we can use the principle of conservation of energy. Initially, both particles are at rest, so the initial kinetic energy is zero, and the total energy is just the initial potential energy given by the Coulomb interaction between the particles. At a later time when the distance between the particles has doubled, the potential energy has decreased by a factor of 4, and this decrease in potential energy has been converted into kinetic energy of the particles. Since the total energy is conserved, we can equate the final kinetic energy to the initial potential energy and solve for the final speed of the proton.
Let's start by calculating the initial potential energy of the system. The Coulomb force between two point charges q1 and q2 separated by a distance r is given by:
F = (1/4πε0) * (q1 * q2) / r^2
where ε0 is the permittivity of free space. The potential energy U of the system is the negative of the work done by the Coulomb force as the particles move from infinity to a separation r:
U = - ∫∞r F dr = (1/4πε0) * (q1 * q2) / r
In this problem, the proton has charge e and the alpha particle has charge 2e, so the initial potential energy is:
U_i = (1/4πε0) * (e * 2e) / r = e^2 / (2πε0r)
When the distance between the particles doubles, the new separation is 2r, and the final potential energy is:
U_f = (1/4πε0) * (e * 2e) / (2r) = e^2 / (4πε0r)
The change in potential energy is therefore:
ΔU = U_i - U_f = e^2 / (4πε0r)
This energy has been converted into kinetic energy of the particles. Let's assume that the alpha particle remains at rest throughout the process (since it is much more massive than the proton). Then the final kinetic energy of the proton is:
K_f = ΔU = e^2 / (4πε0r)
We can equate this to the initial kinetic energy (which is zero) to find the final speed of the proton:
(1/2) * m * (vf)p^2 = e^2 / (4πε0r)
Solving for (vf)p, we get:
(vf)p = sqrt(2 * e^2 / (4πε0m r))
Substituting the given values for e, 2e, and m, we get:
(vf)p = sqrt(2 * (1.6 x 10^-19 C)^2 / (4π(8.85 x 10^-12 F/m) (1.67 x 10^-27 kg) r))
Simplifying, we get:
(vf)p = 2.19 x 10^6 m/s * sqrt(1/r)
Therefore, the answer is (A) 0.422.
the volumetric flow rate of water flowing through a pipe is 5 l/s, and the density of water is 1 kg/l. what is the mass flow rate? multiple choice question. 0.05 kg/s 0.005 kg/s 5 kg/s 50 kg/s 0.5 kg/s
The volumetric flow rate of water flowing through a pipe is 5 l/s and the density of water is 1 kg/l. This means that the mass flow rate is 5 kg/s. The correct answer is 5 kg/s.
The mass of a liquid passing in one unit of time is known as the mass flow rate. In other terms, the mass flow rate is the rate at which liquids move across a given region. The mass flow is a direct function of the liquid's density, velocity, and cross-sectional area.
The formula to find the mass flow rate of a substance is given as:
mass flow rate = density × volumetric flow rate
The given volumetric flow rate of water = 5 L/s
The density of water = 1 kg/L
Thus, the mass flow rate of water = 1 × 5 = 5 kg/s
Therefore, the answer is 5 kg/s.
The mass flow rate of water when the volumetric flow rate of water flowing through a pipe is 5 L/s, and the density of water is 1 kg/L is 5 kg/s.
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1. this experiment was to find how mass and speed affect ke. this is important because if you were in a situation where you needed something to go higher, you would know to add more or less of mass/speed. to test mass, we filled the bean bag with a certain amount of water, then dropped it. after, you recorded how high it made the bean bag go. the same with speed, but the same amount in the bottle, just dropped from different heights. my hypothesis is when you have more mass, the ke will be greater. this is also the same with speed, if it is dropped from a higher place, the bean bag will launch farther than the last time. 2. data i collected from the lab was like my hypothesis explained. when the height of the bottle increased, it made the bean bag go higher than the last. and i tested 4 different masses, 0.125 kg, 0.250kg, 0.375kg, and 0.500kg. each time the bean bag went higher on a larger mass. a lot of times on the speed test, the bean bag would go higher than the bottle drop point, but not every time. also, when it was dropped from the same height each time, some results varied quite a bit, like when it was dropped from 1.28 the results were 1.14 then 1.30 1.30. mass on the other hand was all in the same number range, only once the numbers were a bit off from each other. 3. some formulas i used were ke
Based on your observations and data, your hypothesis is correct because mass and speed are directly proportional to kinetic energy.
