps to compress the vast range of sound intensities that human ears can perceive into a more manageable range of numbers.

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A main difference between gravitational and electric forces is that electrical forces

Answers

Final answer:

The primary difference between gravitational and electric forces is that electric forces can be both attractive and repulsive, whereas gravitational forces are always attractive. Electric forces dominate on small scales, such as within atoms, while gravitational forces are significant on large scales, like between planets.

Explanation:

Main Differences Between Gravitational and Electric Forces:

A main difference between gravitational and electric forces is that electric forces can be either attractive or repulsive, while gravitational forces are always attractive. Gravitational force is proportional to the masses of the interacting objects and always pulls them together, which is important in interactions between large objects like planets and stars. Conversely, electric force is proportional to the charges of interacting objects and plays a crucial role at the microscopic level within atoms and molecules, often keeping them together. It is much stronger than gravity in most systems where both appear.

Another significant difference is that on a small scale, like that of electrons or protons, the electric force is dominant and significantly exceeds the gravitational force. In contrast, on a large scale, gravitational force typically predominates since most objects tend to be electrically neutral; thus, the net Coulomb forces cancel out.

To summarize, gravitational forces are crucial at cosmological scales, handling the structure of galaxies and the dynamics of celestial bodies, whereas electric forces are on a much smaller scale but are instrumental in the binding of atoms, the functioning of everyday materials, and biological processes.

*9. A small loop of area 6.8 mm2 is placed inside a long solenoid that has 854 turns/cm and carries a sinusoidally varying current i of amplitude 1.28 A and angular frequency 212 rad/s . The central axes of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop

Answers

Final answer:

The amplitude of the induced emf can be calculated by first finding the magnetic field within the solenoid using the formula B = µonI, calculating the magnetic flux through the coil, and then applying Faraday's law to find the time derivative of the flux. This result is then multiplied by the number of turns in the coil.

Explanation:

The amplitude of the electromagnetic force (emf) induced in a coil located inside a solenoid can be calculated using Faraday's law. The first step in solving this problem involves obtaining the magnetic field B of the solenoid by using the formula B = µonI, where µo is the vacuum permeability, n represents the number of turns per unit length of the solenoid, and I is the amplitude of the current. Here, n can be obtained by converting the given number of turns per cm into turns per meter. This magnetic field B is then multiplied by the area of the coil to obtain the magnetic flux Φ through the coil.

Next, using Faraday's law, we take the time derivative of the obtained magnetic flux. Since the only variable in time here is the current I, we apply the derivative to it, with the rest of the quantities being treated as constant. Lastly, we multiply the result by the number of turns in the coil to determine the induced emf. Remember that the magnitude of the emf is the peak or 'amplitude' of the sinusoidal function associated with the time-varying current I.


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A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m.What is the magnitude ofthe normal force exerted on the car by the walls of the circle at pointA?

Answers

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, [tex]N - mg = F_c[/tex]

[tex]N-mg=\frac{mv^2}{r}[/tex]

Now substitute the given values, we get

[tex]N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}[/tex]

[tex]N=15.68+46.08[/tex]

N = 61.76  N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

Final answer:

The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In this case, the magnitude of the normal force is equal to the weight of the car, which is the product of its mass and the acceleration due to gravity.

Explanation:

The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In a vertical circle, the net external force equals the necessary centripetal force. The only two external forces acting on the car are its weight and the normal force of the road. Since the car is not leaving the surface and the net vertical force must be zero, the vertical components of the two external forces must balance each other.



The vertical component of the normal force is equal to the car's weight, which is mg. The weight can be calculated by multiplying the mass of the car by the acceleration due to gravity. In this case, the weight is 1.60 kg multiplied by 9.8 m/s². Therefore, the magnitude of the normal force exerted on the car at point A is also 1.60 kg multiplied by 9.8 m/s².

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Suppose that 3.00 g of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the Earth's North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth?

