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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Grade: College
Chapter: Electrostatics
Subject: Physics
Keywords: Potential energy, potential difference, accelerate, kinetic energy, speed, helium ion, charge, significant figures, rest to a speed.
what best describes a lens galaxy
a. a group if stars that serve as a quick reference to different parts of the sky
b. several Billion Stars arranged in a disk shape that has uneven edges
c. millions of planets are arranged in the shape of a disc with spiraling arms
D. billions of stars arranged in the circular pattern with a flattened core
the answer would be D
Answer:
D. billions of stars arranged in the circular pattern with a flattened core
Explanation:
An object that has a small mass and an object that has a large mass have the same momentum. which object has the largest kinetic energy?
When two objects have the same momentum, the one with the smaller mass will have larger kinetic energy, because it must have a higher velocity. Conversely, if they have the same kinetic energy, the object with larger mass will have a higher momentum due to its larger mass.
Explanation:Understanding Momentum and Kinetic Energy.When considering two objects, one with a small mass and another with a large mass, that both have the same momentum, it's important to understand the relationship between momentum (p), mass (m), and velocity (v). Momentum is calculated as the product of mass and velocity (p = mv). To have the same momentum, a small mass must move with a larger velocity compared to a large mass.
When comparing kinetic energy (K), which is given by the formula K = (1/2)mv², the object with the smaller mass but higher velocity will have a larger kinetic energy. This is because kinetic energy is proportional to the square of velocity, and since the smaller mass has a higher velocity to maintain the same momentum, its kinetic energy will be greater. If the scenario is reversed and two objects have the same kinetic energy, then the large mass object must have a larger momentum. Since kinetic energy is the same, the larger mass object compensates for its lower velocity with its greater mass, leading to a greater momentum (p = mv).
How much work must be done on a system to decrease its volume from 20.0 l to 13.0 l by exerting a constant pressure of 4.0 atm?
By exerting the given constant pressure, work done on the system to decrease from initial volume to final volume is 2.84kJ.
Given the data in the question
Initial volume; [tex]v_1 = 20.0L[/tex]Final volume; [tex]v_2 = 13.0L[/tex]Pressure exerted; [tex]P = 4.0 atm[/tex]The magnitude of the work done when a gas expands is equal to the product of the pressure of the gas and the change in the volume of the gas.
For work done by gas against a constant external pressure, we use the
expression:
[tex]Work\ done = -P\ *\ \delta V[/tex]
We substitute our given values into the equation
[tex]Work\ done = -P\ *\ ( v_2 - v_1)\\\\Work\ done = -4.0atm\ *\ ( 13.0L - 20.0L)\\\\Work\ done = -4.0atm\ *\ -7L\\\\Work\ done = 28L.atm\\\\Work\ done = 28L.atm\ *\ [ \frac{101.3J}{L.atm}] \\\\Work\ done = 2836.4J\\\\Work\ done = 2.84 kJ[/tex]
Therefore, by exerting the given constant pressure, work done on the system to decrease from initial volume to final volume is 2.84kJ.
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To decrease the volume of a system from 20.0 l to 13.0 l by exerting a constant pressure of 4.0 atm, a work of 2836.4 Joules must be done on the system.
Explanation:In Physics, the concept of work done in a constant-pressure or isobaric process is defined as the product of pressure and the change in volume (W = PΔV). Since the pressure is given as 4.0 atm and must be converted to Joules using the conversion factor (1 atm = 101.3 J/L), the pressure in usable units becomes 4.0 atm * 101.3 J/L = 405.2 J/L. The change in volume is the initial volume subtracted from the final volume, so ΔV = 20.0L - 13.0L = 7.0L.
Substituting the values into our equation, the work done to decrease the volume of the system becomes W = (405.2 J/L) * (7.0 L) = 2836.4 Joules. So, to decrease the volume of the system from 20.0 l to 13.0 l by exerting a constant pressure of 4.0 atm, a work of 2836.4 Joules must be done on the system.
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1. How is electric potential energy similar to gravitational potential energy? How is it different? Where will an electron bound in an atom and have the largest electrical potential energy?
According to Einstein’s theory, how does an increase in the number of photons affect a beam of light?
The wavelength of the beam of light increases.
The speed of the beam of light increases.
The brightness of the beam of light increases.
