Answer:Which statements describe Rutherford’s model of the atom? Select all that apply.
The atom is mostly empty space.
The atom cannot be divided into smaller particles.
Electrons orbit around the center of the atom.
The atom’s positive charge is located in the atom’s nucleus.
The electrons are located within the positive material of the nucleus.
Electron clouds are regions where electrons are likely to be found.
Explanation:
Sample Response: The one major change that occurred was the placement and organization of the electron. Rutherford’s model identified that the electrons were at a distance from the nucleus, Bohr’s model identified that the electrons occurred at levels that related to their available energy, and the modern atomic model shows that electrons are located in a predicted area but cannot be identified in a specific point.
A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom of a gully. Its momentum upon hitting the ground is
Answer:
2940 kg m/s
Explanation:
First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:
[tex]v_y = v_0+gt[/tex]
where
v0 =0 is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration due to gravity
t = 3 s is the time of the fall
Substituting into the formula, we find
[tex]v_y=0+(9.8 m/s^2)(3 s)=29.4 m/s[/tex]
And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:
[tex]p=mv=(100 kg)(29.4 m/s)=2940 kg m/s[/tex]
The motorcycle's momentum when hitting the ground is calculated using the formulas for momentum and velocity. With a mass of 100 kg and a velocity calculated as 29.4 m/s, the motorcycle's momentum is 2940 kg.m/s.
Explanation:In this case, we are asked to find the momentum of the motorcycle when it hits the ground. The momentum (p) of an object is calculated by the formula p=mv where m is the mass and v is the velocity. The velocity of the motorcycle can be calculated using the formula v=gt, where g is the acceleration due to gravity (9.8 m/s2) and t is the time. So, v=9.8*3 = 29.4 m/s. Therefore, the momentum of the motorcycle is p=100*29.4 = 2940 kg.m/s. Compared to the typical automobile with a momentum of 1400*15 = 21000 kg.m/s, the motorcycle has significantly less momentum due to its smaller mass and velocity.
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The ratio of the output work to input force is the _ of the machine ?
Mechanical advantage
Explanation;Mechanical advantage is the ratio of force output from a machine divided by the force input into the machine.Mechanical advantage measures the machine's force-magnifying effect. It is an advantage gained by using simple machines to accomplish work with less effort.The formula is; M.A = output force/ Input forceAnswer:
Mechanical Advantage
Explanation:
Hermal energy is a form of... A. Potential energy. B. Kinetic energy. C. Chemical energy.
Thermal energy also known as 'heat energy' is a form of kinetic energy.
A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would result in an induced emf in the loop? Choose all that apply.1.Moving the loop outside of the magnetic field region.2.Spin the loop such that its axis does not consistently line up with the magnetic field direction.3.Change the magnitude of the magnetic field.4.Change the diameter of the loop
Answer:
All the 4 options
Explanation:
Electromagnetic induction occurs when the magnetic flux through a coil of wire is changing over time:
[tex]\epsilon = -\frac{\Delta \Phi}{\Delta t}[/tex]
where
[tex]\epsilon[/tex] is the emf induced in the coil
[tex]\Delta \Phi[/tex] is the variation of magnetic flux
[tex]\Delta t[/tex] is the variation of time
The presence of an emf in the coil will generated an induced current.
The magnetic flux through the coil is given by
[tex]\Phi = BA cos \theta[/tex]
where
B is the intensity of the magnetic field
A is the area of the coil
[tex]\theta[/tex] is the angle between the direction of the field and the axis of the coil
We see that any actions that changes one of these 3 variables will change the magnetic flux through the coil, so it will also induce a current.
The 4 options are:
1.Moving the loop outside of the magnetic field region. --> this will decrease the intensity of the magnetic field, B, therefore it will change the flux, and it will induce a current
2.Spin the loop such that its axis does not consistently line up with the magnetic field direction. --> this will change the angle between the direction of the coil's axis and the field B, so this will also change the flux, and therefore will induce a current
3.Change the magnitude of the magnetic field. --> this will change the magnitude of B, so this will also change the flux, and therefore will induce a current
4.Change the diameter of the loop --> this will change the area of the coil A, so this will also change the flux, and therefore will induce a current
Therefore, all 4 options are correct.
An induced emf in a loop of wire can result from moving the loop out of the magnetic field region, spinning the loop to change its orientation, or changing the magnitude of the magnetic field itself.
