What is the electric potential of a 4.5 × 10-5 C charge that has an electric potential energy of 0.027 J?

Answers

Answer 1
The answer is certainly 600 volts, would someone please explain how though?
Answer 2

Answer : Electric potential is 600 V

Explanation :

It is given that,

Electric charge, [tex]q=4.5\times 10^{-5}\ C[/tex]

Electric potential energy, [tex]U=0.027\ J[/tex]

The relation between the electric potential and the electric potential energy is given by :

[tex]U=qV[/tex]

[tex]V=\dfrac{U}{q}[/tex]

[tex]V=\dfrac{0.027\ J}{4.5\times 10^{-5}\ C}[/tex]

[tex]V=600\ V[/tex]

Hence, the electric potential is 600 V.


Related Questions

When populations of two species work together to obtain resources, they are

Answers

cooperation. i think that is it

Answer:

cooperating.

Explanation:

:)

After harvesting their main crops, such as corn and wheat, many farmers plant a second crop called a cover crop. This cover crop grows during the fall and winter and is then plowed over in the spring. What is the main purpose of a cover crop?
A.
to get rid of excess seed rather than storing during the winter
B.
to provide food and shelter for birds and animals during the winter
C.
to prevent water runoff into other parts of the farmers' field
D.
to prevent erosion of the topsoil by wind and water

Answers

it's d. to prevent erosion of the topsoil by wind and water

Answer:

Option (D)

Explanation:

A cover crop is usually defined as a special type of crop that is cultivated in order to increase the productivity of the soil, rather than focusing on the productivity of the crop. These are very commonly used and it helps in the decreasing amount of weeds, controlling pests and diseases, increasing the fertility of the soil, reduction of soil erosion. It also enhances the biodiversity.

Thus, the main purpose of the cover crop is to prevent the erosion of the topsoil by the agents such as water and wind.

Hence, the correct answer is option (D).

for a bohr model,how many protons,electrons,and neutron does Sodium has

Answers

11 protons, 11 electrons, 12 neutrons

To determine an epicentral distance scientists consider the arrival times of what wave types

Answers

 The answer is P-waves and S-waves

An experimenter finds that no photoelectrons are emitted from a particular metal unless the wavelength of light is less than 295 nm. her experiment will require photoelectrons of maximum kinetic energy 2.4 ev. what frequency light should be used to illuminate the metal?

Answers

The frequency of light is approximately [tex]\( 1.016 \times 10^{15} \)[/tex] Hz.

To determine the frequency of light that should be used to illuminate the metal in order to produce photoelectrons with a maximum kinetic energy of 2.4 eV, we can use the relationship between energy, frequency, and wavelength in the context of the photoelectric effect.

The energy of a photon (light particle) is given by Planck's equation:

[tex]\[ E = h \cdot f \][/tex]

Where:

- [tex]\( E \)[/tex] is the energy of the photon in joules (J)

-[tex]\( h \)[/tex] is Planck's constant, approximately [tex]\( 6.626 \times 10^{-34} \)[/tex] J·s

- [tex]\( f \)[/tex] is the frequency of the light in hertz (Hz)

We are given the maximum kinetic energy of the photoelectrons as 2.4 eV. To convert this to joules, we use the conversion factor [tex]\( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \)[/tex]

[tex]\[ E_{\text{max}} = 2.4 \, \text{eV} \times (1.602 \times 10^{-19} \, \text{J/eV}) \]\\\[ E_{\text{max}} = 3.845 \times 10^{-19} \, \text{J} \][/tex]

Now, since the photoelectrons are emitted only when the wavelength of light is less than 295 nm (nanometers), we can use the speed of light equation to relate frequency and wavelength:

[tex]\[ c = f \cdot \lambda \][/tex]

Where:

- [tex]\( c \)[/tex] is the speed of light, approximately [tex]\( 3.00 \times 10^8 \)[/tex] m/s

- [tex]\( f \)[/tex] is the frequency of the light in hertz (Hz)

- [tex]\( \lambda \)[/tex] is the wavelength of the light in meters (m)

First, we convert the wavelength limit of 295 nm to meters:

[tex]\[ \lambda_{\text{max}} = 295 \, \text{nm} \times (1 \, \text{m} / 10^9 \, \text{nm}) \][/tex]

[tex]\[ \lambda_{\text{max}} = 2.95 \times 10^{-7} \, \text{m} \][/tex]

Now, we can rearrange the speed of light equation to solve for frequency:

[tex]\[ f = \frac{c}{\lambda_{\text{max}}} \][/tex]

[tex]\[ f = \frac{3.00 \times 10^8 \, \text{m/s}}{2.95 \times 10^{-7} \, \text{m}} \][/tex]

[tex]\[ f = 1.016 \times 10^{15} Hz[/tex]

Therefore, the frequency of light that should be used to illuminate the metal and produce photoelectrons with a maximum kinetic energy of 2.4 eV is approximately [tex]\( 1.016 \times 10^{15} \)[/tex] Hz.