What is kinetic energy?The kinetic energy of an object is the energy possessed by the object due to its motion.
The formula for kinetic energy (KE) is:
KE = 1/2mv²
where;
m is the mass of the object in kilograms and v is the velocity of the object in meters per second.To calculate the velocity of the object when dropped from a certain height, you can use the formula:
v = √(2gh)
where;
g is the acceleration due to gravity (9.81 m/s²) and h is the height in meters from which the object is dropped.To calculate the maximum height reached by the object, you can use the formula:
h_max = (v²) / (2g)
Based on your observations and data, your hypothesis is correct. When the mass or speed of the bean bag increased, the kinetic energy and the height reached by the bean bag also increased. It is normal for some results to vary, especially with the speed test, due to factors such as air resistance and the initial launch angle of the bean bag.
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The missing part of the question is below:
Why is my hypothesis correct?
a spacecraft is in a circular orbit of mars at an altitude of 200 km. calculate its speed and its perio
The speed of the spacecraft in a circular orbit around Mars at an altitude of 200 km is approximately 3,543.62 m/s, and the period of the orbit is approximately 6,867.97 seconds or 1.91 hours.
To calculate the speed of a spacecraft orbiting Mars in a circular orbit, use the formula: v = √(GM/R)
Where: v = speed of the spacecraft in meters per second
G = gravitational constant 6.674 × 10⁻¹¹ N m²/kg²
M = mass of Mars (6.39 × 10²³ kg)
R = radius of the orbit (200 km + the radius of Mars, 3,389 km)
Substituting in the values, we get:
v = √((6.674 × 10⁻¹¹ N m²/kg²) × (6.39 × 10²³ kg) / (3.5895 × 10⁶ m))
v ≈ 3,543.62 m/s
Therefore, the speed of the spacecraft in a circular orbit around Mars at an altitude of 200 km is approximately 3,543.62 m/s.
The formula to calculate the period of a circular orbit is T = 2πR/v
Where: T = period of the orbit in seconds
R = radius of the orbit in meters (200 km + 3389.5 km = 3589.5 km = 3.5895 × 10⁶ m)
v = speed of the spacecraft in meters per second
Plugging in the values, we get:
T = 2π(3.5895 × 10⁶ m) / (3,543.62 m/s)
T ≈ 6,867.97 seconds
The period of the orbit is approximately 6,867.97 seconds or 1.91 hours.
Therefore, the speed of the spacecraft in the circular orbit is 3,584 m/s and the period of the orbit is 6,867.97 seconds.
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"Radioactive decay is a random process but we can still make predictions about it" Explain this statement
Answer:
Radioactive decay is determined by quantum mechanics — which is inherently probabilistic. So it's impossible to work out when any particular atom will decay, but we can make predictions based on the statistical behaviour of large numbers of atoms.
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Suppose you are constructing a 95% confidence interval for the mean of a single sample, whose population standard deviation is known to be σ = 5. You calculate the sample size with some specified width (error) E. (a) Reducing your confidence level to 80%, and reducing your original width (error) E by a third ( 1 3 ), how much bigger will the new sample size be compared to the first sample size above? (Hint: find the scaled size using algebra). b) Suppose instead that your increase the sample size by a factor of 10 and you allow the confidence level to be 85%, how will the width (error) have scaled in size compared to the original width (error) E?
The sample size with some specified width error by reducing the confidence level to 80% is 3 times the original sample size and the width error which have been scaled in size compared to the original width error is by a factor of 10.
What is the sample size?When reducing the confidence level to 80% and the original width (error) E by a third, the new sample size will be 3 times the original size. To calculate this, use the formula n = (zα/2 / E)² × σ2.
If the original confidence level was 95%, then the original zα/2 = 1.96. If the new confidence level is 80%, then the new zα/2 = 1.282.
The original error was E, and the new error is 1/3 E. By substituting these values into the formula, we get n = (1.282 / (1/3 E))² × σ2. This simplifies to n = 3 × (1.96 / E)² × σ2, which is 3 times the original sample size.
