Which statement accurately compares the weight of an object on these two planets

Answers

Answer 1
"An object weighs about two times as much on Jupiter as on Neptune." Hope this helps :)

Related Questions

A force of 75 N is applied to a spring, causing it to stretch 0.3 m. What is the spring constant of the spring?

0.004 N/m
22.5 N/m
75.3 N/m
250 N/m

Answers

the answer is D. 250 N/m 

Answer:

D 250

Explanation:

EDGE

HELP PLZ!!!
Which describes the difference between a bowling ball sitting in the rack waiting to be used and the bowling ball knocking pins down? A. In the rack, the bowling ball has only potential energy. As it knocks pins down, its potential energy has decreased, while its kinetic energy has increased. B. In the rack, the bowling ball has no energy at all. As it knocks down the pins the energy is 75% kinetic and 25% potential. C. In the rack and when knocking down pins, the potential energy and the kinetic energy of the bowling ball are equal. D. In the rack, the bowling ball has only kinetic energy. As it knocks pins down, its potential energy has increased, while its kinetic energy has decreased.

Answers

The answer would be A. In the rack, the bowling ball has only potential energy. As it knocks down pins, it’s potential energy has decreased while it’s kinetic increased.

Kinetic energy is the energy of motion and potential energy is stored energy when and object is not in motion

I hope this helped! :-)

An object is places 20.0 cm from the front of a converging lens of focal length 10.0 cm. What is the image distance and magnification?

Answers

1/f=1/object distance + 1/image distance so if we flip this around and get 1/f - 1/object distance which when we plug in values we get 1/10cm - 1/20 we get 1/20 for image which means the image appears 20 cm away from the lens

The image distance and magnification is mathematically given as

i2= - 24.6 cmM= -2.69

What are the image distance and magnification?

Question Parameter(s):

An object is placed 20.0 cm from the front of

a converging lens of focal length 10.0 cm.

Generally, the equation for the lens equation   is mathematically given as

1 / i + 1 / o = 1 / f

Therefore

i = o f / (o - f)

i1 = 30 * 15.2 / (30 - 15.2)

i1= 30.8 cm

Therefore

i2 = 9.4 * 15.2 / (9.4 - 15.2)

i2= - 24.6 cm

In conclusion, magnification

M = m1 * m2

M= (-30.8 / 30) * (24.6 / 9.4)

M= -2.69

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The mass of a bucket full of water is 15 Kg. It is being pulled up from a 15m deep well. Due to a hole in bucket 6 Kg of water flows out. Find the work done in pulling it out of the well.,

Answers

Solution:
workdone = change in potential energy 
W = Force times distance

= ( m -dm)g • ( x)
= mgx -(dm(gx))
= (15 × 10 × 15) -(6/15 gx.dx)
=2250 - (6/15 x10x 225/2)
=2250 -  (3 x 15x 10)
= 2250 - 450
= 1800 Joule
The answer is  1800 Joules.

The work done in pulling a bucket full of water with a weight of 15 kg that lost 6 kg of it from the hole from a well 15 m deep is 1350 Joules

Further Explanation;Work done Work done refers to the measure of energy when a body is moved by a force through a certain distance.Work done is therefore equivalent to energy transferred.Work done is given by the product of force acting on an object and distance moved by the object.Therefore; Work done = Force × distanceForce refer to a pull or a push and is measured in Newtons, N.Distance is measured in meters, m.

Thus; Work done is measured in Joules, J, or Nm

In this case;

Mass of the bucket is 15 Kg

However, 6 kg of water flew out due to the whole, therefore the remaining mass is 9 kg.

Force or Weight = Mass (kg) × 10 N/kg

Therefore;

Force = 90 Newtons

Distance = 15 m

Therefore;

Work done = 90 N × 15 m

                  = 1350 Joules

Hence; Work done pulling out the bucket from the well is 1350 Joules.

Key words: Work done, Force, Distance

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Level: High school

Subject: Physics

Topic: Work, power and simple machines

With a quarter-turn flick of the wrist, a student sets the frisbee rotating at 570 rpm . what is the magnitude of the torque, assumed constant, that the student applies?

Answers

Due to missing information such as the time of the wrist flick or the frisbee's moment of inertia, we cannot calculate the exact torque that the student applied to the frisbee. To find the torque, we would need to determine the angular acceleration based on the provided angular velocity and the additional data.