The frequency of the light beam increases.
The energy of each photon in the beam of light increases.
Answer:
The brightness of the beam of light increases.
Explanation:
As per Einstein's experimental verification he proves that light is of particle nature and his equation for energy balance is given as
[tex]h\nu = \phi + KE[/tex]
so incident photons on a photosensitive plate will eject the electrons from the plate with some sufficient KE.
So here light is considered as particles which means one photon will eject one electron. So here intensity of light is considered as
Intensity = (number of photons per second) (energy of one photon)
so it is given as
[tex]I = \frac{N}{t}(\frac{hc}{\lambda})[/tex]
now if the intensity of light is increased for same wavelength of light then it means the number of photon must have to be increase.
so correct answer will be
The brightness of the beam of light increases.
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). if a particular disk is spun at 792.7 rad / s while it is being read, and then is allowed to come to rest over 0.234 seconds, what is the magnitude of the average angular acceleration of the disk?
If you want to know how energy will move between two objects, what do you need to know about the objects?
Answer:
Need to know THEIR TEMPERATURES
Explanation:
Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a 4.0-kilogram ball rolling along the same path at a speed of 1.0 m/sec?
The force require to stop the second object is greater because it has bigger momentum when compared to the first object.
The given parameters;
mass of the first object, m₁ = 8 kgspeed of the first object, u₁ = 0.2 m/smass of the second object, m₁ = 4 kgspeed of the second object, u₂ = 1 m/sThe force required to stop each object in the given time is determined by applying Newton's second law of motion as shown below;
[tex]F = ma = \frac{mv}{t} \\\\F_1 = \frac{8 \times 0.2}{10} \\\\F_1 = 0.16\ N[/tex]
For the second object;
[tex]F_2 = \frac{4 \times 1}{10} \\\\F_2 = 0.4 \ N[/tex]
Thus, the force require to stop the second object is greater because it has bigger momentum when compared to the first object.
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Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what is the x position of the center of mass? m 2) what is the y position of the center of mass? m 3) what is the speed of the center of mass? m/s 4) when a fifth mass is placed at the origin, what happens to the horizontal (x) location of the center of mass? it moves to the right. it moves to the left. it does not move. it can not be determined unless you know the mass. 5) when a fifth mass is placed at the center of mass, what happens to the vertical (y) location of the center of mass? it moves up. it moves down. it does not move. it can not be determined unless you know the mass.
The center of mass and speed can be calculated using the formulas provided. Adding a fifth mass at either the origin or the current center of mass will affect the location of the center of mass, but it can only be determined with knowledge of the mass of the new particle.
Explanation:The center of mass for a system of particles can be calculated using the following formulas: Xcm = ∑mi xi/mi and Ycm = ∑mi yi / ∑ mi, where 'mi' denotes the mass of the ith particle and 'xi' and 'yi' are the x and y coordinates respectively. This principle extends not just to two dimensions but also to three dimensions.
The speed of the center of mass of a group of particles can be calculated by the formula Vcm = ∑mi vi / ∑ mi, where 'vi' is the vector sum of the velocities of the particles, and 'mi' is the mass of the ith particle.
When a fifth mass is placed at the origin, the location of the center of mass can only be determined if the mass is known. Similarly, when a fifth mass is placed at the center of mass, the new location depends on the mass of the fifth particle and cannot be determined otherwise.
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A 15 kg block is attached to a very light horizontal spring of spring constant 500 N/m and is resting on a frictionless horizontal table as shown in the figure. Suddenly it is struck by a 3 kg stone traveling horizontally at 8 m/s to the right; the stone rebounds at 2 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.
0.35 m is the maximum distance that the block will compress the spring after the collision.
Given :
Mass, [tex]\rm m_b = 15 \;Kg\;and\; m_s = 3\;Kg[/tex]
Spring constant, K = 500 N/m
Solution :
Before the collision the total momentum is,
[tex]\rm p_i = m_s \times v_s[/tex]
Whare [tex]m_s[/tex] is mass of stone and [tex]\rm v_s[/tex] is the velocity of stone beforethe collision.
After the collision stone and block both are moving therefore total momentum is,
[tex]\rm p_f = m_sv_s' + m_bv_b[/tex]
where, [tex]\rm v_s'[/tex] is the velocity of stone after collision, [tex]\rm m_b[/tex] is the mass of block and [tex]\rm v_b[/tex] is the speed of the block after collision.