Which labels correctly identify the layers most closely associated with gamma rays and visible light? Z: Gamma rays X: Visible light X: Gamma rays Z: Visible light Z: Gamma rays Y: Visible light Y: Gamma rays Z: Visible light

Answers

The answer is Z: Gamma rays X: Visible light. Gamma ray is the shortest and has the greatest frequency and power. That is why it is in letter Z.

Answer:

A on edge

Explanation:

The Doppler effect is a shift in the _______________ of an oscillation caused by the  motion of the source of the oscillation, and occurs at speeds below the speed of sound.  a) amplitude     b) frequency      c) speed 

Answers

b. frequency is the answer 
The question was written by a teacher who isn't quite clear
on the concepts.  Both the question and the list of answers
are misleading.  If you're serious about learning this subject,
you're better off ignoring this question, or at least not working
too hard on it.

... The Doppler shift is an APPARENT shift in the wavelength and
frequency of the waves that come from a source of oscillations. 

... The shift is measured by the observer.  He sees and hears
different frequency and wavelength, compared to what's actually
coming out of the source.

... The shift can be the result of EITHER the source OR the observer.
Any time the distance between them is changing, the shift occurs.
It doesn't matter which one is moving.

A water bed weighs 1025 N, and is 1.5 m wide by 2.5 m long. How much pressure does the water bed exert if the entire lower surface of the bed makes contact with the floor?

a.

3.6 x 10^2 Pa

b.

2.7 x 10^4 Pa


c.

1.2 x 10^3 Pa

d.

2.7 x 10^2 Pa

Answers

By definition we have to:
 [tex]P = F / A [/tex]
 Where,
 F: Force
 A: area
 For the area we have:
 [tex]A = (w) * (l) [/tex]
 Where,
 w: width
 l: long
 Substituting values we have:
 [tex]A = (1.5) * (2.5) A = 3.75 m ^ 2[/tex]
 Substituting the values we have:
 [tex]P = 1025 / 3.75 P = 273.3333333 N / m ^ 2[/tex]
 Rewriting we have:
 [tex]P = 2.7 * 10 ^ 2 Pa [/tex]
 Answer:
 
d.
 
2.7 x 10 ^ 2 Pa
Answer:
D) 2.7 * 10² Pa

Explanation:
To answer this question, we will use the following equation:
pressure = [tex] \frac{force}{area} [/tex]

1- getting the area:
The water bed is in the form of a rectangle. Therefore, the area can be calculated as follows:
area = length * width
We are given that:
lenth = 2.5 m
width = 1.5 m
This means that:
area = 2.5 * 1.5 = 3.75 m²

2- getting the pressure:
We noted that:
pressure = [tex] \frac{force}{area} [/tex]

We have the force = 1025 N
We calculated the area = 3.75 m²

Substitute in the above equation to get the pressure as follows:
pressure = [tex] \frac{1025}{3.75} [/tex]
pressure = 2.7 * 10² Pa

Hope this helps :)

A heat engine takes in 840 kJ per cycle from a heat reservoir. Which is not a possible value of the engine's heat output per cycle?

Answers

We have by the first law of thermodynamics tha energy is preserved, hence we cannot have over 840kJ per cycle. We have by the laws of thermodynamics (the 2nd one in specific) that the entropy of a system cannot increase. We cannot have an output of  840 kJ per cycle from a heat engine because then that would mean that the entropy would stay the same, while any heat engine increases it. Hence, any value [tex] \geq 840 kJ [/tex] is acceptable.

A rock is thrown vertically upward from ground level at time t = 0. at t = t, it passes the top of the tower and ât later (i.e., when t = t + ât) it reaches the maximum height. what is the height of the tower? leave g as g â do not substitute with numbers

Answers

...........................................................