To answer the question about the magnitude of the torque applied by a student to a frisbee rotating at 570 revolutions per minute (rpm), we need to consider several factors, including the angular velocity, the moment of inertia, and the time during which the torque was applied. However, the problem statement does not provide all the necessary information (like the time duration and the moment of inertia of the frisbee).

Typically, to calculate the torque, you would use the formula: torque (\(\tau\)) = moment of inertia (I) \(\times\) angular acceleration (\(\alpha\)). The angular acceleration could be found if we had the time duration for the quarter-turn and the final angular velocity.

Since these key pieces of data are missing in the student's problem, we cannot calculate the exact torque without additional information. If the time duration of the wrist flick or the moment of inertia of the frisbee were provided, we could find the angular acceleration using \(\alpha = \Delta\omega / \Delta t\), where \(\Delta\omega\) is the change in angular velocity and \(\Delta t\) is the change in time. From there, the torque could be calculated.

The torque applied by the student on the frisbee can be calculated using the formula torque = moment of inertia x angular acceleration. The primary force acting on the student when flicking the frisbee is the force applied perpendicular to the frisbee's radius.

Calculating the torque: To find the torque applied by the student on the frisbee, we can use the formula for torque, which is given by torque = moment of inertia × angular acceleration. First, convert the frisbee's initial angular velocity of 570 rpm to radians per second by multiplying by 2π/60.

Forces acting on the student: The primary force acting on the student when flicking the frisbee is the force applied perpendicular to the frisbee's radius. This force creates a torque and initiates the rotational motion of the frisbee.

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the tension in the rope and the angle the rope makes with the horizontal

Answers

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
[tex]W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N[/tex]
the force of the wind F, acting horizontally, with intensity
[tex]F=12 N[/tex]
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
[tex]T \cos \alpha -F=0[/tex]
[tex]T \sin \alpha -W=[/tex]
By dividing the second equation by the first one, we get
[tex]\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04 [/tex]
From which we find
[tex]\alpha = 63.8 ^{\circ}[/tex]
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
[tex]T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N [/tex]




True or false A physical change occurs when matter changes state as from a liquid to a gas Becuase

Answers

the answer will be false
False.

During chemical changes, atoms are rearranged and different kids of matter form. Liquids and gases are not solid, solid is a whole other phase. It cannot be a physical change, because it would have to involve a solid. An example of a physical change is ice melting. This would not be a physical change, but it would qualify as a chemical change.

Which has more thermal energy: lake or a cup of hot chocolate?

Answers

Though the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.

Answer:

A cup of hot chocolate

Explanation:

A cup of hot chocolate will have more thermal energy has the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.

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Two very quick questions!!

1) If a pulley system with an actual mechanical advantage of 2 is used in lifting a 2,000 lb. car, what effort would be needed to move the car?


If a pulley system with an actual mechanical advantage of 2,000,000 is used in lifting a 2,000 lb. car, what effort would be needed to move the car?,

Answers

The weight lifted by a machine to the applied force on a machine is called mechanical advantage. This is written as Mechanical advantage, M. A, = load(weight)/effort. So for 1) M.A = 2 and load = 2, 000lb = 8896.446N. So 2 = 8896.446/ effort Effort = 8896.446/2 = 4448.48 Similarly for M.A of 2, 000, 000 we have Effort = 8896.446/ 2, 000, 000 = 0.004448

Explanation:

1. Mechanical advantage of a pulley system, m = 2

Mechanical advantage of a machine is defined as the ratio of load to the effort force i.e.

[tex]m=\dfrac{F_L}{F_E}[/tex]

Here, [tex]F_L=2000\ lb[/tex]

[tex]F_E=\dfrac{F_L}{m}[/tex]

[tex]F_E=\dfrac{2000\ lb}{2}[/tex]

[tex]F_E=1000\ lb[/tex]

Hence, 1000 lb effort will be needed to move the car.

2. Mechanical advantage of a pulley system, m = 2,000,000

Value of load, [tex]F_L=2000\ lb[/tex]

Mechanical advantage, [tex]m=\dfrac{F_L}{F_E}[/tex]

[tex]F_E=\dfrac{F_L}{m}[/tex]

[tex]F_E=\dfrac{2000\ lb}{2000000}[/tex]

[tex]F_E=0.001\ lb[/tex]

Hence, 0.001 lb effort would be needed to move the car.