Answers

Answer:

Explanation:

Atomic mass of hydrogen is 1 . so 1 g of hydrogen will have 6.02 x 10²³ atoms

3 g of hydrogen will have 3 x 6.02 x 10²³ atoms . This will give 3 x 6.02 x 10²³  protons and same number of electrons

So number of protons at north pole = 3 x 6.02 x 10²³

charge on these protons = 1.6 x 10⁻¹⁹ x  3 x 6.02 x 10²³ C

= 28.9 x 10⁴ C  

similarly total charge on electrons at south pole = 28.9 x 10⁴ C

distance between them = diameter of the earth = 2 x 6356 x 10³ m

= 12712 x 10³ m

Attractive force between these charges

= k q₁q₂ / r² , q₁ ,q₂ are charges and r is distance between charges.

= 9 x 10⁹ x (28.9 x 10⁴)² / (12712 x 10³ )²

= 4.6517 x 10⁻⁵ x 10¹¹

= 4.6517 x 10⁶ N

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means that unrealistically large currents are needed to produce noticeable torques. Suppose a 20-cm-diameter loop of wire is oriented for maximum torque in the earth's field.What current would it need to carry in order to experience a very modest 1.0?

Answers

Answer:

636.619772368 A

Explanation:

[tex]\tau[/tex] = Torque = [tex]1\times 10^{-3}\ N/m[/tex]

B = Magnetic field of Earth = [tex]5\times 10^{-5}\ T[/tex]

A = Area

d = Diameter = 20 cm

Current is given by

[tex]I=\dfrac{\tau}{BA}\\\Rightarrow I=\dfrac{1\times 10^{-3}}{5\times 10^{-5}\times \dfrac{\pi}{4}\times 0.2^2}\\\Rightarrow I=636.619772368\ A[/tex]

The current is 636.619772368 A

The energy required to initiate an exergonic reaction is called

Answers

Answer:

energy of activation is the correct answer.

Explanation:

All the chemical reactions including an exothermic reaction require activation energy to initiate a reaction.Activation energy is the minimum amount of energy required to activate the molecules to the situation that they can experience a chemical change. If the activation energy is lower the faster the reaction occurs.Joules, kilocalories/mol, kilojoules per mole are the units for activation energy.
Final answer:

The activation energy is the energy needed to initiate an exergonic reaction. It is required to kickstart the process, despite the reaction being one that eventually releases energy.

Explanation:

The energy required to initiate an exergonic reaction is called the activation energy. This is the minimum amount of energy required for a chemical reaction to occur. In an exergonic reaction, the activation energy is critical even though these reactions release energy overall, because it's needed to start the process. To envision this, think of pushing a boulder off a hill; the initial push (activation energy) is required for the action (the exergonic reaction) to begin, after which gravity (the reaction's natural tendency) takes over and the boulder rolls down (energy is released).

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Shannon is playing tennis. She swings her racket to hit the tennis ball and sends the ball flying in the same direction.

Answers

Answer:

she used force to swing her ball all way back

Explanation:

Final answer:

Shannon swinging her racket to hit a tennis ball illustrates key physics concepts, including momentum and energy transfer, crucial for propelling the ball forward. The energy transferred from the racket to the ball depends on the racket's motion, enhancing the ball's speed when the racket moves toward it. Physics principles, such as Newtonian mechanics, are essential for understanding and improving tennis skills.

Explanation:

When Shannon swings her racket to hit a tennis ball, sending it flying in the same direction, several principles of physics come into play, particularly related to momentum and energy transfer. First, when the racket contacts the ball, momentum is transferred from the racket to the ball, propelling it forward. According to Newton's third law, for every action, there is an equal and opposite reaction. Therefore, as the racket exerts force on the ball, the ball exerts an equal and opposite force on the racket, though the effects are more noticeable on the ball due to its much smaller mass.

Energy transfer during the collision between the racket and the ball is significant. A moving racket transfers kinetic energy to the stationary ball, which then moves with a velocity dependent on the imparted energy. If the racket is moving towards the ball, the speed of the ball post-collision is greatly increased. This results from the addition of the racket's velocity to that imparted to the ball, showcasing an efficient energy transfer.