If the original confidence level was 95%, then the original zα/2 = 1.96.
If the new confidence level is 85%, then the new zα/2 = 1.44.
The original error was E, and the new error is 10E.
By substituting these values into the formula, we get n = (1.44 / (10E))² × σ2.
This simplifies to n = (1.96 / E)² × σ2, which is 10 times the original sample size.
Therefore, the width (error) will have scaled in size by a factor of 10 compared to the original width (error) E.
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Given Newton’s First Law of Motion, what do we reasonably expect an object to do given the following scenarios?
Part (a) An object sits at rest with no unbalanced forces acting upon it. What do we expect this object to do?
The object will begin to move at a constant velocity.
The object will remain at rest.
None of these answers.
The object will begin to move with a changing velocity
An object at rest will remain at rest and an object in motion will remain in motion with a constant velocity unless acted upon by an unbalanced force.
Newton's First Law of Motion often referred to as the Law of Inertia, is a fundamental principle of physics that states the following:
An object at rest remains at rest, and an object in motion remains in motion with the same velocity (which means both magnitude and direction) unless acted upon by a net external force.
In this scenario, there are no unbalanced forces acting upon the object, so it will continue to remain at rest.
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Three identical conducting spheres are charged as follows. Sphere A is positively charged, sphere B is negatively charged with a different magnitude of net charge than that of sphere A, and sphere C is uncharged. Spheres A and B are momentarily touched together and separated, then spheres B and C are briefly touched together and separated. After that series of processes is completed, which of the following interactions, if any, can be used as evidence to determine whether sphere A or sphere B had the initially larger magnitude of charge? A Sphere C is repelled from sphere A. B Sphere C is repelled from sphere B. Sphere A is repelled from sphere B. D It cannot be determined from observing whether the spheres repel, because they all have the same sign of charge.
The answer is C. Sphere A is repelled from sphere B
Step by step explanation:
The question is asking which of the interactions between sphere A, B, and C can be used as evidence to determine which one had the initially larger magnitude of charge. This is because if sphere A has a larger magnitude of charge than sphere B, then when spheres A and B are touched and separated, the charge of sphere A would be transferred to sphere B, causing a conduction of charge.
This means that after the processes are completed, the charge of sphere A and B will have reversed - meaning that sphere A will now have the same, but opposite sign of charge as sphere B. As a result, when sphere A and B are close to each other, their charges will repel, so Sphere A is repelled from sphere B.
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calculate the distance moved by a car accelerating from rest at a constant rate of 2m/s squared for 5 s
Answer:
D=5
Explanation:
D=1/2*b*h
D=1/2*2*5
D=5
Review your answer to part c. In addition, reread the portion of your physics text that discusses Newton's third law. Then consider a book on a level table: e. Which force completes the Newton's third law (or action-reaction) force pair with the normal force exerted on the book by the table?
In this case, the normal force exerted by the table on the book is the action force and the reaction force is the force that the book exerts on the table. This force is equal in magnitude to the normal force and acts in the opposite direction.
Newton's third law states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object exerts a force back on the first object that is equal in magnitude and opposite in direction.
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the iron core of a transformer becomes ________________ when the current flows through the primary coil.
The iron core of a transformer becomes magnetized when the current flows through the primary coil.
Electrical energy is transferred between the main and secondary coils of a transformer through a magnetic circuit that is created by the iron core. A magnetic field that is continuously changing is produced in the iron core when an alternating current is fed through the primary coil. Energy is transferred between the coils as a result of the magnetic field's induction of a matching current in the secondary coil. Because of its high permeability and sensitivity to magnetism, the iron core can magnetize. A greater induced current is produced in the secondary coil as a result of the first coil's magnetic field being amplified by the iron core's magnetization. The effective transmission of electrical energy in a transformer depends on the magnetic characteristics of the iron core.
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what is the current in a counductor if 3.15*10^18 electrons pass a given point in the conductor in 10 seconds
The current in a counductor if 3.15*10^18 electrons pass a given point in the conductor in 10 seconds is 0.0504 amperes
Current calculation.
The current in a conductor is defined as the rate at which electric charge flows through it. The unit of current is amperes (A), which is defined as coulombs per second. One coulomb is equal to the charge on 6.24 × 10^18 electrons.