"The height of the tower is given by the equation:

[tex]\[ h = \frac{1}{2} g (t + \hat{t})^2 - \frac{1}{2} g t^2 \][/tex]

 This equation represents the difference in height between the maximum height the rock reaches and the height it reaches at time \( t \) when it passes the top of the tower. Here, \( g \) is the acceleration due to gravity, \( t \) is the time it takes for the rock to reach the top of the tower, and \( \hat{t} \) is the additional time it takes for the rock to reach the maximum height after passing the top of the tower.

To find the height of the tower, we can simplify the equation:

[tex]\[ h = \frac{1}{2} g ((t + \hat{t})^2 - t^2) \][/tex]

[tex]\[ h = \frac{1}{2} g (t^2 + 2t\hat{t} + \hat{t}^2 - t^2) \][/tex]

[tex]\[ h = \frac{1}{2} g (2t\hat{t} + \hat{t}^2) \][/tex]

[tex]\[ h = g t\hat{t} + \frac{1}{2} g \hat{t}^2 \][/tex]

 Since [tex]\[ u = g \hat{t} \][/tex]is the time it takes for the rock to reach the maximum height from the top of the tower, and at the maximum height the velocity of the rock is zero, we can use the kinematic equation for the final velocity \( v \) at the maximum height:

[tex]\[ v = u + g t \][/tex]

Here, \( u \) is the initial velocity (which is the velocity of the rock at the top of the tower), \( g \) is the acceleration due to gravity, and \( t \) is the time taken to reach the maximum height, which is \( \hat{t} \). At the maximum height, \( v = 0 \), so:

[tex]\[ 0 = u - g \hat{t} \][/tex]

[tex]\[ u = g \hat{t} \][/tex]

 Now, we can substitute \( u \) back into the height equation:

[tex]\[ h = g t\hat{t} + \frac{1}{2} g \hat{t}^2 \][/tex]

[tex]\[ h = g t\left(\frac{u}{g}\right) + \frac{1}{2} g \left(\frac{u}{g}\right)^2 \][/tex]

[tex]\[ h = u t + \frac{1}{2} \frac{u^2}{g} \][/tex]

Since \( u t \) represents the distance the rock would have traveled in time \( t \) if it continued at a constant velocity \( u \), and \( \frac{1}{2} \frac{u^2}{g} \) represents the total height the rock would reach if it continued to move upward with initial velocity \( u \) and was only acted upon by gravity, the term \( u t \) is actually the height of the tower. Thus, the height of the tower is:

[tex]\[ h = u t \][/tex]

 This is the final answer, and it shows that the height of the tower is the product of the initial velocity of the rock at the top of the tower and the time it takes to reach the top of the tower. The term[tex]\( \frac{1}{2} \frac{u^2}{g} \)[/tex]is not needed for the height of the tower, as it represents the additional height the rock reaches after passing the top of the tower until it reaches the maximum height."

a 13 kg sled is moving at a speed of 3.0 m/s. at which of the following speeds will the sled have tace as mucb as the kinetic

Answers

a 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speed will the sled have twice as much kinetic energy?

4.2 m/s

A child goes down a playground slide with an acceleration of 1.12 m/s2 .find the coefficient of kinetic friction between the child and the slide if the slide is inclined at an angle of 27.0 ∘ below the horizontal.

Answers

Let's consider all the forces acting on the direction parallel to the surface of the slide. We have in total two forces in this direction:
1) the component of the weight parallel to the surface, which is equal to
[tex]W_{//}= mg \sin \alpha[/tex]
where [tex]\alpha =27.0 ^{\circ}[/tex], directed toward the bottom of the slide;
2) The frictional force, which is given by
[tex]F_f = \mu m g \cos \alpha[/tex]
where [tex]\mu [/tex] is the coefficient of kinetic friction that we want to find, and this force is directed towards the top of the slide, so in the opposite direction of the previous force.

For Newton's second law, the algebraic sum of the forces acting in this direction must be equal to the product of the mass and the acceleration of the child:
[tex]\sum F = ma[/tex]

So, if we explicite the forces, this becomes
[tex]mg \sin \alpha - \mu mg \cos \alpha = ma[/tex]
from which we can isolate [tex]\mu[/tex], and by using the information about the acceleration ([tex]a=1.12 m/s^2[/tex]), we find:
[tex]\mu = \frac{g \sin \alpha -a}{g \cos \alpha}= \frac{(9.81 m/s^2)(\sin 27^{\circ})-1.12 m/s^2}{(9.81 m/s^2)(\cos 27^{\circ})} =0.38 [/tex]
Final answer:

The coefficient of kinetic friction is obtained by considering the forces acting on the child as they slide. These are the force of gravity (mg sin θ) and the force of kinetic friction (μk mg cos θ). Setting these forces equal gives us an equation that can be solved to find the coefficient of kinetic friction, μk.