Calculate the total force on the Earth due to Venus, Jupiter, and Saturn, assuming all four planets are in a line. The masses are Mv=0.815ME, MJ=318ME, Msat=95.1ME, Msun=1.99x10^30kg, ME=5.98x10^24kg and the mean distances of the four planets from the Sun are 108, 150, 778, and 1430 million km. Apparently the answer is 9.56x10^17 N but I'm not sure how to get to that .-.,

Answers

You need to consider the following:
Me (mass of Earth) = 5.98 x 10^24 kg 
Ms (mass of Sun) = 1.99 x 10^30 kg 
G = 6.67 x 10^-11 N 

Formula:
 F = G * M1M2/r^2 
The ratio FT/F = 4.02x10^-4 / 14.8
 = 2.72x10^-5 

Since,
1/2.72x10^-5 = 36800
The fraction ratio is  1/36800
9.56x10^17 N

Answer: Total Force = [tex]9.56*10^{17}[/tex]

Explanation:

Line points are:  Sun - Venus - Earth - Jupiter - Saturn

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This means:

[tex]F=G\frac{m_{1} m_{2}}{r^{2} }[/tex]

Where,

G is the gravitational constant,

m1 and m2 are the masses of the objects,

and r is the distance between the centers of their masses.

So, if G value is [tex]6.674*10^{-11}  [\frac{m^{3}}{kg*s^{2}}][/tex], then we replace the equation with the corresponding values:

[tex]F=6.674*10^{-11} (-\frac{0.815ME^{2}}{(4.2*10^{10})^{2}} + \frac{318ME^{2}}{(6.28*10^{11})^{2}} + \frac{95.1ME^{2}}{(1.28*10^{12})^{2}})[/tex]

To get the distances we subtract the distances between the sun and earth and the distances between the other planets and the sun.

Which of these observations of galileo refuted ptolemy's epicycles? the craters on the moon the rotation of sunspots across the sun's surface the revolution of jupiter's moons around it the complete cycle of venus' phases the visibility of many more stars with the telescope?

Answers

The answer is : the complete cycle of venus' phases. This is the observation of Galileo that refuted ptolemy's epicycles. . Using his telescope, Galileo found that Venus went through phases, just like our Moon. But, the nature of these phases could only be explained by Venus going around the Sun, not the Earth.

Final answer:

Galileo's observations that refuted Ptolemy's epicycles were the complete cycle of Venus' phases and the revolution of Jupiter's moons. These observations showed that these bodies were orbiting the Sun and Jupiter respectively, not the Earth as Ptolemy's model suggested.

Explanation:

The observation Galileo made that refuted Ptolemy's epicycles was the complete cycle of Venus' phases. Ptolemy's model suggested that all planets revolved in small circles, or 'epicycles', around a point that in turn orbited Earth. However, Galileo's observation showed that Venus goes through a set of phases similar to the Moon's, which could only occur if Venus orbits the Sun, not Earth. This discovery was incompatible with Ptolemy's model which largely placed Earth as the center of the universe. Galileo's discovery of the revolution of Jupiter's moons around it, instead of Earth, also offered further proof against Ptolemy's model.

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What is the frequency of a microwave that has a wavelength of 0.050 m?

Answers

Answer:

[tex]6x10^9Hz[/tex]

Explanation:

we use a formula that relates the frequency and the wavelength:

[tex]f=\frac{c}{\lambda}[/tex]

where [tex]f[/tex] is the frequency of the wave, [tex]c[/tex] is the speed of light [tex]c=3x10^8m/s[/tex], and [tex]\lambda[/tex] is the wavelength.

We know that the wavelength is: [tex]\lambda=0.050m[/tex], so substituting the known values in the equation for the frequency we get:

[tex]f=\frac{3x10^8m/s}{0.05m} \\f=6x10^9s^{-1}=6x10^9Hz[/tex]

The frequency is:

[tex]6x10^9Hz[/tex]

Final answer:

The frequency of a microwave with a wavelength of 0.050 m is [tex]6*10^9[/tex] Hz, or 6000 MHz, calculated using the speed of light and the wavelength-to-frequency formula.

Explanation:

To calculate the frequency of a microwave that has a wavelength of 0.050 m, you can use the formula

c = λf, where c is the speed of light in a vacuum (approximately [tex]3*10^8[/tex] m/s), λ is the wavelength, and f is the frequency. Solving for f, the equation becomes

f = c / λ.