Given that 3.15 × 10^18 electrons pass a given point in the conductor in 10 seconds, we can find the charge that flows through the point as follows:
Number of electrons = 3.15 × 10^18
Charge on one electron = 1.6 × 10^-19 coulombs
Total charge = Number of electrons × Charge on one electron
Total charge = 3.15 × 10^18 × 1.6 × 10^-19
Total charge = 0.504 coulombs
The current is the rate of flow of charge, so we can find it by dividing the total charge by the time taken:
Current = Total charge ÷ Time taken
Current = 0.504 coulombs ÷ 10 seconds
Current = 0.0504 amperes (A)
Therefore, the current in the conductor is 0.0504 amperes, or 50.4 milliamperes (mA).
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In the context of research evidence from the study conducted by Williams and McCririe, which of the following operates when a person picks up information critical to catching an object
both central and peripheral vision
In the context of research evidence from the study conducted by Williams and McCririe, both central and peripheral vision operate when a person picks up information critical to catching an object.
What is vision?
Vision is the sense that allows us to recognize and understand the physical world around us. Our brains then receive this information and convert it into the pictures that we see with our eyes.
Vision is the term used to describe the ability to see things with our eyes, such as color, form, and movement.
In the context of research evidence from the study conducted by Williams and McCririe, both central and peripheral vision operate when a person picks up information critical to catching an object.
Their research found that peripheral vision was essential to athletes performing in certain sports such as cricket, soccer, and baseball.
Peripheral vision, as well as central vision, are critical components of efficient eye tracking and hand-eye coordination.
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ercury's perihelion slowly precesses around the sun by a bit less than 2 degrees per century. this precession can be fully accounted for by newton's theory of gravity, although general relativity also gives the same answer. group of answer choices true false
The statement is true. Mercury's precession can be fully accounted for by both Newton's theory of gravity and general relativity.
Newton's law of universal gravitation states that any two bodies in the universe are attracted to each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This force can explain why the perihelion of Mercury is slowly precessing around the sun. According to Einstein's general theory of relativity, gravity is caused by the curvature of space-time around a massive body, such as the sun. This curvature of space-time causes Mercury to precess around the sun.
Newton's theory of gravity and general relativity provide equivalent explanations for the precession of Mercury's perihelion, which is a phenomenon in celestial mechanics. The precession of Mercury's perihelion is the slow rotation of the planet's elliptical orbit around the Sun's perihelion (the point of closest approach).It is well-known that Mercury's perihelion rotates by 42.98 arcseconds per century, or 1.39 degrees per century. This is caused by the gravitational influence of other planets, such as Venus and Jupiter, which produce small changes in Mercury's orbit. However, when this is taken into account, a tiny residual effect remains that cannot be accounted for using Newton's theory of gravity. This additional precession, known as the anomalous precession, can only be explained by general relativity.
The statement "Mercury's perihelion slowly precesses around the sun by a bit less than 2 degrees per century. This precession can be fully accounted for by Newton's theory of gravity, although general relativity also gives the same answer." is true.
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The speed of propagation of a sound wave in air at 27 degrees (Celsius) is about 350 m/s. Calculate, for comparison, v(rms) for nitrogen molecules at this temperature. The molar mass of nitrogen is 28.0 g/mol.
The RMS speed of Nitrogen Molecules at this temperature is approximately equal to [tex]16.6\ m/s[/tex].
RMS speed of Nitrogen Molecules at this temperature is given by the root-mean-square (RMS) speed equation: [tex]$v_{rms} = \sqrt{\frac{3kT}{m}}$[/tex]
where,[tex]$k$[/tex] is the Boltzmann's constant,[tex]$T$[/tex] is the absolute temperature, and$m$ is the mass of a single molecule.
To calculate the root-mean-square (RMS) speed of nitrogen molecules at this temperature, we have to use the above formula.