Explanation:

To figure out the coefficient of kinetic friction, we need to consider the forces acting on the child as they slide down the playground. There are two key forces to consider, the force caused by gravity which is pulling the child down the slide, and the force of kinetic friction which is trying to slow the child down.

The force of gravity working to pull the child down the slide, or the component of the weight down the slope is given as mg sin θ. Here m is the mass of the child, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the slide. The force of kinetic friction, which opposes the motion, is expressed as μk mg cos θ, where μk is the coefficient of kinetic friction we're trying to find.

Since the child is sliding down the slide with an acceleration of 1.12 m/s², the force due to gravity exceeds the force of friction. Setting up the equation μk mg cos θ = mg sin θ - ma, where a is the acceleration of 1.12 m/s², and solving for μk gives us the coefficient of kinetic friction between the child and the slide.

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A ball is thrown at an angle of 45° to the ground. if the ball lands 87 m away, what was the initial speed of the ball? (round your answer to the nearest whole number. use g â 9.8 m/s2.)

Answers

just try your best best friend everyone

A sinusoidal wave travels with speed 250 m/s . its wavelength is 3.5 m . part a what is its frequency

Answers

The relationship between the frequency f of a wave, its wavelength [tex]\lambda[/tex] and its speed v is given by
[tex]v=\lambda f[/tex]
The wave of our exercise has a speed of [tex]v=250 m/s[/tex] and a wavelength of [tex]\lambda=3.5 m[/tex], so re-arranging the previous equation we can find its frequency:
[tex]f = \frac{v}{\lambda}= \frac{250 m/s}{3.5 m}=71.4 Hz [/tex]

Five metal samples, with equal masses, are heated to 200oC. Each solid is dropped into a beaker containing 200 ml 15oC water. Which metal will cool the fastest?
A) aluminum
B) copper
C) gold
D) platinum

Answers

B) copper is the answer       shoot im late 

Answer:

aluminum...

Explanation:

What potential difference is needed to accelerate a he+ ion (charge +e, mass 4u) from rest to a speed of 1.0×106 m/s ? express your answer using two significant figures?

Answers

For the law of conservation of energy, the loss in potential energy of the He+ ion should be equal to the gain in kinetic energy:
[tex]-\Delta U=\Delta K[/tex]
which can be rewritten as
[tex]-q \Delta V = \frac{1}{2}mv^2 [/tex]
where
[tex]q=+e = 1.6 \cdot 10^{-19}C[/tex] is the charge of the ion,
[tex]m=4u=4\cdot 1.67 \cdot 10^{-27}kg[/tex] is the mass of the ion,
[tex]v=1.0 \cdot 10^6 m/s[/tex] is the speed of the ion.
By using these values, we find the potential difference needed:
[tex]\Delta V = \frac{1}{2} \frac{mv^2}{-q}= \frac{1}{2} \frac{(4\cdot 1.67 \cdot 10^{-27}kg)(1.0 \cdot 10^6 m/s)^2}{-1.6 \cdot 10^{-19}C}= -20875 V=-21 kV [/tex]
and the negative sign means the final point is at lower voltage than the initial point, and this is correct, because the ion has positive charge and a positive charge travels naturally from higher voltages to lower voltages.

The potential difference required to accelerate the helium ion to a speed of   [tex]1.0 \times {10^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] is [tex]\boxed{21\,{\text{kV}}}[/tex] or [tex]\boxed{20875\,{\text{V}}}[/tex].

Further Explanation:

The potential difference, through which the helium ion passes, provides the potential energy and this potential energy provided by the potential difference to the Helium ion leads to the increase in the kinetic energy of the ion.

The charge on the [tex]H{e^ + }[/tex] ion is equal to the charge of one electron and the mass of the helium atom is equal to the mass of [tex]4\,{\text{amu}}[/tex].

The expression for the conservation of the energy of the Helium ion in the potential difference is:

[tex]\Delta PE = \Delta KE[/tex]  

Substitute [tex]q\Delta V[/tex] for [tex]\Delta PE[/tex] and [tex]\dfrac{1}{2}m{v^2}[/tex] for [tex]\Delta KE[/tex] in above expression.