An induced electromotive force (emf) occurs in a loop of wire when the magnetic flux through the loop changes. This can happen in several ways, such as:
Moving the loop outside the magnetic field region, would change the amount of magnetic flux through the area of the loop.Spinning the loop to change its orientation relative to the magnetic field direction, thus altering the flux through the loop.Changing the magnetic field's magnitude directly changes the flux through the loop.Changing the diameter of the loop does not induce an EMF unless the magnetic field or the loop's orientation relative to the field changes at the same time.
A series combination of two resistors, 7.25 ω and 4.03 ω, is connected to a 9.00 v battery.
a. calculate the equivalent resistance of the circuit and the current.
b. what is the potential difference across each resistor?
a. [tex]11.28\Omega[/tex]
The equivalent resistance of a series combination of two resistors is equal to the sum of the individual resistances:
[tex]R_{eq}=R_1 + R_2[/tex]
In this circuit, we have
[tex]R_1 = 7.25 \Omega\\R_2 = 4.03 \Omega[/tex]
Therefore, the equivalent resistance is
[tex]R_{eq}=7.25 \Omega + 4.03 \Omega=11.28 \Omega[/tex]
b. 5.8 V, 3.2 V
First of all, we need to determine the current flowing through each resistor, which is given by Ohm's law:
[tex]I=\frac{V}{R_{eq}}[/tex]
where V = 9.00 V and [tex]R_{eq}=11.28 \Omega[/tex]. Substituting,
[tex]I=\frac{9.00 V}{11.28 \Omega}=0.8 A[/tex]
Now we can calculate the potential difference across each resistor by using Ohm's law again:
[tex]V_1 = I R_1 = (0.8 A)(7.25 \Omega)=5.8 V[/tex]
[tex]V_2 = I R_2 = (0.8 A)(4.03 \Omega)=3.2 V[/tex]
The equivalent resistance of the circuit is 11.28 Ω, and the current through the circuit is approximately 0.798 A. The potential differences across the 7.25 Ω resistor and the 4.03 Ω resistor are approximately 5.79 V and 3.21 V, respectively.
To solve the problem, we need to brreak it down in the following steps:
Part a):
Calculate the equivalent resistance (R_eq) of the circuit:Part b):
Calculate the potential difference (V) across each resistor:Thus, the potential differences across the resistors are approximately 5.79 V and 3.21 V, respectively.
A chemical symbol represents the ____ of an element.
Choices
name
reaction
group
structure
Answer:
Explanation:
its the name of an element
A chemical symbol represents the name of an element. These symbols are abbreviations used for chemical elements or compounds in chemistry.
Explanation:A chemical symbol in Chemistry stands for the name of an element. Chemical symbols are abbreviations used in chemistry for chemical elements, functional groups or chemical compounds. These symbols consist of letters - for example, the symbol for Hydrogen is H, Oxygen is O, and Carbon is C.
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How does density affect refraction?
Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond.
Explain where you observe reflection, refraction, and absorption of light in your everyday activities.
Answer:
1. When light passes from a more dense to a less dense substance the light is refracted.
2. The diamond would act like a prism and make rainbows.
3. You observe all of these pretty much every day. Like if you notice how on a glass building it reflects the sunlight and some other things around it. Then another example could be just shining a flash light in a room.
I do FLVS and I got a 100 on this assignment so I hope it helped a little :)
PLZ HELP FAST! Which graph BEST shows the relationship of kinetic energy to potential energy as a book sits on the edge of a desk?
Answer:
both energy will remain constant and potential energy must be more than the kinetic energy of the block
So correct Graph will be
A) option
Explanation:
As we know that kinetic energy of the block is given as
[tex]K = \frac{1}{2}mv^2[/tex]
potential energy is given as
[tex]U = mgh[/tex]
so when a block is stationary at the surface of table so due to height of the table the potential energy of the block is given as
[tex]U = mgh[/tex]
but as we know that the speed of the block is zero
so kinetic energy must be
[tex]K = 0[/tex]
so here in the graph both energy will remain constant and potential energy must be more than the kinetic energy of the block
For a batter to increase momentum at the plate, he/she can? Choose a lighter bat Choose a heavier bat Choose an aluminum bat Choose a wooden bat
Choose a heavier bat
Explanation;Momentum is a vector which is often defined as the product of an object’s mass with its velocity.Two factors affecting momentum is the mass and velocity of the object. An object that has a low velocity and a small mass produce minimal momentum because it would take a small force and/or time to stop it.Therefore, choosing a heavier bat would increase the momentum of the plate.Final answer:
To boost momentum, a batter is suggested to choose a heavier bat due to its larger moment of inertia and the greater force it can impart on the ball, according to the principles of moment of inertia, angular momentum, and Newton's third law of motion.