Now, momentum should be conserve
[tex]\rm p_i = p_f[/tex]
[tex]\rm m_sv_s= m_sv_s'+ m_bv_b[/tex]
[tex]\rm v_b = \dfrac{m_s}{m_b}(v_s - v_s')[/tex] ------- (1)
[tex]\rm v_s = 8\;m/sec[/tex] (Given)
[tex]\rm v_s' = 2 \;m/sec[/tex] (Given)
Now putting the values in equation (1) we get,
[tex]\rm v_b = \dfrac{3}{15}(8-2)[/tex]
[tex]\rm v_b = 2\;m/sec[/tex]
Therefore, block slides with the speed of 2 m/s to the right after the collision.
We know that the block starts to move with speed 2 m/s, so its kinetic energy is
[tex]\rm K = \dfrac{1}{2}m_bv_b^2[/tex]
As the spring is compress, block speed decreases and its kinetic energy is converted into elastic potential energy of the spring and the elastic potential energy of the spring is,
[tex]\rm U = \dfrac{1}{2} Kx^2_m_a_x[/tex]
Now through conservation of energy we have,
[tex]\rm K = U[/tex]
[tex]\rm m_bv_b^2 = Kx^2_m_a_x[/tex]
[tex]\rm x_m_a_x = \sqrt{\dfrac{m_bv_b^2}{K}}[/tex]----- (2)
Putting the values of [tex]\rm m_b,\;v_b\;and\; K[/tex] in equation (2),
[tex]\rm x_m_a_x= \sqrt{\dfrac{15\times2^2}{500}}[/tex]
[tex]\rm x_m_a_x=0.35\;m[/tex]
0.35 m is the maximum distance that the block will compress the spring after the collision.
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The flower of the species Rosa verdus can be either green or red. in the species a single gene with two alleles determines flower color. Small g is the recessive allele that results in red flowers in the homozygous condition. Which two genotypes will result in plants with green flowers.
Answer:
SS and Ss
I did the test!!!!
A 1.10 kg block is attached to a spring with spring constant 13.5 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 36.0 cm/s . what are
The amplitude of the block's oscillation is 0.103 A.
The speed of the block at the given position is 0.34 m/s.
How to calculate the amplitude of the oscillation?
The amplitude of the block's oscillation is calculated by applying the principle of conservation of energy as follows.
K.E = U
¹/₂mv² = ¹/₂kA²
mv² = kA²
A² = mv²/k
A = √ (mv² / k)
where;
m is the massv is the speedk is spring constantA = √ (1.1 x 0.36² / 13.5)
A = 0.103 A
The speed of the block at the given position is calculated as follows;
¹/₂kA² = ¹/₂mv² + ¹/₂kx²
kA² = mv² + kx²
mv² = kA² - kx²
v² = (kA² - kx²) / m
v = √ (13.5 x 0.103² - 13.5(0.3 x 0.103)² ) / 1.1
v = 0.34 m/s
The complete question is below:
A 1.10 kg block is attached to a spring with spring constant 13.5 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 36.0 cm/s .
What are the amplitude of the subsequent oscillations and the block's speed at the point where x = 0.3 A?
A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. light waves arrive at the two slits in phase, and a fringe pattern is observed on a screen 4.8 m from the slits. if there are 6.0 complete bright fringes per centimeter on the screen near the center of the pattern, what is the wavelength of the monochromatic light?
Final answer:
The wavelength of the monochromatic light in the double slit experiment is calculated to be 625 nm, using the provided measurements and the double slit interference formula.
Explanation:
The student is asking to determine the wavelength of monochromatic light based on observations from a double slit interference experiment. Given that the slit separation is 1.8 mm, the distance to the screen is 4.8 m, and there are 6.0 complete bright fringes per centimeter on the screen, we can calculate the wavelength using the formula for double slit interference, λ = Δy × d / D, where λ is the wavelength of light, Δy is the distance between adjacent bright fringes (the fringe spacing), d is the separation between the slits, and D is the distance from the slits to the screen.