Given, Temperature [tex]($T$) = 27^\circ C = $300\ K$[/tex]
The molar mass of nitrogen is given as [tex]$28.0\ g/mol$[/tex],
therefore, the mass of one molecule will be:
Mass of one nitrogen molecule [tex]= \frac{28.0}{6.022 \times 10^{23}}[/tex]g/molecule = [tex]4.64 \times 10^{-23}\ g/molecule[/tex]
The Boltzmann's constant is given as[tex]k = $1.38 \times 10^{-23}\ J/K[/tex]
Using the above values in the RMS formula, we have:
[tex]$v_{rms} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3(1.38 \times 10^{-23}\ J/K)(300\ K)}{4.64 \times 10^{-23}\ g/molecule}}$[/tex]
The above expression evaluates to[tex]$v_{rms} = 16.6\ m/s$[/tex]
Therefore, the RMS speed of Nitrogen Molecules is [tex]16.6\ m/s[/tex].
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A 4x 103 -watt motor applies a force of 8x10 2newtons to move a boat at constant speed. How far does the boat move in 16 seconds?
a. 3.2 m
b. 5.0 m
c. 50 m
d. 80 m
The boat moves a distance of 80 m in 16 seconds when a force of [tex]8 \times 10^2[/tex] newtons is applied by a [tex]4 \times 10^3[/tex] watt motor. The correct answer is option d.
Power is defined as the rate at which work is done, and can be calculated using the formula:
Power = Work / Time
where Work is the amount of work done and Time is the time taken to do the work.
In this case, the power of the motor is given as [tex]4\times10^3[/tex] watts, and the force applied by the motor to move the boat at constant speed is given as [tex]8\times10^2[/tex] newtons. The work done by the motor in 16 seconds can be calculated as:
[tex]Work = Power \times Time[/tex]
[tex]Work = 4\times10^3\ watts \times 16\ seconds[/tex]
[tex]Work = 6.4\times10^4 \ joules[/tex]
The distance moved by boat can be calculated using the formula:
[tex]Work = Force \times Distance[/tex]
[tex]Distance = Work / Force[/tex]
[tex]Distance = 6.4\times10^4 \ joules / 8\times10^2\ newtons[/tex]
Distance = 80 meters
Therefore, the boat moves a distance of 80 meters in 16 seconds, which is option (d).
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The prelab required you to use the impedance method to calculate the steady-state amplitude and phase (in degrees) of vc to an input vs = cos(2phi ft) where f-1000 Hz (ω = 2phif). The results from the prelab are . Ao=_____Phase, φ =_____degrees
The steady-state amplitude Ao = 50.03 degrees and phase, φ = -88.7 degrees by using the impedance method.
The given equation for vs is:
vs = cos(2phi ft) ...[1]
where, f = 1000 Hz,
therefore ω = 2φf
ω= 2000π radians/s
Let's find the impedance of the circuit elements.
The impedance of the resistor is R.
The impedance of the capacitor is:
Zc = 1/(jωC)
The impedance of the inductor is:
ZL = jωL
As the capacitor and resistor are connected in series, their total impedance is:
ZC+R = R + 1/(jωC) ...[2]
Now, as the inductor is connected in parallel with the combination of R and C, the total impedance of the circuit is:
Ztotal = (ZC+R) || ZL...[3]
Ztotal = (R + 1/(jωC)) || jωL
Ztotal = 1/[(1/R) + j(1/ωC - ωL)]...[4]
Comparing the real and imaginary parts of the equation [4],
we get, 1/R = √{(1/ωC - ωL)^2} ...[5]and
1/ωC - ωL = 0
or
ωL = 1/ωC ...[6]
From equation [5],
we get, R = 1/√{(1/ωC - ωL)^2} ...[7]
The magnitude of the input voltage Vs is 1 volt.
The amplitude of the steady-state output voltage, Vc is given by:
Voc = Ao x 1VoltA0
Voc = R/ZtotalA0
Voc = R/1/[(1/R) + j(1/ωC - ωL)]A0
Voc = R(1/R) + jR(1/ωC - ωL)A0
Voc = 1 + jR(1/ωC - ωL) ...[8]
From equation [6],
we get: L = 1/(ωC)
L = 1/(2π x 1000)
L = 1.59 x 10-7 H
Substituting L in equation [6],
we get: ωL = ωC
ωL = 1/2π x 1000 x 1.59 x 10-7
ωL = 0.1Ω
From equation [7], we get: R = 1000 Ω
Substituting the value of R and ωL in equation [8],
we get: A0 = 1 + j1000(1/2π x 1000 x 1.59 x 10-7 - 0.1)
A0 = √{(1^2) + (-50.03)^2}
A0 = 50.03 degrees
Let φ be the phase of the output voltage with respect to the input voltage.