Let’s plug in the values:

f = ([tex]3*10^8[/tex] m/s) / (0.050 m)

After calculating, the frequency f is [tex]6*10^9[/tex] Hz, or 6000 MHz.

What is the relationship between atomic radius, ionization energy, and electronegativity? What is the relationship between atomic radius, ionization energy, and electronegativity? @Physics,

Answers

Ionization energy is the energy required to remove an electron from a given chemical species. Ionization energy is related to the atomic radiii; such that they are inversely related, therefore ionization increases, atomic radius decrease. Atoms with many electrons get bigger as the electrons become more shielded  from the nucleus. Electronegativity is the atom's chemical ability to attract and bind with electrons. It is the tendency of the atom to attract and form bonds with electrons. Electronegativity increases across the period of elements with decrease in atomic radii, while electronegativity decreases down the period with increase in atomic radii.

The relationship between electronegativity, ionization energy, and atomic radius is that larger atoms are less attracted by nuclear force, which impacts their ability to retain and attract electrons.

Atomic Radius:

It is generally defined as the distance from the nucleus to the outermost electron cloud. Atomic radius tends to increase as you move down a group on the periodic table due to the addition of electron shells.Conversely, it decreases as you move across a period from left to right because of the increasing positive charge in the nucleus, which pulls electrons closer.

Ionization Energy:

The energy needed to extract an electron from an atom in the gas phase is known as ionization energy.The first ionization energy refers to the energy needed to remove one electron. This energy increases across a period and decreases down a group. Higher ionization energy indicates that an atom holds its electrons more tightly.

Electronegativity:

An atom's capacity to draw in and form bonds with electrons is measured by its electronegativity.It goes higher periodically and down within a group.When an atom has a high electronegativity, it draws electrons toward itself in a chemical connection.

Relationship Between the Three Properties:

As the atomic radius increases, ionization energy and electronegativity decrease. This is because larger atoms have more electron shells, resulting in weaker attraction between the nucleus and the outermost electrons.Conversely, as the atomic radius decreases, ionization energy and electronegativity increase. This is due to stronger nuclear attraction in smaller atoms, making it harder to remove electrons and easier to attract electrons from other atoms.

in a lab investigation. one group of students (group A) measures the speed of a 0.1-kilogram car at 2.5 m/sec at the bottom of a hill. Another group of students (group B) measures the speed of the car at 3 m/sec at the bottom of the hill. the car's starting position at the top of the hill is one-meter high. what is the potential energy of the car at the beginning of the experiment before its speed is measured?

Answers

The potential energy of the car at the top of the hill, before the beginning of the experiment, is given by
[tex]U=mgh[/tex]
where [tex]m=0.1 kg[/tex] is the mass of the car, [tex]g=9.81 m/s^2[/tex] is the gravitational acceleration and [tex]h=1.0 m[/tex] is the height of the hill. Using these data, we have
[tex]U=(0.1 kg)(9.81 m/s^2)(1.0 m)=0.98 J[/tex]

The potential energy of the car at the beginning of the experiment, before its speed is measured, is approximately 0.98 Joules.

To calculate the potential energy of the car at the beginning of the experiment, we can use the formula for gravitational potential energy:

Potential energy (PE) = mass (m) × gravitational acceleration (g) × height (h)

Given:

Mass of the car (m) = 0.1 kg

Height of the hill (h) = 1 meter

Gravitational acceleration (g) = 9.8 m/s² (approximately)

The potential energy (PE) is as follows:

PE = m × g × h

= 0.1 × 9.8 × 1

= 0.98 Joules

Therefore, the potential energy of the car at the beginning of the experiment, before its speed is measured, is approximately 0.98 Joules.

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a body of mass 1.5kg, traveling along the positive x axis with speed 4.5m/s,collides with another body B of mass 3.2kg which,initially is at rest. A is deflected and moves with a speed of 2.1m/s in a direction which is 30 degrees below the x axis. B Is set in motion at angle b above the x axis. calculate the velocity of B after collision.,

Answers

Data:
m₁ = 1.5kg
m₂ = 3.2kg
α = -30° (negative because it is below the x-xis)
[tex] v_{1i} [/tex] = initial speed of object 1 = 4.5m/s
[tex] v_{2i} [/tex] = initial speed of object 2 = 0m/s
[tex] v_{1f} [/tex] = final speed of object 1 = 2.1m/s
[tex] v_{2f} [/tex] = ?
β = ?