First, we find the fringe spacing by noting that there are 6 bright fringes per centimeter, so Δy = 1 cm / 6 = 0.1667 cm = 1.667 mm. We can then use the given values to calculate the wavelength:
[tex]λ = (1.667 \times 10^{-3} m) \times (1.8 \times 10^{-3} m) / (4.8 m) = 6.25 \times 10^{-7} m = 625 nm.[/tex]
Therefore, the wavelength of the monochromatic light is 625 nm.
Airbags will deploy no matter from what angle your car is hit.
Answer:
Airbags:
" As the airbags comes in inbuilt form these days in most of the cars, as they fulfill the safety concerns of the passengers and make it sure that non of the individuals gets hurt in any accident faced during the travel."
Angle of Deployment:As the airbags will only deploy in the specific conditions provided, as the car must be speed over about the 25 km/hr and the accident must be more like a head-on-collision making the airbags to deploy. Or else if the car experiences an side wise crash with any other entity then it will never deploy in that case.
Is a higher value for k result in a stiffer/less stretchy spring, or a less stiff/more stretchy spring?
Answer:
if value of k is high then it results less stretchy spring
Explanation:
As we know by formula of spring force
F = kx
here we know
k = spring constant
x = stretch in the spring
Now we know that when same force is applied on two types of spring then the stretch of the spring is given as
[tex]x = \frac{F}{k}[/tex]
so if the value of k is higher then the stretch in the spring will be less as it is inversely depends on the spring constant.
So here we can say that if value of k is high then it results less stretchy spring
Which of the following situations represents a negative displacement? (Assume positive position is measured vertically upward along a y-axis.) a. A cat stands on a tree limb. b. A cat jumps from the ground onto a tree limb. c. A cat jumps from a lower tree limb to a higher one. d. A cat jumps from a tree limb to the ground. Please select the best answer from the choices provided A B C D
Answer:
Option-(D): "A cat jumps from a tree limb to the ground."
Explanation:
Negative displacement(Assume positive position is measured vertically upward along a y-axis):
The displacement can be visualized going from the initial point been supposed as the ground level. While, the limb is considered more elevated as compared to the ground level. So, going from the lower level to the higher position is considered as positive displacement while the movement from the higher position to the lower point is considered as negative displacement.
Determine the energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom. determine the energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom. -1.82 × 10-19 j +2.69 × 10-19 j +5.51 × 10-19 j +3.03 × 10-19 j -3.03 × 10-19 j
The energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom is about -3.03 × 10⁻¹⁹ J
[tex]\texttt{ }[/tex]
Further explanationThe term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :
[tex]\large {\boxed {E = h \times f}}[/tex]
E = Energi of A Photon ( Joule )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
f = Frequency of Eletromagnetic Wave ( Hz )
[tex]\texttt{ }[/tex]
The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
E = Energi of A Photon ( Joule )
m = Mass of an Electron ( kg )
v = Electron Release Speed ( m/s )
Ф = Work Function of Metal ( Joule )
q = Charge of an Electron ( Coulomb )
V = Stopping Potential ( Volt )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
initial shell = n₁ = 3
final shell = n₂ = 2
Unknown:
ΔE = ?
Solution:
We will use this following formula to solve this problem:
[tex]\Delta E = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]
[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{2^2} - \frac{1}{3^2})[/tex]
[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{4} - \frac{1}{9} )[/tex]
[tex]\Delta E = -2.18 \times 10^{-18} \times \frac{5}{36}[/tex]
[tex]\Delta E \approx -3.03 \times 10^{-19} \texttt{ J}[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: College
Subject: Physics
Chapter: Quantum Physics
What marine life zones are warm well lit and have lots of marine life
The marine life zones that are warm, well-lit, and have an abundance of marine life are known as the photic zone of the ocean. This zone includes both the Sunlight Zone and the Twilight Zone, which extends to a depth of about 200 meters below the sea surface. The photic zone is characterized by sufficient sunlight to support photosynthesis, which enables the growth of phytoplankton, the primary producers in marine ecosystems. Consequently, this zone supports a diverse array of marine life.
The photic zone is the uppermost layer of the marine biome where sunlight penetrates and enables photosynthesis. Phytoplankton, tiny photosynthetic organisms, form the base of the marine food web in this area, supporting a complex and diverse ecosystem. Other marine organisms, such as fish, marine mammals, and various invertebrates, depend on the photic zone directly or indirectly for sustenance. The availability of sunlight, warmth, and nutrients makes this zone the most lively and productive part of the ocean, contrasting with the aphotic zone, where the lack of light limits life. The health of the photic zone is vital for marine biodiversity and for global ecological processes such as oxygen production and carbon dioxide absorption.