Understanding the dynamics of hitting a tennis ball, including the effects of the racket's speed, angle of swing, and point of impact, requires knowledge of Newtonian mechanics and is essential for improving one's tennis playing skills. Moreover, accounting for the arc of the ball and its spin are crucial for predicting how the ball will move, enhancing the player's ability to plan their shots strategically.

The angular speed of an automobile engine is increased at a constant rate from 1150 rev/min to 2680 rev/min in 14.2 s. (a) What is its angular acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this 14.2 s interval?

Answers

Answer:

(A) [tex]\alpha =11.277rad/sec^2[/tex]

(B) [tex]\Theta =2846.30rad[/tex]

Explanation:

We have given initial angular speed

[tex]\omega _i=1150rpm=\frac{2\times 3.14\times 1150}{60}=120.366rad/sec[/tex]

[tex]\omega _f=2680rpm=\frac{2\times 3.14\times 2680}{60}=280.506rad/sec[/tex]

Time t = 14.2 sec

(a) From first equation of motion

[tex]\omega _f=\omega _i+\alpha t[/tex]

[tex]280.506=120.366+\alpha \times 14.2[/tex]

[tex]\alpha =11.277rad/sec^2[/tex]

(b) From third equation of motion

[tex]\omega _f^2=\omega _i^2+2\alpha \Theta[/tex]

[tex]280.506^2=120.366^2+2\times 11.277\times \Theta[/tex]

[tex]\Theta =2846.30rad[/tex]

The gravitational potential energy of an object is equal to its weight multiplied by its

Answers

Answer:

Height above a surface

Explanation:

Gravitational potential energy is the energy which an object possesses due to its position above a surface.

It is also the amount of work a force has to do in order to bring an object from a particular position to a point of reference.

It is given mathematically as:

P. E. = m*g*h

where m = mass of the body

g = acceleration due to gravity

h = height above a surface

m*g represents the weight of the object.

Hence, Gravitational potential energy is the product of an object's weight and its height above a surface/reference point.

You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you slowly pull down with a spring scale so the block is now at position y bottom, below the equilibrium position y0 where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
a. Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is a positive direction.
b. Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
c. What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
d. How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy   ⇒  k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and  Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy   ⇒   Δy = f/k

⇒   Δy = 2.019 N/14 N/m

⇒   Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m  and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

 

Each of the two wires in a particular 16-gauge extension cord has an electrical resistance of 0.04 V. You're using this extension cord to operate a toaster oven, so a current of 15 A is fl owing through it. What is the voltage drop across each wire in this extension cord?

Answers

The voltage drop across each wire in the extension cord with a resistance of 0.04 ohms and a current of 15 A flowing through it is calculated using Ohm's Law (V = I x R) to be 0.6 volts.

To calculate the voltage drop across each wire in the extension cord, we can apply Ohm's Law, which states that voltage (V) is the product of current (I) and resistance (R). Given that each wire has a resistance of 0.04 Ω and a current of 15 A is flowing through it, we can calculate the voltage drop by multiplying these two figures:

V = I × R

V = 15 A × 0.04 Ω = 0.6 V

Therefore, the voltage drop across each wire in the extension cord is 0.6 volts.

Which statement about asteroids and meteoroids is true?
Most meteoroids orbit between Mars and Jupiter.
Asteroids frequently crash into Earth
© Meteoroids are pieces of leftover comets.
© Asteroids are pieces of meteors.

Answers

Answer:

there pieces of leftover comets :)

Explanation:

An open container holds ice of mass 0.550 kg at a temperature of -15.3 âC . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 900 J/minute .The specific heat of ice to is 2100 J/kgâK and the heat of fusion for ice is 334Ã103J/kg.a) How much time tmelts passes before the ice starts to melt?tmelts=___minutesb)From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0âC?trise=____minutes

Answers

Final answer:

The ice in the open container begins to melt after 19.82 minutes of heating, and the temperature of the system starts to rise above 0°C after a total heating time of 223.93 minutes.