Since the motion after the collision is in 2 dimentions, it is better to write the speeds with their components along the x and the y-axis:
[tex] v_{1ix} [/tex] = initial speed of object 1 along x-axis = 4.5m/s
[tex] v_{1iy} [/tex] = initial speed of object 1 along y-axis = 0m/s
[tex] v_{2ix} [/tex] = initial speed of object 2 along x-axis = 0m/s
[tex] v_{2iy} [/tex] = initial speed of object 2 along y-axis = 0m/s

[tex] v_{1fx} [/tex] = final speed of object 1 along x-axis = 2.1 cos(-30) = 1.82m/s
[tex] v_{1iy} [/tex] = final speed of object 1 along y-axis = 2.1 sin(-30) = -1.05m/s 

In this kind of collision, we have the conservation of momentum, therefore we can write the system:
[tex] \left \{ {{m_{1} v_{1ix} + m_{2} v_{2ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop { m_{1} v_{1iy} + m_{2} v_{2iy} = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right. [/tex]

Considering the terms that are zero, it becomes:
[tex]\left \{ {{m_{1} v_{1ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop {0 = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right.[/tex]

Let's face first the y-component:
[tex]m_{2} v_{2fy}[/tex] = [tex]-m_{1} v_{1fy}[/tex]

therefore:
[tex]v_{2fy}[/tex] = [tex]\frac{-m_{1} v_{1fy}}{m_{2}}[/tex]=[tex]\frac{-(1.5)(-1.05)}{3.2}[/tex] = 5.04m/s

Now, let's face the x-component:
[tex]v_{2fx}[/tex]=[tex]\frac{m_{1} v_{1ix} - m_{1} v_{1fx}}{m_{2}}[/tex] =
[tex]\frac{m_{1} (v_{1ix} - v_{1fx})}{m_{2}}[/tex] = [tex]\frac{(1.5)(4.5-1.82)}{3.2}[/tex] = 1.26m/s

Now that we have the two components, we can find:

[tex]v_{2f} [/tex] = [tex] \sqrt{ v_{2fx}^2 + v_{2fy}^2 } [/tex] = [tex] \sqrt{5.04^{2} + 1.26^{2} } [/tex] = 6.35m/s

Lastly, the angle can be found with trigonometry:

β = tan⁻¹([tex] \frac{ v_{2fy} }{ v_{2fx} } [/tex]) = tan⁻¹([tex] \frac{ 1.26} }{ 5.04} } [/tex]) = 14°



A positive charge of 6.0 × 10-4 C is in an electric field that exerts a force of 4.5 × 10-4 N on it. What is the strength of the electric field

Answers

The magnitude of the electric force exerted on a charge in an electric field is given by

[tex] F=qE [/tex]

where

q is the charge

E is the magnitude of the electric field


In this problem, we have a charge of [tex] q=6.0 \cdot 10^{-4} C [/tex], while the force exerted on it is [tex] F=4.5 \cdot 10^{-4}N [/tex], so we can rearrange the previous formula to calculate the magnitude of the electric field:

[tex] E=\frac{F}{q}=\frac{4.5 \cdot 10^{-4} N}{6.0 \cdot 10^{-4} C}=0.75 N/C [/tex]


Answer:

0.75

Explanation:

A laser beam of wavelength 439.4 nm is inci- dent on two slits 0.306 mm apart.
How far apart are the bright interference fringes on a screen 6.08 m away from the double slits?
Answer in units of cm.

Answers

Constructive interference happens when this condition is satisfied:
[tex]dsin(\theta)=m\lambda; m=1,2,3,4,5,...[/tex]
This means that the difference in the distance traveled by the waves, from the slit to the screen, is equal to the whole multiple of the wavelength.
If we say that the distance between two interference fringes is much smaller than the distance from the slit to the screen, we can use the following approximation:
[tex]sin(\theta)=\frac{y}{L}[/tex]
Finally for the bright spots we have:
[tex]y=\frac{m\lambda L}{d}[/tex]
The spacing between bright spots is:
[tex]y=\frac{\lambda L}{d}=\frac{6.08\cdot 439.4\cdot 10^{-9}}{0.306 \cdot 10^{-3}}=8.73mm[/tex]



Please help! A net force of 2.0 N acts on a 2.0-kg object for 10 seconds. What is the object’s kinetic energy after that 10 seconds (assuming the object starts from rest and there is no friction)?