Only moving things have energy
As per the question only moving objects have energy.
The statement given here is not always true.
we know that every moving object has kinetic energy.But in the universe there are other kinds of energy also.
We may take a simple example of potential energy which is the energy possessed by a body due to its configuration and position.
A stone present on the top of a mountain has zero kinetic energy but here the total mechanical energy is in the form of potential energy.
The electrostatic potential energy is not due to the motion of objects also.
There are also various other kinds of energy existing in the nature which are not due to the motion of objects.
While scuba diving with some friends, Deshawn notices that the air bubbles exhaled by his friends increase in size as they get closer to the surface of the water.
Why does this occur and which gas law explains it?
hurrry plzz/////
Answer:
Sample Response: As the air bubbles rise to the surface, the pressure on them decreases. This decrease in pressure allows the air to expand and each bubble increases in size. Boyle’s law explains why this occurs.
Explanation:
Describe the relationship between the force between two magnetic poles and the distance of separation.
The magnetic force is directly proportional to the distance of separation.
The magnetic force is inversely proportional to the distance of separation.
The magnetic force is directly proportional to the square of the distance of separation.
The magnetic force is inversely proportional to the square of the distance of separation.
The relationship between the force between two magnetic poles and the distance of separation is The magnetic force is inversely proportional to the square of the distance between them.
What is Coloumb's law of magnetism?Coloumb's law of magnetism states the relationship between the magnetic strength of two magnets. The force of attraction or repulsion between two magnets is directly proportional to the strength of magnetic poles and inversely proportional to the square of the distance between them.
Coloumb's law of magnetism describes the electric field between two charged objects. The force of attraction or repulsion is inversely proportional to the absolute permeability of the surrounding medium.
The attractive and repulsive force, F∝ m₁×m₂ / μ₀r², where m₁, and m₂ are the magnetic strength or magnetic poles and r is the distance between two magnets.
The attractive or repulsive force, F = k (m₁×m₂) / μ₀r², where μ₀ is the permeability of the medium and k is the constant of proportionality. The unit is Newton/Tesla. The magnetic force is inversely proportional to the square of the distance of separation.
Hence, the ideal solution is option D.
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According to Newton's second law of motion, force can be calculated by multiplying an object's mass by its acceleration. If weight is a measure of an object's force due to gravity, what is the weight of a student who has a mass of 65 kg? (Assume that the acceleration due to gravity equals 9.8 m/s2.)
Final answer:
The weight of a student with a mass of 65 kg on Earth is 637 N (newtons).
Explanation:
According to Newton's second law of motion, force (F) is the product of an object's mass (m) and its acceleration (a). The formula for this relationship is F = ma. When talking about weight, we refer to the force of gravity acting on an object's mass. The acceleration due to gravity (g) on Earth is approximately 9.8 m/s2. Therefore, to calculate the weight of an object on Earth, we use the aforementioned relationship but with g as the acceleration, leading to W = mg. Substituting the given mass (65 kg) and the acceleration due to gravity, we get W = 65 kg × 9.8 m/s2 = 637 N, which is the student's weight on Earth.
An athlete with mass m running at speed v grabs a light rope that hangs from a ceiling of height h and swings to a maximum height of 1 h . in another room with a lower ceiling of height 2 h , a second athlete with mass 2m running at the same speed v grabs a light rope hanging from the ceiling and swings to a maximum height of 2 h . how does the maximum height reached by the two athletes compare, and why?
Both athletes reach the same maximum height. This is because the kinetic energy they start with (due to their initial speed) is the same, which is fully transformed into potential energy when they reach their maximum heights.
Explanation:The problem is solved by relating the kinetic energy and potential energy of the athletes.
Before the athletes grab the rope, they have kinetic energy due to their motion (1/2mv2). Once they reach the maximum height, this kinetic energy is fully transformed into gravitational potential energy (mgh).
In the first case, the kinetic energy is (1/2)m(v2) and this is equal to the potential energy h at the maximum height (mgh), thereby, h = (v2)/2g.