[tex]\begin{aligned}q\Delta V &= \frac{1}{2}m{v^2} \hfill \\\Delta V &= \frac{{m{v^2}}}{{2q}} \hfill \\\end{aligned}[/tex]  

Here, [tex]\Delta V[/tex] is the potential difference, [tex]m[/tex] is the mass of the ion, [tex]q[/tex] is the charge over the ion.

Substitute [tex]1.6 \times {10^{ - 19}}\,{\text{C}}[/tex] for [tex]q[/tex], [tex]\left( {4 \times 1.67 \times {{10}^{ - 27}}\,{\text{kg}}} \right)[/tex] for [tex]m[/tex] and [tex]1.0 \times {10^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]v[/tex] in above expression.

[tex]\begin{aligned}\Delta V &= \frac{{\left( {4 \times 1.67 \times {{10}^{ - 27}}\,{\text{kg}}} \right){{\left( {1.0 \times {{10}^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}} \right)}^2}}}{{2\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)}} \\&= \frac{{6.68 \times {{10}^{ - 15}}}}{{3.2 \times {{10}^{ - 19}}}}\,{\text{V}} \\&= 2{\text{0875}}\,{\text{V}} \\&\approx {\text{21}}\,{\text{kV}}\\\end{aligned}[/tex]  

Thus, the potential difference required to accelerate the helium ion to a speed of [tex]1.0 \times {10^6}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] is [tex]\boxed{21\,{\text{kV}}}[/tex] or [tex]\boxed{20875\,{\text{V}}}[/tex].

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords:  Potential energy, potential difference, accelerate, kinetic energy, speed, helium ion, charge, significant figures, rest to a speed.

A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5 × 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction. the mass of a proton is 1.673 × 10−27 kg. find the magnitude of the field. answer in units of t.

Answers

The magnetic force experienced by the proton is given by
[tex]F=qvB \sin \theta[/tex]
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and [tex]\theta[/tex] the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so [tex]\sin \theta=1[/tex] and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
[tex]F=ma[/tex]

So we have
[tex]ma=qvB[/tex]
from which we can find the magnitude of the field:
[tex]B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T [/tex]

You are asked to design a horizontal curve with a 40-degree central angle (' = 40) for a two-lane road with 11-ft lanes. the design speed is 70 mi/h and superelevation is limited to 0.06 f
a. True
b. Falset. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h

Answers

The answer is b. False. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h. Give it a try man
 
Final answer:

To design a horizontal curve for a two-lane road, you can calculate the radius, degree of curvature, and length of the curve using formulas.

Explanation:

To design a horizontal curve, we need to calculate the radius, degree of curvature, and length of the curve. First, we can calculate the radius using the formula: R = (V2) / (15 * f), where R is the radius in feet, V is the design speed in miles per hour, and f is the superelevation. Plugging in the values, we get: R = (702) / (15 * 0.06) = 43,333.33 feet. The degree of curvature can be calculated using the formula: (180 * L) / (π * R), where L is the length of the curve in feet. Since the central angle is given as 40 degrees, the degree of curvature is also 40 degrees. Finally, we can calculate the length of the curve using the formula: L = (π * R * d) / 180, where d is the degree of curvature. Plugging in the values, we get: L = (π * 43,333.33 * 40) / 180 = 9,632.87 feet.

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what is air resistance?

Answers

Basically, the "push back" of the air against the object moving forward.  It acts similar to friction, except the resistance is based on the shape of the object instead of the interaction of the object and the ground

Air resistance is the pushing of air against an object that is moving. Both the air and the object rub together and this slows the down the object and making it use more energy to reach a required speed. If an object moves faster, the greater the air resistance it encounters.

A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential difference? Recall that Fe = qE.

Answers

The electric potential difference is 7.5 V.

Electric potential difference between two points is defined as the work done in moving unit positive charge between the two points. Electric potential difference can also be defined as the work done W per unit charge q in an electric field.

[tex]\Delta V=\frac{W}{q}[/tex]......(1)

Work done by a force is the product of the force F and the displacement s made in the direction of the force. Assuming that the displacement is parallel to the line of action of the force,

[tex]W=F.s[/tex]......(2)

Rewrite equation (1) using equation (2).

[tex]\Delta V=\frac{Fs}{q}[/tex]......(3)

Substitute the given values of force, displacement and charge in the equation (3).