Answers

For Newton's second law, the force F applied to the object of mass m will cause an acceleration a of the body:
[tex]F=ma[/tex]
So, the acceleration is 
[tex]a= \frac{F}{m}= \frac{2.0 N}{2.0 kg}=1 m/s^2 [/tex]

The object undergoes through this acceleration for 10 seconds, t=10 s. Since it is an accelerated motion, we can find its final velocity after 10 seconds:
[tex]v_f = v_i + at=0 m/s+(1m/s^2)(10 s)=10 m/s[/tex]
where [tex]v_i[/tex] is the initial velocity of the object, which is zero since it starts from rest.

Finally we can calculate the final kinetic energy of the object, which is given by
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(2.0 kg)(10 m/s)^2=100 J [/tex]

As you add or remove neutrons from the nucleus, isotopes are formed

Answers

That's true: isotopes are chemical elements with same atomic number (i.e. same number of protons) but different mass number (i.e. different number of protons+neutrons, so if the number of protons is the same, the number of neutrons should be different). 
For instance, hydrogen has three different isotopes: 
- protium, with only one proton in the nucleus
- deuterium, with one proton and one neutron
- tritium, with one proton and two neutrons

A 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equilibrium position and released, what is its maximum speed while it oscillates?

Answers

The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
[tex]E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2 [/tex]
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
[tex]E=U_{max}= \frac{1}{2}k(x_{max})^2 [/tex]

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
[tex]E=K_{max}= \frac{1}{2}m(v_{max})^2 [/tex]

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
[tex]U_{max}=K_{max}[/tex]
[tex] \frac{1}{2}k(x_{max})^2 = \frac{1}{2}m(v_{max})^2 [/tex]
[tex]v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s [/tex]

The maximum speed of the 0.40 kg ball while it oscillates is approximately 1.10 m/s. This is found using the conversion of potential energy into kinetic energy at the equilibrium position. The mechanical energy principles of simple harmonic motion are applied.

To determine the maximum speed of a ball oscillating on a spring, we can use principles from simple harmonic motion. Given:

Mass of the ball, m = 0.40 kgSpring constant, k = 12 N/mDisplacement from equilibrium, x = 0.20 m

We start by calculating the total mechanical energy when the ball is at maximum displacement. The potential energy at maximum displacement is given by:

U = 0.5 * k * x2 = 0.5 * 12 N/m * (0.20 m)2 = 0.24 J

Since there is no damping, this total mechanical energy converts entirely into kinetic energy at the equilibrium position where the speed is maximum. The kinetic energy K at this point is:

K = 0.5 * m * [tex]v_{max}^2[/tex] = 0.24 J

Solving for [tex]v_{max}[/tex] :

[tex]v_{max}[/tex] = √((2 * K) / m) = √((2 * 0.24 J) / 0.40 kg) = √(1.2) ≈ 1.10 m/s

Therefore, the maximum speed of the ball while it oscillates is approximately 1.10 m/s.

When red light and green light shine on the same place on a piece of white paper, the spot appears t obe?

Answers

When red light and green lights shine on the same place on a piece of white paper, the spot appears to be Yellow in color. Complementary colors are always located directly across from each other on the graphic. For example cyan is located across from red, magenta across from green, and yellow across from blue. When primary color mix the produce secondary colors , for example a red light and a green light will combine to form a yellow color. The production of various colors of light by the mixing of the three primary colors of light is known as color addition.

Answer:

yellow

Explanation:

red + green light makes yellow

Which has a higher acceleration: a 10-kg object acted upon with a net force of 20 N or an 18-kg object acted on by a net force of 30 N?

A. Both accelerate at the same rate.

B. the 10-kg object

C. the 18-kg object

Answers

the first one accelerate at 2m/s
the second one accelerates at 1.6666m/s

therefore statement b. the 10kg object is your answer

hope this helps :)

Answer:

the answer is B, 10-kg object

Explanation:

Because of Newton we know that if a force is applied to an object, it will have an acceleration on the same direction of the force applied.