Explanation:
To increase momentum at the plate, a batter can choose a heavier bat. This is because a more massive bat has more inertia and therefore, when swung, will impart a larger force on the ball for the same amount of time compared to a lighter bat. This is evident from the concepts of moment of inertia and angular momentum. When a bat is swung, the moment of inertia plays a role in determining the bat's resistance to changes in its rotational speed. A heavier bat, generally, has a larger moment of inertia, allowing it to maintain its speed through the batting zone and thus create a more forceful impact, enhancing the ball's exit velocity.
Considering the batter swinging a Wiffle ball bat, swinging at the end, furthest from the pivot point, increases the moment of inertia and thus requires more torque to achieve the same angular acceleration as grabbing the middle. However, the increased distance from the pivot allows for a higher linear speed at the end of the bat, creating a larger angular momentum on contact which could result in a more powerful impact if the batter can successfully swing the bat quickly.
In a collision, such as a bat hitting a ball, both objects exert forces on each other, as explained by Newton's third law of motion. This interaction also informs us that not only the ball is affected by the bat but the force exerted by the ball also impacts the bat, as any baseball player who has felt the sting of a ball hit off the end of the bat can attest.
A potential difference of 12v produces a current of 0.30A in a piece of cooper wire. What is the resistance of the copper wire?
Well , its a easy problem
By ohms law,
we have V= IR
Where V is the voltage applied across the wire and R is the resistance and I is the current
=> R = V/I
=> R = 12/0.3 = 40 ohm
The resistance of the copper wire will be 400 ohms. Resistance is found as the ratio of the voltage and electric current.
What is resistance?
Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
The given data in the problem is;
V is the voltage= 12 V
I is the current = 0.30A
R is the resistance
The resistance of the circuit from the Ohm's law is found as;
[tex]\rm V= IR \\\\ R = \frac{V}{I} \\\\\ R = \frac{12}{0.30} \\\\\ R =400 \ ohm[/tex]
Hence, the resistance of the copper wire will be 400 ohms.
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A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m. a)Find the force constant of the spring. Express your answer with the appropriate units. b)The fish is now pulled down 5.00 cm and released. What is the period of oscillation of the fish? Express your answer with the appropriate units. c)What is the maximum speed it will reach? Express your answer with the appropriate units.
(a) 3539 N/m
Hook's law states that:
[tex]F=kx[/tex]
where F is the force applied on the spring, k is the spring constant, x is the stretching of the spring.
In this problem, we have:
[tex]F=mg=(65.0 kg)(9.8 m/s^2)=637 N[/tex] is the force applied (the weight of the fish)
[tex]x=0.180 m[/tex] is the stretching of the spring
Solving the equation for k, we find the spring constant:
[tex]k=\frac{F}{x}=\frac{637 N}{0.180 m}=3539 N/m[/tex]
(b) 0.85 s
The period of oscillation of a spring-mass system is
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
In this case,
m = 65.0 kg
k = 3539 N/m
Substituting into the formula,
[tex]T=2 \pi \sqrt{\frac{65.0 kg}{3539 N/m}}=0.85 s[/tex]
(c) 0.37 m/s
The initial elastic potential energy of the spring when the fish is pulled down is:
[tex]U=\frac{1}{2}k\Delta x^2[/tex]
where
[tex]\Delta x = 5.00 cm=0.05 m[/tex] is the stretching of the spring with respect to the initial position
Substituting,
[tex]U=\frac{1}{2}(3539 N/m)(0.05 m)^2=4.4 J[/tex]
The spring reaches its maximum speed when it crosses the equilibrium position, for which [tex]\Delta x=0[/tex], so when all the elastic potential energy has been converted into kinetic energy:
[tex]E=K=\frac{1}{2}mv^2[/tex]
where v is the speed of the fish. Solving for v, we find
[tex]v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(4.4 J)}{65.0 kg}}=0.37 m/s[/tex]
The force constant of the spring is 3.53 * 10^3 N/m, the period of oscillation of the fish is 0.115 s and the maximum speed it will reach is 0.869 m/s
Explanation:A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m.
By using the amount of the spring is stretched by the weight of the fish, we can calculate the force constant [tex]k[/tex] of the spring.