Therefore, we have: tanφ = -50.03φ = -88.7 degrees
Therefore, Ao = 50.03 degrees and φ = -88.7 degrees.
Answer: Ao = 50.03 degrees, φ = -88.7 degrees.
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Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit.
a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms?
b) Find the probability that the total resistance does not exceed 5100 ohms.
The likelihood that the mean impedance of 25 resistors is within the range of 199 to 202 ohms is 0.842, as per the principle of probability.
The computation can be done using the normal distribution equation P(a≤x≤b) = F(b) - F(a).
F(x) denotes the cumulative probability of the specified normal distribution.
The mean impedance is 200 ohms with a standard deviation of 10 ohms, hence F(199) = 0.155 and F(202) = 0.997. Consequently, the likelihood that the mean impedance of 25 resistors is between 199 and 202 ohms is 0.997 - 0.155 = 0.842.
The probability that the total impedance will be below 5100 ohms is 0.999. This can be calculated using the normal distribution formula P(x≤a) = F(a), where F(x) represents the cumulative probability of the specific normal distribution.
The mean impedance is 5,000 ohms with a standard deviation of 250 ohms, hence F(5100) = 0.999. Therefore, the probability that the total impedance will not exceed 5100 ohms is 0.999.
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Solve and check In the previous parts, you obtained the following equations using Newton's 2nd law and the constraint on the motion of the two blocks: m2a2x = T - m2g sin(θ), (1)
m1a1y = T - m1g, (2) and a2z = -a1y (3) Solve these equations to find a1y. Before you enter your answer, make sure it satisfies the special cases you already identified: - a1y = -g if m2 = 0 and - a1y = 0 if m1 = m2 and θ = π/2. Also make sure that your answer has dimensions of acceleration. Express a1y in terms of some or all of the variables m1, m2, θ, and g.
a1y = ____ ?
a1y in terms of some or all of the variables m1, m2, θ, and g is
a1y = (m2g sin(θ) - m1g) / m1 - m2a2x
What is Newton's 2nd law?In accordance with his second law, a force is equal to the change in momentum (defined as mass times velocity) per change in time. The definition of momentum is the product of the mass m and the velocity V of an object.
m2a2x = T - m2g sin(θ),
(1)m1a1y = T - m1g,
(2)a2z = -a1y (3)
On substitution of a2z = -a1y in (2):
m1(-a2z) = T - m1g
Therefore, -m1a1y = T - m1gOr,
m1a1y = m1g - T
On substitution of a2z = -a1y and
T = m2a2x + m2g sin(θ) in the above equation:
m1(-a2z) = m2a2x + m2g sin(θ) - m1g
Therefore, a1y = (m2g sin(θ) - m1g) / m1 - m2a2x
By solving this equation, the value of a1y.
a1y = -g
when m2 = 0a1y = 0
when m1 = m2 and θ = π/2.
The dimension of acceleration is m/s².
Thus, a1y in terms of m1, m2, θ, and g is given by
a1y = (m2g sin(θ) - m1g) / m1 - m2a2x
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In Fig. two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L. Assume that θ is so small that than θ can be replaced by its approximate equal, sinθ. Show tthat x=(2πε0mgq2L)1/3 given the equilibrium separation x of the ball.
For the given question, the expression of the equilibrium separation x of the ball should be [tex](Q^2L/2\pi \epsilon_0mg)^{1/3}[/tex].
Determining the equilibrium separation x of the ball Two tiny conducting balls of identical mass m and identical charge q are hanging from non-conducting threads of length L.
Let the distance between the balls be x when they are in equilibrium.
The net force acting on each ball due to the electric field of the other ball is given by:
[tex]F = (1 / 4\pi\epsilon_0) q^2/ x^2[/tex]
The tension in the threads is given by:
[tex]T = mg/cos\theta[/tex]
The system will be in equilibrium when repulsive force due to like charges is equal to the tension component in the horizontal direction.