[tex]F=m*a\\\\where:\\m=mass\\a=acceleration[/tex]

in order to obtain the acceleration we have to reordenate the formula:

[tex]a=\frac{F}{m}[/tex]

The acceleration of the 10 kg object is:

[tex]a=\frac{20N}{10kg}=2\frac{m}{s^2}[/tex]

The acceleration of the object of 18 kg is:

[tex]a=\frac{30N}{18kg}=1.67\frac{m}{s^2}[/tex]

The biggest acceleration was for the object with less mass.

Calculate δe, if the system absorbs 7.24 kj of heat from the surroundings while its volume remains constant (assume that only p−v work can be done). express your answer using three significant figures.

Answers

I believe the answer is 7.24 kJ.
From the equation ΔE = dW + dQ; where W is the work done on/by the system and Q is the heat the system absorbs/loses.
Therefore; ΔE = 72.4 kJ since the system has bot done any p-v work (dV= zero) and has absorbed heat.

Answer:

[tex]\delta Q = \delta E = 7.24 kJ[/tex]

Explanation:

Heat absorbed by the system is given as

[tex]\delta Q = 7.24 kJ[/tex]

now from first law of thermodynamics we know that

[tex]\delta Q = \delta E + W[/tex]

here

W = work done by the system

[tex]\delta E[/tex] = change in internal energy

also we know that when volume of the system remains same then work done by the system must be zero

[tex]W = 0[/tex]

so from above equation

[tex]\delta Q = \delta E = 7.24 kJ[/tex]

High-voltage power lines are a familiar sight throughout the country. the aluminum (resistivity = 2.82 x 10-8Ï m) wire used for some of these lines has a cross-sectional area of 4.1 x 10-4 m2. what is the resistance of 8.2 kilometers of this wire?

Answers

The formula that relates the resistance R of a wire with the resistivity [tex]\rho[/tex] of the material of the wire is
[tex]R= \frac{\rho L}{A} [/tex]
where L is the length of the wire (in our problem, [tex]L=8.2 km=8200 m[/tex]) and A is the cross sectional area (in our problem, [tex]A=4.1 \cdot 10^{-4} m^2[/tex]).
For the aluminim wire ([tex]\rho =2.82 \cdot 10^{-8} \Omega \cdot m[/tex]) of the problem, the resistance is
[tex]R= \frac{(2.82 \cdot 10^{-8}\Omega m)(8200 m)}{4.1\cdot 10^{-4}m^2}=0.564 \Omega [/tex]

Two Jupiter-size planets are released from rest 1.40×10^11m apart. What is their speed as they crash?,

Answers

To solve this problem, we derive Newton’s Law of Universal Gravitation as the basis of computation

Where: M₁ = mass of planet #1

M₂ = mass of planet #2

M = total mass

R₁ = radius of planet #1

R₂ = radius of planet #2

d₁ = initial distance between planet centers

d₂ = final distance between planet centers

a = semimajor axis of plunge orbit

v₁ = relative speed of approach at distance d₁

v₂ = relative speed of approach at distance d₂

 

To determine velocity during the impact of two heavenly bodies, the solution is as follows:

M₁ = M₂ = 1.8986e27 kilograms

M = M₁ + M₂ = 3.7972e27 kg

G = 6.6742e-11 m³ kg⁻¹ sec⁻²

GM = 2.5343e17 m³ sec⁻²

d₁ = 1.4e11 meters

a = d₁/2 = 7e10 meters

R₁ = R₂ = 7.1492e7 meters

d₂ = R₁ + R₂ = 1.42984e8 meters

v₁ = 0

v₂ = √[GM(2/d₂−1/a)]

v₂ = 59508.4 m/s

The final answer is 59508.4 m/s.

Answer:

this verified kid is way too smart for his own good

Explanation:

Which best describes the current atomic theory?
A since it is only a theory it shouldn’t be used in practice
B it has not been tested enough to be useful for developing technology
C it is the most powerful explanation scientist have to offer at this time
D there is very lil change it will be changed in the future

Answers

Choice-C is a correct statement.

C. it is the most powerful explanation scientists have to offer at this time. I am 100% right, I just finish the test and got it right.

If a material is a good conductor of electricity, it is a ______
A. Metal
B. Nonmetal
C. Metalloid

Answers

Answer:

Metal

Explanation:

If a material is a good conductor of electricity, it is a metal, therefore the correct answer is option A.

What are metalloids?