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex] with [tex]v_{max}=\omega A = 2 \pi f A[/tex]
When the fish hangs at rest the upward spring force [tex]|F_x| = kx[/tex] equals the weight [tex]mg[/tex]of the fish [tex]f = \frac{1}{T}[/tex]. Therefore the amplitude of the SHM is [tex]0.0500 m.[/tex]
a)Find the force constant of the spring. Express your answer with the appropriate units.
[tex]mg=kx\\k = \frac{mg}{x} \\k = \frac{65*9.8}{0.180} = 3.53 * 10^3 N/m\\[/tex]
b)The fish is now pulled down 5.00 cm and released. What is the period of oscillation of the fish? Express your answer with the appropriate units.
[tex]T = 2\pi \sqrt{\frac{m}{k}[/tex]
[tex]T = 2\pi \sqrt{\frac{65 kg}{ 3.53*10^3 N/m} [/tex]
[tex]T =0.115 s[/tex]
c)What is the maximum speed it will reach?
[tex]v_{max}=\omega A = 2 \pi f A\\v_{max}=\omega A = \frac{ 2 \pi A}{T} \\v_{max}=\omega A = \frac{ 2 \pi 0.05 m}{0.115}\\v_{max}=\omega A = 0.869 m/s[/tex]
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When the distance between two stars decreases by one-third, the force between them
Answer:
the force will increase by a factor 2.25
Explanation:
The gravitational force between the two stars is given by:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
G is the gravitational constant
m1, m2 are the masses of the two stars
r is the distance between the stars
If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance
[tex]r'=\frac{2}{3}r[/tex]
So the new force will be
[tex]F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F[/tex]
So, the force will be 2.25 times the previous value.
A vector is 0.888 m long and points in a 205 degree direction.
Find the x-component of the vector.
Answer:
-0.805 m
Explanation:
The x-component of a vector is given by:
[tex]v_x = v cos \theta[/tex]
where
v is the magnitude of the vector
[tex]\theta[/tex] is the angle of the vector with respect to the positive x-direction
In this problem we have
v = 0.888 m
[tex]\theta=205^{\circ}[/tex]
so we have
[tex]v_x = (0.888 m)(cos 205^{\circ})=-0.805 m[/tex]
A car engine changes chemical potential energy into the blank energy of the moving car?
Kinetic (mechanical) energy
Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?A) both exert equal non-zero torquesB) the first at the midpointC) both exerts zero torquesD) the second at the doorknobE) additional information is needed
Answer:
D) the second at the doorknob
Explanation:
The torque exerted by a force is given by:
[tex]\tau = Fdsin \theta[/tex]
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation
[tex]\theta[/tex] is the angle between the direction of the force and d
In this problem, we have:
- Two forces of equal magnitude F
- Both forces are perpendicular to the door, so [tex]\theta=90^{\circ}, sin \theta=1[/tex]
- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force
--> therefore, the 2nd force exerts a greater torque
A fan rotating with an initial angular velocity of 1000 rev/min is switched off. In 2 seconds, the angular velocity decreases to 200 rev/min. Assuming the angular acceleration is constant, how many revolutions does the blade undergo during this time.
A) 10
B) 20
C) 100
D) 125
E) 1200
The average angular acceleration of the fan is
[tex]\alpha_{\rm avg}=\dfrac{\Delta\omega}{\Delta t}=\dfrac{200\frac{\rm rev}{\rm min}-1000\frac{\rm rev}{\rm min}}{2\,\rm s}=-6.67\dfrac{\rm rev}{\mathrm s^2}[/tex]
The number of revolutions after time [tex]t[/tex] is given by
[tex]\theta=\omega_0t+\dfrac{\alpha}2t^2[/tex]
Acceleration is assumed to be constant, so [tex]\alpha=\alpha_{\rm avg}[/tex] and over the 2 second interval we have
[tex]\theta=\left(1000\dfrac{\rm rev}{\rm min}\right)(2\mathrm s)+\dfrac{\alpha_{\rm avg}}2(2\,\mathrm s)^2=20\,\mathrm{rev}[/tex]
so the answer is A.
No of revolution is defined as the no of loops taken from the starting to the end to the end from the same point started. The number of the revolution blade experiences will be 20.
what is angular acceleration?Angular acceleration is defined as the pace of change of angular velocity with reference to time.