[tex]\Delta V=\frac{Fs}{q}\\ =\frac{(25*10^-^6N)(15m)}{(50*10^-^6C)} \\ =7.5V[/tex]

The potential difference between the two points is 7.5 V


You are at home in your air conditioned garage. You are planning a family road trip from New York to Florida. You are lnfating the tires of your suv. You fill each car-tire with about 10 Liters of air. The temperature inside your air conditioned garage is about 10 degrees Celsius. Your family packs the car, and you begin your trip to Florida. The temperature in New York is about 21 degrees Celsius, and the happen temperature in Florida is 32 degrees Celsius. Assuming the pressure stays constant from the garage all the way to Florida, what could to your car tires during your road trip.

Write a paragraph that makes a claim about what happen to the car tires

Answers

The tires deflated and so that means that you won’t be able to travel

If electrons of energy 12.8 ev are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photons that are emitted by the excited gas?

Answers

Final answer:

When 12.8 eV energy electrons excite hydrogen atoms, the atoms are elevated to an energy level just below ionization. To determine the specific energy state, the energy formula for hydrogen atoms is used, then the energies of emitted photons are found by calculating the energy difference between the excited and final states.

Explanation:

If electrons with an energy of 12.8 eV are incident on hydrogen atoms in their ground state, we first need to determine the energy level to which the hydrogen atoms are excited. The ground state energy of hydrogen is -13.6 eV. If 12.8 eV is provided, the electron reaches an energy level just below 0 eV (ionization energy), since 12.8 eV is not sufficient to completely ionize the atom (13.6 eV needed).

To find out which energy level the electron will jump to, we use the formula for the total energy of an electron in a hydrogen atom, En = (- 13.6 eV/n²), and solve for n.

Once the electron is excited, it will eventually drop back to a lower energy state, emitting a photon in the process.

The energy of the emitted photon can be calculated using the difference in energy levels, ΔE = E1 - Ef.

Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sound waves?

Answers

Ans: Beat frequency = [tex]f_b[/tex] = 4Hz

Explanation: 
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. According to this interference, sound will be soft or loud.

Hence. the formula is:
Beat frequency = [tex]f_b = |f_2 - f_1|[/tex]

Since,
[tex]f_1 = 480Hz[/tex]
[tex]f_2 = 484Hz[/tex]

Therefore,
Beat frequency = [tex]f_b = |484 - 480|[/tex]

=> Beat frequency = [tex]f_b = 4Hz[/tex]
-i

A force of 35 n acts on an object which has a mass of 5.4 kg. what acceleration (in m/s2) is produced by the force

Answers

∑F = ma
a = ∑F/m
a = 35 N / 5.4 kg
a = 6.5 m/s²

A grindstone, initially at rest, is given a constant angular acceleration so that it makes 20.0 rev in the first 8.00 s. what is its angular acceleration?

Answers

The angle covered by the grindstone during this time is, converting into radians,
[tex]\theta = 20 rev = 20 rev \cdot \frac{2 \pi rad}{rev} =125. 6rad[/tex]

This is a rotational motion with constant angular acceleration [tex]\alpha[/tex], and with initial angular speed [tex]\omega _0 =0[/tex]. The angle covered in a time t is given by
[tex]\theta (t)= \omega_0 t+ \frac{1}{2} \alpha t^2 = \frac{1}{2} \alpha t^2 [/tex]
because the initial angular speed is zero.
Using t=8.00 s, we can find the value of angular acceleration:
[tex]\alpha = \frac{2 \theta}{t^2}= \frac{2 \cdot 125.6 rad}{(8.0 s)^2}=3.93 rad/s^2 [/tex]

The angular acceleration of a grindstone that makes 20 revolutions in 8 seconds from rest, convert the revolutions to radians and use the rotational kinematic equation, resulting in an angular acceleration of approximately 3.93 radians/s².

The angular acceleration of a grindstone that makes 20 revolutions in 8 seconds, starting from rest.

To find the angular acceleration, we can use the kinematic equation for rotational motion:

θ = ω₀t + 1/2αt²

where:

θ is the angular displacement in radians,

ω₀ is the initial angular velocity (which is 0 because it starts from rest),

t is the time in seconds,

and α is the angular acceleration in radians per second squared.

First, we convert the revolutions to radians:

20 revolutions * 2π radians/revolution = 40π radians

We then have:

40π radians = 0 + 1/2α(8²)

Solving for α gives:

α = (40π radians) / (1/2 * 64 s²)

α = 80π / 64 radians/s²

α ≈ 3.93 radians/s²

Therefore, the angular acceleration of the grindstone is approximately 3.93 radians/s².