Explanation:

To answer both parts of this question, we need to calculate the time for two processes: the heating of the ice to 0°C (melting point), and the melting of the ice into water at 0°C.

Firstly, we find the heat (Q) required to raise the temperature of the ice to 0°C using Q=mcΔT, which is 0.55kg * 2100 J/kg°C * (0 - (-15.3°C)) = 17842.5 J. As the heater adds 900J per minute, the time for this is 17842.5J ÷ 900J/min = 19.82 minutes.

Next, we find the heat required to melt ice into water at 0°C using Q=mLf, where Lf = 334000 J/kg. This is 0.55kg * 334000 J/kg = 183700J. The time for this can be found by 183700J ÷ 900J/min = 204.11 minutes.

tmelts (the time before ice begins to melt) is 19.82 minutes, and trise (the total time before the temperature begins to rise above 0°C) is 19.82 min + 204.11 min = 223.93 minutes.

Learn more about Heat Transfer Calculations here:

https://brainly.com/question/31080599

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Final answer:

The time before the ice starts to melt (tmelts) is 19.63 minutes, and the total time before the temperature begins to rise above 0 °C (trise) is 223.74 minutes.

Explanation:Calculating Time Until Ice Starts to Melt

To calculate the time tmelts before the ice starts to melt, you firstly need to determine how much heat is necessary to raise the ice from -15.3 °C to 0 °C using the formula Q = m * c * ΔT, where Q is the heat, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature. In this case, Q would be:

Q = (0.550 kg) * (2,100 J/kg·K) * (15.3 K) = 17,661.5 J.

Since heat is supplied at 900 J/minute, the time tmelts can be calculated as:

tmelts = Q / (heat rate) = 17,661.5 J / (900 J/min) = 19.63 minutes.

Calculating Time Until Temperature Rises Above 0 °C

To calculate the time trise until the temperature begins to rise above 0 °C, we must add the time it takes to melt the ice completely at 0 °C. The heat of fusion (∆Hfus) formula is used here: Q = m * Lf where Lf is the heat of fusion for ice. Considering that the heat of fusion for ice is 334,000 J/kg:

Q = (0.550 kg) * (334,000 J/kg) = 183,700 J.

For melting the ice at 0 °C, it will take:

trise = Q / (heat rate) = 183,700 J / (900 J/min) = 204.11 minutes.

Therefore, the total time before the temperature begins to rise above 0 °C is the sum of the time to warm the ice to 0 °C and the time to melt it completely, which will be 19.63 minutes + 204.11 minutes = 223.74 minutes.

Learn more about Heat Transfer in Ice here:

https://brainly.com/question/18940221

#SPJ3

By what factor will the electrostatic force between two charged objects change when the amount of charge on one object doubles?

Answers

Answer:

The electrostatic force between two charge doubles when when the amount of charge on one object doubles.

Explanation:

Force between two charge particle [tex]F=\frac{1}{4\pi\epsilon } \frac{q_1\cdot\ q_2}{r^{2} }[/tex]

Where,  [tex]\frac{1}{4\pi\epsilon } = constant[/tex]

[tex]q_1,q_2\ magnitude\ of\ charges.\\r\ is\ the\ distance\ between\ the\ particles.[/tex]

Now suppose as per the question

[tex]q_1= 2 q_1\ with\ q_2\ and\ r\ with\ same\ value.\\Hence\ force\ between\ the\ particle\ F_2 =2 F[/tex]

Hence, electrostatic force between two charge doubles when when the amount of charge on one object doubles.

Si te dijera que la energía es una cualidad o capacidad que tienen los cuerpos materiales ¿para qué sirve o emplean los cuerpos esa cualidad o capacidad?

Answers

Answer:

La energía es una cualidad o capacidad que tienen los cuerpos materiales y posee diversas funciones. Hay distintos tipos de energía, pero ésta permite que un cuerpo se mueva por ejemplo, cambie de temperatura o de estado.