[tex]\rm{\alpha _{avg}= \frac{w_f-w_i}{t_f-t_i}}[/tex]
given ,
initial angular velocity = 1000 rev/min = 16.67 rev /sec
final angular velocity = 200 rev/min = 3.33 rev/sec
time taken = 2 second
[tex]\rm{\alpha _{avg}= \frac{w_f-w_i}{t_f-t_i}}\\\\\\\rm{\alpha _{avg}= \frac{3.33-16.67}{0.3}}\\\\\\\rm{\alpha _{avg}= -6.67 rev/ sec^2[/tex]
According to newtons second law of motion,
No of revolution after time t
[tex]\rm{\theta= \omega_it+\frac{1}{2} \alpha t^{2} }\\\\\\\rm{\theta= \ 16.67\times2+\frac{1}{2}(-6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 16.67\times2-\frac{1}{2} (6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 16.67\times2-\frac{1}{2} (6.67)\times{2}^{2} }\\\\\\\rm{\theta= \ 33.3- 13.3}\\\\\\\rm{\theta= \ 20rev }\\\\\\[/tex]
Therefore the number of revolutions the blade undergoes will be 20 revolutions.
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What unit is commonly used when measuring the energy rating of an electrical appliance
usually it is watts
The much more common rating for an appliance is Power.
Power is the rate at which energy is used. Small appliances ... like toasters and blow-driers ... are rated in watts. Large appliances ... like industrial electric water boilers, electric construction cranes, and elevator motors ... are rated in Kilowatts (1 kilowatt is 1,000 watts). This is the RATE at which the appliance uses energy. You leave it running longer, you use more energy, and you owe more on your electric bill.
Some appliances are advertised and sold with an energy rating. This is the amount of energy it would consume if you used it the way a normal standard person uses it, for some standard amount of time like 1 month or 1 year. If the appliance does have such a rating, it'll be in units of Kilowatt-hours. But it'll be no better than a rough estimate, because you may use the appliance more often or less often than the normal standard person.
Which is the electric potential energy of a charged particle divided by its charge?Electric fieldelectric field lineelectric potentialelectric potential difference
electric potential
Explanation;Electric potential is the electric potential energy per unit charge.
Mathematically; V =PE/q
Where; PE is the electric potential energy, V is the electric potential and q is the charge.
Electric potential is more commonly known as voltage. If you know the potential at a point, and you then place a charge at that point, the potential energy associated with that charge in that potential is simply the charge multiplied by the potential.
Answer:
c. electric potential
Explanation:
on edg
What elements compose the stars and planets in the universe
Well hydrogen would be the main element, as a process called nuclear fusion with both helium and hydrogen atoms occurs within stars. And planets are the products of dead stars that have burned through their supplies of hydrogen, helium, and carbon. Planets are a product of this.
Stars and planets are mainly composed of hydrogen and helium, making up from 96 to 99% of their total mass. They also contain heavier elements, the concentrations of which vary among different stars. This fundamental composition was first discovered by Cecilia Payne-Gaposchkin in 1925.
Explanation:The stars and planets in the universe are composed mainly of hydrogen and helium, as determined through analyses of stellar spectra. These elements constitute from 96 to 99% of the total mass of most stars. However, they also contain heavier elements in varying quantities. For instance, Population I stars, like the Sun, contain around 1-4% of heavy elements. However, Population II stars found in the outer galactic halo and in globular clusters have much lower concentrations.
These elements must have been fused somewhere. The stars, having incredibly high temperatures, are speculated to be the source of this element formation. Therefore, stars are believed to play a critical role in the chemical richness that characterizes our world.
The discovery that stars are primarily composed of light gases like hydrogen and helium was first made in a pioneering thesis by Cecilia Payne-Gaposchkin in 1925, making a significant contribution to our understanding of the universe. This composition is also found in the material between the stars, with about 99% of it being a gas composed of individual atoms or molecules of hydrogen and helium.
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As the ambulance got closer, Marge noticed that the pitch of the siren got higher. This happened because
A) the sound waves were pushed closer together.
B) the human ear increases pitch over time.
C) the ambulance was further away.
D) the volume increased.
A) the sound waves were pushed closer together
Answer:
The correct choice is
A) the sound waves were pushed closer together.
Explanation:
As the ambulance gets closer, the distance between the ambulance and marge decreases and so does the wavelength of the siren observed by marge. As the wavelength decrease and speed remains the same for the siren, the frequency of the siren increases because wavelength, frequency and speed are related as
Frequency = Speed/wavelength
Clearly, the frequency is inversely related to wavelength.
ANSWER TRUE OR FALSE: earth is much farther away from the sun in winter, making it colder
I see TWO statements here:
#1). Earth is much farther away from the Sun in Winter.