"a metal sphere of radius 5.00 cm is initially uncharged. how many electrons would have to be placed on the sphere to produce an electric field of magnitude 1.53 ✕ 105 n/c at a point 8.64 cm from the center of the sphere?"

Answers

Outside the radius of the sphere, the electric field generated by a charged sphere (with charge Q on its surface) at a distance r from the centre is equivalent to the electric field generated by a single-point charge with total charge Q:
[tex]E= k_e \frac{Q}{r^2} [/tex]
The problem asks to find the electric field at r=8.64 cm, while the radius of the sphere is R=5.00 cm, so r>R and we are exactly in this condition.

Re-arranging the previous formula, we can solve to find the total charge Q on the sphere. Using [tex]r=8.64 cm=8.64 \cdot 10^{-2}m[/tex] and [tex]E=1.53 \cdot 10^5 N/C[/tex], we find
[tex]Q= \frac{Er^2}{k_e} = \frac{(1.53 \cdot 10^5 N/C)(8.64 \cdot 10^{-2}m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} =1.27 \cdot 10^{-7}C[/tex]

This is the total charge on the surface. If we want to find the number of electrons composing this charge, we should divide the total charge by the charge of a single electron, e:
[tex]N= \frac{Q}{e} = \frac{1.27 \cdot 10^{-7}C}{1.6 \cdot 10^{-19}C}=7.9 \cdot 10^{11} [/tex]
and this is the number of electrons.

At the moment the acceleration of the blue ball is 9.40 m/s2, what is the ratio of the magnitudes of the force of air resistance to the force of gravity acting on the blue ball? use 9.81 m/s2 for the standard value of g at the location of the experiment.

Answers

Let us see. We have that Newton's Law holds in this case, so [tex]\sum{F}=m*a[/tex] where a is the acceleration of the ball, F the forces acting on it and m its mass. There are 2 forces acting on it, the force of gravity W and air resistance R. We can easily see that these two act in opposite directions (gravity towards the ground, resistance upwards), so |ΣF|=W-R (1) . We have that W=m*g where g=9.81 m/s2. We also have that ΣF=9.40*m. From this relationship, substituting into (1) all the known values, we get that R=m(9.81*-9.40)=m*0,41. We take then the ratio R/W. This is equal to:[tex] \frac{0.41 m }{9.81m} =0.042=4.2\%[/tex]. This is our final answer.

The ratio of the magnitudes of the force of air resistance to the force of gravity acting on the ball is approximately 1:24.

The student is asking about the ratio of the force of air resistance to the force of gravity acting on a blue ball, when the ball has an acceleration of 9.40 m/s2. To calculate this ratio, we can use Newton's second law, F = ma, where F is the force, m is the mass of the ball, and a is the acceleration of the ball due to the net force acting on it. If we denote the force of air resistance as Fair and the force of gravity as Fgrav, we can write Fnet = Fgrav - Fair, since the air resistance works in the opposite direction of gravity.

Given that the acceleration due to gravity is approximately 9.81 m/s2, the force due to gravity (weight) can be expressed as Fgrav = mg. Here, the acceleration is less than g due to air resistance, thus we can write Fnet = ma = mg - Fair. Since the acceleration of the ball is given as 9.40 m/s2, we can now express Fair as mg - ma. Taking the ratio of Fair to mg (the weight), we get:

(mg - ma) / mg = (g - a) / g = (9.81 - 9.40) / 9.81.

Therefore, the ratio of the magnitudes of the force of air resistance to the force of gravity is equal to 0.0418, which simplifies to approximately 1:24.

Does someone have this??

Answers


Let's now answer number 1:

Momentum is mass x velocity

p = mv

Block 1 is 10.00kg and is moving at a speed of 4.50m/s. Just put in what you know to get the answer:


p= 10.00kg x 4.50 m/s = 45 kg.m/s

The momentum of block 1 is 45 kg.m/s.

2. In a collision if one object bumps into another object and they both travel the same direction, the scenario would be considered as a PERFECT inelastic collision. 