Explanation:

La Energía de los cuerpos puede ser definida de distintas maneras según la ciencia en la cual estemos trabajando. Para la física, la energía es la capacidad para realizar un trabajo. Esa energía puede ser cinética, térmica, química, y cada una de ellas define distintas propiedades de los cuerpos, que hacen que puedan cambiar de posición, temperatura, estado.

. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At the same instant the balloon is released; a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon

Answers

Answer:

Explanation:

Height of building

H = 6m

Horizontal speed of first balloon

U1x = 2m/s

Second ballot is thrown straight downward at a speed of

U2y = 2m/s

Time each gallon hits the ground

Balloon 1.

Using equation of free fall

H = Uoy•t + ½gt²

Uox = 0 since the body does not have vertical component of velocity

6 = ½ × 9.8t²

6 = 4.9t²

t² = 6 / 4.9

t² = 1.224

t = √1.224

t = 1.11 seconds

For second balloon

H = Uoy•t + ½gt²

6 = 2t + ½ × 9.8t²

6 = 2t + 4.9t²

4.9t² + 2t —6 = 0

Using formula method to solve the quadratic equation

Check attachment

From the solution we see that,

t = 0.9211 and t = -1.329

We will discard the negative value of time since time can't be negative here

So the second balloon get to the ground after t ≈ 0.92 seconds

Conclusion

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

Answer:

Second balloon hits ground Δt = 0.185 seconds sooner than first balloon

Explanation:

Given:-

- The first balloon is thrown horizontally with speed, u1 = 2.0 m/s

- The second balloon is thrown down with speed, u2 = 2.0 m/s

- The height from which balloon are thrown, si = 6.0 m (above ground)

Find:-

Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon

Solution:-

- We will first determine the time taken (t1) for the first balloon thrown horizontally with speed u1 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.

- Using the second kinematic equation of motion in vertical direction:

                         si = 0.5*g*t1^2

Where,     g: The gravitational constant = 9.81 m/s^2

                        6.0 = 4.905*t1^2

                        4.905*t1^2 - 6.0 = 0

- Solve the quadratic equation:

                        t 1 = 1.106 s

- Similarly, the time taken (t2) for the second balloon thrown down with speed u2 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.

- Using the second kinematic equation of motion in vertical direction:

                         si = u2*t2 + 0.5*g*t1^2

Where,     g: The gravitational constant = 9.81 m/s^2

                        6.0 = 4.905*t1^2 + 2*t2

                        4.905*t1^2 + 2*t2 - 6.0 = 0

- Solve the quadratic equation:

                        t 2 = 0.9208 s

- We see that the second balloon thrown down vertically hits the ground first. The second balloon reaches ground, t1 - t2 = 0.185 seconds, sooner than first balloon.

Which factors are most likely to affect whether a galaxy will evolve by merging with a second galaxy? Select the two correct answers.(1 point)

its proportion of gas compared to that of the other galaxy

its distance from the other galaxy

its size compared to that of the other galaxy

its age compared to that of the other galaxy

Answers

Answer:

ii. its distance from the other galaxy.

iii. its size compared to that of the other galaxy

Explanation:

A galaxy is a vast constituent of interstellar medium of gas, dust and billions of stars bounded by the gravitational pull. There are tens of billions of  galaxies existing in the whole universe. Various numbers of galaxies can merge to form an inter-cluster, which is a more massive and bigger galaxy.

The constituent of an individual galaxy are held together by mutual gravitational pull, but may interact with other galaxies depending on; its distance from other galaxies and its size compared to that of the other galaxy. This process would result into the formation of a giant elliptical galaxy and a number of stars.

Answer:

1.Gravity acts on existing material, shaping it into a new form.

2.a much larger galaxy, a nearby galaxy

3.low gas content, little star formation

Explanation:

i just did the test and its 100%. dont believe me? just do the test and see for your self

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