#2). That's what makes it cold in the Winter.
Let's use our heads on #1 for just a little moment:
-- When it's Winter in Canada, the USA, and France, it's Summer in Australia, Paraguay, and Namibia. This is the end of December, all of January and February, and most of March.
-- When it's Winter in Tasmania, South Africa, and Botswana, it's Summer in Germany, Israel, and Mexico. This is the end of June, all of July and August, and most of September.
So WHOSE Winter is the question asking about ? ? It doesn't say.
Earth is FARTHEST away from the Sun during the first week in July, and it's NEAREST to the Sun during the first week in January.
The difference between the nearest distance and the farthest distance is about 3%. It has almost ZERO effect on how hot or cold the days are. Anywhere !
So BOTH of the statements in the question are FALSE.
You drop a 3.00 kg book to a friend who stands on the ground at distance D = 14.0 m below. If your friend's outstretched hands are at distance d = 1.30 m above the ground (see the figure), (a) how much work Wg does the gravitational force do on the book as it drops to her hands? (b) What is the change ΔU in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy U of that system is taken to be zero at ground level, what is U (c) when the book is released and (d) when it reaches her hands? Now take U to be 100 J at ground level and again find (e) Wg, (f ) ΔU, (g) U at the release point, and (h) U at her hands.
(a) 373.4 J
The displacement of the book is
[tex]d=14.0 m - 1.30 m=12.7 m[/tex]
While the gravitational force acting on the book is
[tex]F=mg=(3.00 kg)(9.8 m/s^2)=29.4 N[/tex]
And the force (downward) is parallel to the displacement (downward), so the work done by gravity is
[tex]W=Fd=(29.4 N)(12.7 m)=373.4 J[/tex]
(b) -373.4 J
The work done by the Earth on the book has been converted into kinetic energy of the book (because the book accelerates as it approaches the ground). SInce the total mechanical energy of the Earth-book system must be conserved, this means that potential energy U has been converted into kinetic energy K: therefore, the loss in potential energy U is exactly -373.4 J.
(c) 411.6 J
The gravitational potential energy U is given by:
[tex]U=mgh[/tex]
where m = 3.00 kg is the mass of the book, g = 9.8 m/s^2 is the gravitational acceleration, h=14.0 m is the height of the book above the reference level (the ground). Substituting,
[tex]U=(3.00 kg)(9.8 m/s^2)(14.0 m)=411.6 J[/tex]
(d) 38.2 J
When the book reaches the hand, its height above the ground level is
h = 1.30 m
therefore, the gravitational potential energy this time is
[tex]U=mgh=(3.00 kg)(9.8 m/s^2)(1.30 m)=38.2 J[/tex]
(e) 373.4 J
The work done by gravity does not change if we change the value of the potential energy at ground level. In fact, the work done by gravity is still calculated as before:
[tex]W=Fd=(29.4 N)(12.7 m)=373.4 J[/tex]
(f) -373.4 J
As we did in point b), the work done by the Earth on the book has been converted into kinetic energy of the book (because the book accelerates as it approaches the ground), so the loss in potential energy is equal to the work done by gravity, and this value does not depend on the value of the potential energy at ground level, so it is still -373.4 J.
(g) 511.6 J
The gravitational potential energy U is given by:
[tex]U=mgh+U_0[/tex]
where m = 3.00 kg is the mass of the book, g = 9.8 m/s^2 is the gravitational acceleration, h=14.0 m is the height of the book above the reference level (the ground), and [tex]U_0 = 100 J[/tex] is the potential energy at ground level, which must be added into the formula. Substituting,
[tex]U=(3.00 kg)(9.8 m/s^2)(14.0 m)+100 J=511.6 J[/tex]
(h) 138.2 J
As before, we can calculate the potential energy of the book at a height of h=1.30 m, adding 100 J of energy which is the value of the potential energy at ground level. we find:
[tex]U=mgh+U_0=(3.00 kg)(9.8 m/s^2)(1.30 m)+100 J=138.2 J[/tex]
The work done by gravity, Wg, is calculated to be 372.66 J. The gravitational potential energy, U, decreases by this amount during the fall, from 411.60 J at the release point to 38.94 J at the friend's hands. If the ground level has an arbitrary gravitational potential energy of 100 J, the final values are adjusted accordingly to 511.60 J and 138.94 J.