In the second case, the kinetic energy is (1/2)(2m)(v2), and this is equal to the potential energy at the maximum height (2m)(g)(2h). In this scenario, the maximum height reached is 2h = (v2)/2g.
So, comparing the two scenarios, we clearly see that the maximum height reached by both athletes is identical. Although the second athlete is twice as heavy, they both start with the same speed and so the initial kinetic energy is the same.
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At approximately what frequency do most people perceive the least intense sounds? enter your answer numerically in hertz to two significant figures.
Most people perceive the least intense sounds at a frequency around 2000 Hz with an intensity level at the threshold of hearing, which is about 0 dB or 10^-12 W/m².
Explanation:The frequency at which most people perceive the least intense sounds is typically around the threshold of human hearing. This threshold is at about 0 dB, which corresponds to an intensity level of approximately 10-12 W/m². However, frequency and intensity are not the only factors that contribute to the perception of sound, as the sensitivity of the human ear varies across different frequencies. The human ear has the maximum sensitivity to frequencies ranging from 2000 to 5000 Hz. Sounds at these frequencies are perceived as louder compared to those at lower or higher frequencies, even if they have the same intensity level. Therefore, the least intense sounds that humans can perceive are at the lower end of this range, around 2000 Hz, at the absolute threshold of hearing of 0 dB.
If two oscillators connected to speakers create frequencies of 594 hz and 600 hz, what is the beat frequency when you hear both at the same time? hz
In the video, you saw NASA launch a rocket carrying a rover to Mars. What is the mission of the Mars land rover, and why is it important?
The rover sends photographs of Mars back to Earth. These pictures show scientists the landscape of the planet. The rover also collects specimens and looks for evidence of water so we can determine whether life could ever exist on Mars.
Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 kg/s with a speed of 4.5 × 104 m/s.
This question involves the concept of Newton's Second Law of Motion.
The force exerted on the rocket is "5.85 x 10⁷ N".
NEWTON'S SECOND LAW OF MOTIONAccording to Newton's Second Law of Motion, whenever an unbalanced force is applied to an object, it produces an acceleration in the direction of the force. Mathematically,
[tex]F=ma[/tex]
where,
F = force = ?a = acceleration = [tex]\frac{\Delta v}{t}[/tex]Therefore,
[tex]F = \frac{m\Delta v}{t}[/tex]
where,
[tex]\frac{m}{t}[/tex] = mass flow rate of gases = 1300 kg/sΔv = change in velocity = 4.5 x 10⁴ m/sTherefore,
F = (1300 kg/s)(4.5 x 10⁴ m/s)
F = 5.85 x 10⁷ N
Learn more about Newton's Second Law of Motion here:
https://brainly.com/question/23845187
The chart shows characteristics of stars. Sirius is a blue-white star with a surface temperature of about 9800 K. What is most likely the mass of Sirius? 1.0 solar mass 1.4 solar mass 2.0 solar mass 12 solar mass
Answer:
The answer would be 2 solar masses.
A fan blade rotates with angular velocity given by ωz(t)= γ − β t2, where γ = 4.90 rad/s and β = 0.750 rad/s3 . part a calculate the angular acceleration as a function of time.
Answer:
Angular acceleration, α = -2βt
Explanation:
Angular velocity of fan is [tex]\omega_{z(t)}=\gamma -\beta t^2[/tex]
[tex]\gamma=4.90\ rad/s[/tex]
[tex]\beta=0.750\ rad/s^3[/tex]
Angular acceleration is given by :
[tex]\alpha=\dfrac{d\omega}{dt}[/tex]
[tex]\alpha=\dfrac{d(\gamma -\beta t^2)}{dt}[/tex]
[tex]\alpha=-2\beta t[/tex]
Hence, the above equation is the angular acceleration as a function of time.
What change occurs to the mass of an object when an unbalanced force is applied to it? The mass increases. The mass decreases. The mass remains the same. The mass reduces to zero. The mass changes by an indeterminate amount
Third option is correct. The mass of an object remains same when an unbalanced force is applied to it.
the mass on an object doesn't depend on the applied force. it is constant.When an unbalanced force is applied to an object, it will get accelerated. The velocity of the object will increase or decrease depending on the direction of the force but the mass of the object remains same. That doesn't change with the applied force.