We can solve for unknowns using this formula:

[tex]m_{1}v_{1i}+m_{2}v_{2i} = (m_{1}+m_{2})v_{f}[/tex]
Where:
m1 = mass of object 1
m2 = mass of object 2
v1i = velocity of object 1 before collision
v2i = velocity of object 2 before collision

So before we solve for what you need, let's list down what you know based on the problem. 
m1 = 10.00kg (Block1)
m2 = 25.0kg   (Block2)
v1i = 4.50 m /s (Block1)
v2i = 0 m/s       (Block2)

You may be wondering why the velocity of object 2 is 0m/s. Well, the problem says that Block 2 was initially at rest, then that means it is not moving, so it's speed would be 0 m/s.

Now use the formula and put in what you know and derive what you do not know. 

[tex]m_{1}v_{1i}+m_{2}v_{2i} = (m_{1}+m_{2})v_{f}[/tex]
[tex](10.00kg)(4.50m/s)+(25.0kg)(0m/s) = (10.00kg + 25.00kg)v_{f}[/tex]
[tex](45kg.m/s)+(0) = (35kg)v_{f}[/tex]
[tex](45kg.m/s)= (35kg)v_{f}[/tex]
[tex] \frac{45kg.m/s}{35kg} = v_{f} [/tex]
[tex]1.29m/s = v_{f}[/tex]

Final velocity or vf of both blocks after collision is 1.29m/s.

3. To get the momentum after collision,just keep in mind that momentum is conserved even after collision. So your answer would be 45kg.m/s. If it is necessary for you to solve for it, just solve for it using the equation on the right side, which is (m1+m2)vf.

(m1+m2)vf
=(10.00kg+25.0kg)(1.29m/s)
=(35kg)(1.29m/s)
=45.15kg.m/s or 45 kg.m/s

As for the next problem, a cannon being fired is also considered a collision and this type of collision is called explosion, where the momentum before and after collision is zero. Why? Because before firing, the ball is not moving and neither is the cannon. 

So now that we know this, using the formula of explosion we can solve for what we need.

[tex]m_{1}v_{1}+m_{2}v_{2}=0[/tex]

Where:
m1 = mass of object 1
m2 = mass of object 2
v1= velocity of object 1
v2 = velocity of object 2

[tex]m_{1}v_{1}+m_{2}v_{2}=0[/tex]
[tex](450kg)(v1) + (2.75kg)(23.2m/s)=0[/tex]
[tex](450kg)(v1) + (63.8kg.m/s)=0[/tex]
[tex](450kg)(v1) =0 - 63.8kg.m/s[/tex]
[tex](450kg)(v1) =- 63.8kg.m/s[/tex]
[tex]v1 = \frac{-63.8k.m/s}{450kg} [/tex]
[tex]v1 = -0.14 m/s[/tex]

The cannon travelled -0.14m/s. 
Notice that the value is negative, this means that it went the opposite direction it initially traveled or it traveled backwards. 

What is the energy of the spring-mass system when the mass first passes through the equilibrium position? (you may wish to include a logical test to help you find when this occurs)?

Answers

The spring-mass system moves by simple harmonic motion, where there is a continuous conversion from elastic potential energy to kinetic energy and viceversa.

The total mechanical energy of the system at any moment of the motion is
[tex]E=U+K= \frac{1}{2}kx^2 + \frac{1}{2}mv^2 [/tex]
where the first term U is the elastic potential energy, with k being the spring constant and x the displacement of the spring with respect to its rest position, and the second term K is the kinetic energy, with m being the mass of the object attached to the spring and v its speed.

The total energy E is constant during the oscillation of the spring, but the values of U and K change. In fact, when the displacement of the spring is maximum (x is maximum), then all the energy is potential energy U, because the speed of the object is zero (it's the moment when the mass is changing direction). On the contrary, when the mass crosses the equilibrium position (rest position) of the spring, then the potential energy is zero (U=0) because the displacement is zero (x=0), and so all the energy is kinetic energy of the motion, and so K is maximum.

A man drags a 12.0 kg bag of mulch at a constant speed applying a 39.5 N force at 41°. What is the normal force acting on the bag?

Answers

The normal force is perpendicular to the surface. Its intensity is equal to the force pressing the object against the surface, but it has opposite direction.
We need to calculate the force pushing the bag down in order to calculate normal force.
The force that pushes the bag down is equal to:
[tex]F_d=mg-sin(41)\cdot F[/tex]
Where F is the force of a man dragging the bag and mg is gravity pulling the bag down.
The resulting force is:
[tex]F_d=12\cdot9.81-sin(41)\cdot 39.5=91.8N[/tex]
Normal force has the same intensity but the opposite direction of this force.


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