Explanation:To calculate the work Wg done by the gravitational force as the book drops to the friend's hands, we use the formula W = mgh, where m is the mass of the book, g is the acceleration due to gravity (9.8 m/s²), and h is the height through which the book falls. The change in height is the total distance D minus the distance d, which the friend's hands are above the ground. This gives us the effective height the book falls through.
(a) The work done by gravity, Wg, is:
Wg = mgh
Wg = (3.00 kg)(9.8 m/s²)(14.0 m - 1.30 m)
Wg = (3.00 kg)(9.8 m/s²)(12.7 m)
Wg = 372.66 J
(b) The change in the gravitational potential energy ΔU is equal to the work done by gravity, so ΔU = 372.66 J.
(c) Initially, the gravitational potential energy U at the release point is U = mgh, which is:
U = (3.00 kg)(9.8 m/s²)(14.0 m)
U = 411.60 J
(d) When the book reaches the friend's hands, the gravitational potential energy is:
U = (3.00 kg)(9.8 m/s²)(1.30 m)
U = 38.94 J
If the gravitational potential energy at ground level is 100 J instead of 0 J, the values for Wg and ΔU do not change because these are quantities that depend only on the change in height. However, the absolute values of U at the initial and final points will be 100 J more than previously calculated:
(e) Wg remains 372.66 J
(f) ΔU remains 372.66 J
(g) U at the release point now is U + 100 J:
U = 411.60 J + 100 J
U = 511.60 J
(h) U at the friend's hands now is U + 100 J:
U = 38.94 J + 100 J
U = 138.94 J
Calculate the volume of 5250 kilograms of wood if the density of wood is 750kg/m?
5250/750 m³ or 7 m³
formula is density x volume = mass
Answer:
Volume of wood, [tex]V=7\ m^3[/tex]
Explanation:
It is given that,
Mass of wood, m = 5250 kg
The density of wood, [tex]d=750\ kg/m^3[/tex]
We need to calculate the volume of wood. It can be calculate using the formula of density. It is given by :
[tex]d=\dfrac{m}{V}[/tex]
[tex]V=\dfrac{m}{d}[/tex]
[tex]V=\dfrac{5250\ kg}{750\ kg/m^3}[/tex]
[tex]V=7\ m^3[/tex]
So, the volume of wood is 7 cubic meter. Hence, this is the required solution.
Which of these illustrates that sound waves can travel through a liquid?A. Voices can be heard as people speak to one another B. Whales and porpoises signal each other by making sounds C. Astronauts can’t hear each other in space unless they use a radio D. Footsteps can be heard through the ceiling as someone walks on the roof
Answer:
B. Whales and porpoises signal each other by making sounds
Explanation:
Sound is a type of mechanical wave, which consists of oscillations of the particles in a medium. It is also classified as longitudinal wave, which means that the direction of the oscillation of the particles in the medium is parallel to the direction of propagation of the wave, forming alternating regions of higher particle density (compressions) and lower particle density (rarefactions).
Since sound is a mechanical wave, it just need a medium to propagate through. Therefore, it can travel through solids, liquids and gases.
In particular, the option
B. Whales and porpoises signal each other by making sounds
is an example of this: these animals emit ultrasound waves (a type of sound waves with high frequency), that travel through the water, and then are reflected back to the animal, allowing the animal to understand the distance of the object from which the wave has been reflected back.
What are the three ways in which voltage can be induced in a loop of wire?
Answer:
1. Changing the magnetic field intensity
2. Changing the area enclosed by the loop
3. Changing the orientation of the loop of wire with respect to the magnetic field
Explanation:
Electromagnetic induction occurs when there is a variation of magnetic flux through a coil of wire: as a result, an emf (electromotive force) is induced in the coil, according to the equation
[tex]\epsilon=-\frac{\Delta \Phi}{\Delta t}[/tex]
where
[tex]\Delta \Phi[/tex] is the variation of magnetic flux through the coil
[tex]\Delta t[/tex] is the time elapsed
The magnetic flux through a coil of wire is given by:
[tex]\Phi = BA cos \theta[/tex]
where
B is the magnetic field intensity
A is the area enclosed by the coil
[tex]\theta[/tex] is the angle between the direction of B and the perpendicular to the area enclosed by the coil
As we can see, the magnetic flux depends on these three factors, so changing any of them will change the magnetic flux, and an electromotive force will be induced in the coil as a result.
Voltage can be induced in wire loops by changing the magnetic field passing through the loop, physically moving the wire within a stationary magnetic field, or altering its electrical resistance.