The elements of the periodic tables that behave as metal, as well as the nonmetal in some chemical or physical aspects, are known as metalloids. Some examples of metalloids are Boron, silicon, germanium, arsenic, antimony, tellurium, etc.

Metals are very good conductors of heat and electricity while on the other hand nonmetals are very poor conductors of heat and electricity.

Metalloids are semiconducting materials that behave as the intermediate behavior between metal and nonmetal.

Thus, If a material is a good conductor of electricity, it is a metal, therefore the correct answer is option A.

Learn more about metalloids from hee the link given below;

brainly.com/question/2548493

#SPJ2

What is the velocity of the object at 20.0s

Answers

So notice that at the beginning of the motion, the object moves to 16m in a span of about 10 seconds. You should know that the velocity in a pvt graph is the slope, so here the v = 1.6 m/s more or less. At 20 seconds, the object doesn't seem to be moving and you see it stays still from 10-26ish seconds. From observing you can tell that there is no velocity, but the slope of this segment is zero so you can be sure of it. 

Which of the following describes the role of a refrigerant in an air conditioner? to absorb heat flow from air molecules circulating around the refrigerant tubing cools the air molecules by adding cool energy to the air as it is circulated around the refrigerant tubing increases the specific heat of the air molecules and allows them to release more heat once cooled decreases the specific heat of the air molecules and allows them to absorb more heat before cooling

Answers

to absorb heat flow from air molecules circulating around the refrigerant tubing i hope this helps you

Answer:

to absorb heat flow from air molecules circuling around the refrigerant tubing

Explanation:

I know this is correct because I just had this question and this was the correct answer.

Highway curves are marked with a suggested speed. If this speed is based on what would be safe in wet weather, estimate the radius of curvaure for a curve marked 50km/h?The coefficient of static friction of rubber on wet concrete is .7, the coefficient of kinetic friction of rubber on wet concrete is .5 .

Answers

Radial (aka centripetal) acceleration keeps cars on an unbanked road in a curve. This acceleration is given by

[tex]a_{rad}= \frac{v^2}{r} [/tex]

The force required to keep in going in a circle is given by Newton's second law, F=ma.  Here a is the radial acceleration from the above equation, so we get a radial force:

[tex]F_{rad}=m \frac{v^2}{r} [/tex]

This force is mediated by the friction force between the tires of the car and the road.  For friction force we use the coefficient of friction and the normal force of the road on the car (which will also be the force of the car on the road).  On an unbanked road this is just the weight of the car, mg.  

The question now is, which coefficient of friction to use?  It might surprise you to find that although the car is moving relative to the road, we should use the static coefficient.  To see why this is the case, think about what the wheel's surface is doing as it meets the road.  As a wheel rotates around and comes into contact with the road surface, it's bottom section is moving in a direction opposite to the car's forward movement.  The point that actually touches the road is moving backward at the same speed that the car moves forward and so it is at rest with the street.  If this were not the case, your tires would always be skidding on the road.  (If you're having trouble seeing this, there are plenty of animations of this phenomenon on the internet). Once a car is skidding it is no longer in control and cannot turn like one that is not skidding.

So now the friction force is just

[tex]F_{fric}= \mu_s F_n=\mu_smg[/tex]

We now have to set these equal and solve for r, the radius of the circle we are traveling in.  Note that mass will cancel and so we don't need to worry about the mass of the cars involved:

[tex]m \frac{v^2}{r}=\mu_smg \\ \\ \frac{v^2}{r}=\mu_sg \\ \\ \frac{v^2}{\mu_sg}=r \\ \\ \frac{(13.89m)^2}{0.7*9.8 \frac{m}{s^2} } =28.12m[/tex]

Where I have converted the speed to m/s for unit consistency. 

The radius of curvature for wet weather is required.

The radius of curvature would be 28.1 m.

[tex]\mu_s[/tex] = Coefficient of static friction for wet concrete = 0.7

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

m = Mass of car

v = Velocity of car = 50 km/h

The force balance is

[tex]\dfrac{mv^2}{r}=\mu_s mg\\\Rightarrow r=\dfrac{v^2}{\mu_s g}\\\Rightarrow r=\dfrac{(50\times \dfrac{1000}{3600})^2}{0.7\times 9.81}\\\Rightarrow r=28.1\ \text{m}[/tex]

The radius of curvature would be 28.1 m.

Learn more:

https://brainly.com/question/11859809?referrer=searchResults

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