Explanation:There are three primary ways in which voltage can be induced in a loop of wire:
Magnetic change: Voltage can be induced in a wire loop by changing the magnetic field passing through the loop. This is a principle of electromagnetism known as Faraday's law of electromagnetic induction.Physical movement: Voltage can also be induced by physically moving the wire loop within a stationary magnetic field. This phenomenon is exploited in the operation of dynamos and generators.Variable resistance: Finally, voltage can be induced in a wire loop by changing its electrical resistance. This is less common but can happen in certain scenarios such as in a thermocouple.Learn more about Voltage Induction here:https://brainly.com/question/35297457
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How are magnetic fields and electric fields similar
electric fields and magnetic fields both exhibit similar properties. where magnetic fields are natural electric fields are not but man made and can do the same things as a magnetic field.
An astronaut has landed on planet n-40 and conducts an experiment to determine the acceleration due to gravity on that planet. She uses a simple pendulum that is 0.640 m long and measures that 10 complete oscillations 26.0 s. What is the acceleration of gravity on planet n-40?
Answer:
3.73 m/s^2
Explanation:
The period of a simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] (1)
where
L is the length of the pendulum
g is the gravitational acceleration on the planet
The pendulum in the problem has length
L = 0.640 m
and makes 10 oscillations in 26.0 s; it means that its frequency is
[tex]f=\frac{10}{26 s}=0.385 Hz[/tex]
And so its period is
[tex]T=\frac{1}{f}=\frac{1}{0.385 Hz}=2.6 s[/tex]
So now we can solve equation (1) using L=0.640 m and T=2.6 s, so we can find the value of g on the planet:
[tex]g=(\frac{2\pi}{T})^2L=(\frac{2\pi}{2.6 s})^2 (0.640 m)=3.73 m/s^2[/tex]
If a car goes from 20 miles per hour to 10 miles per hour in 5 seconds, find its acceleration
A= -2 due to is a desacceleration
Answer:
[tex]a = -0.89 m/s^2[/tex]
Explanation:
initial speed of the car is given as
[tex]v_i = 20 miles/hour[/tex]
here we know that
[tex]v_i = 20 \times \frac{1609}{3600}[/tex]
[tex]v_i = 8.94 m/s[/tex]
final speed of the car is given as
[tex]v_f = 10 mile/hour[/tex]
[tex]v_f = 10 \times \frac{1609}{3600}[/tex]
[tex]v_f = 4.47 m/s[/tex]
now we have
[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]
[tex]a = \frac{4.47 - 8.94}{5}[/tex]
[tex]a = -0.89 m/s^2[/tex]
How big are protons compared to electrons
Protons are about 1836 times bigger than electrons.
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Suppose that the strength of the electric field about an isolated point charge has a certain value at a distance of 4 m. How will the electric field strength compare at a distance of 8 m from the point charge? 1. At twice the distance the field strength will be twice the original value. 2. At twice the distance the field strength will be 1 3 of the original value. 3. At twice the distance the field strength will be 1 2 of the original value. 4. At twice the distance the field strength will be 1 4 of the original value. 5. At twice the distance the field strength will be the same.
Answer:
4. At twice the distance the field strength will be 1/4 of the original value
Explanation:
The magnitude of the electric field produced by an isolated poitn charge is
[tex]E=k\frac{q}{r^2}[/tex]
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, the distance from the charge is doubled from 4 m to 8 m:
[tex]r'=2 r[/tex]
Therefore, the new electric field will be
[tex]E'=k\frac{q}{(r')^2}=k\frac{q}{(2r)^2}=k\frac{q}{4r^2}=\frac{1}{4}E[/tex]
So, the field strength is 1/4 of the original value.
By doubling the distance between two charges the field strength will be [tex]\dfrac{1}{4} th[/tex] of the original value.
What will be the value of the new electric field when the distance is doubled?
The formula for the magnitude of the electric field is given by
[tex]E=k\dfrac{q}{r^2}[/tex]
Here
k is the Coulomb's constant
q is the charge
r is the distance from the charge
It is given in the problem that the distance from the charge is doubled from 4 m to 8 m:
[tex]r'=2r[/tex]
So to find the new electric field
[tex]E'=k\dfrac{q}{r^2} = k\dfrac{q}{(2r)^2} =\dfrac{1}{4} k\dfrac{q}{r^2} =\dfrac{1}{4} E[/tex]
Thus By doubling the distance between two charges the field strength will be [tex]\dfrac{1}{4} th[/tex] of the original value.
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