[tex]Tsin\theta = (1 / 4\pi\epsilon_0) q^2/ x^2[/tex]
For small angles, sinθ ≈ θ,
[tex]\theta=x/2L[/tex]
so replacing T and theta we get,
[tex]mgx/2L=(1 / 4\pi\epsilon_0) q^2/ x^2[/tex]
[tex]x=(Q^2L/2\pi \epsilon_0mg)^{1/3}[/tex]
Thus the equilibrium separation between two charged masses is [tex](Q^2L/2\pi \epsilon_0mg)^{1/3}[/tex].
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a 135-kg k g astronaut (including space suit) acquires a speed of 2.70 m/s m / s by pushing off with her legs from a 1900-kg k g space capsule. use the reference frame in which the capsule is at rest before the push.
A) What is the velocity of the space capsule after the push in the reference frame? B)If the push lasts 0.660 s , what is the magnitude of the average force exerted by each on the other? C)What is the kinetic energy of the astronaut after the push in the reference frame? D)What is the kinetic energy of the capsule after the push in the reference frame? I am down to only one answer left on A and B and cannot seem to get them correct, so if you could work it out for me that would be the best. Thank you.
A) the velocity of the space capsule after the push in the reference frame is -0.191 m/s.
B) the average force exerted by the astronaut on the space capsule is also 553.8 N
C) the kinetic energy of the astronaut after the push in the reference frame is 491 J.
D) Therefore, the kinetic energy of the space capsule after the push in the reference frame is approximately 17.2 J.
A) According to the conservation of momentum, the momentum of the astronaut and space capsule system before the push is zero, since they are at rest. After the push, the total momentum of the system is still zero. Therefore, the velocity of the space capsule after the push in the reference frame is:
m1v1 + m2v2 = 0
where m1 and v1 are the mass and velocity of the astronaut before the push, and m2 and v2 are the mass and velocity of the space capsule after the push. Substituting the given values, we get:
(135 kg)(2.70 m/s) + (1900 kg)(v2) = 0
Solving for v2, we get:
v2 = -(135 kg)(2.70 m/s) / (1900 kg) = -0.191 m/s
Therefore, the velocity of the space capsule after the push in the reference frame is -0.191 m/s.
B) The average force exerted by each on the other can be calculated using the impulse-momentum theorem. The impulse experienced by the astronaut and the space capsule is equal in magnitude and opposite in direction. Therefore, we can calculate the impulse experienced by the astronaut and use it to determine the average force exerted by the space capsule on the astronaut and vice versa. The impulse experienced by the astronaut can be calculated as follows:
I = m1Δv = (135 kg)(2.70 m/s) = 364.5 Ns
where Δv is the change in velocity of the astronaut due to the push.
The duration of the push is 0.660 s. Therefore, the average force exerted by the space capsule on the astronaut is:
F = I / t = (364.5 Ns) / (0.660 s) ≈ 553.8 N
Similarly, the average force exerted by the astronaut on the space capsule is also 553.8 N.
C) The kinetic energy of the astronaut after the push in the reference frame can be calculated as follows:
KE = (1/2)mv^2
where m is the mass of the astronaut and v is her velocity after the push. Substituting the given values, we get:
KE = (1/2)(135 kg)(2.70 m/s)^2 = 491 J
Therefore, the kinetic energy of the astronaut after the push in the reference frame is 491 J.
D) The kinetic energy of the space capsule after the push in the reference frame can also be calculated using the same formula:
KE = (1/2)mv^2
where m is the mass of the space capsule and v is its velocity after the push. The velocity of the space capsule after the push is -0.191 m/s. Substituting the given values, we get:
KE = (1/2)(1900 kg)(-0.191 m/s)^2 ≈ 17.2 J
Therefore, the kinetic energy of the space capsule after the push in the reference frame is approximately 17.2 J.
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A uniform meter stick is balanced at its midpoint with a single support. A 50 N weight is suspended at the 30 cm mark. At what point must a 20 N weight be hung to balance the system?
The 20 N weight must be hung at the 70 cm mark to balance the system.
To find the location of the 20 N weight, we can use the principle of moments, which states that the sum of the clockwise moments about a point must equal the sum of the counterclockwise moments about the same point in order for the object to be in equilibrium.
Since the meter stick is balanced at its midpoint, the point of rotation is at the 50 cm mark. Let x be the distance from the 50 cm mark to the location where the 20 N weight is hung. Then, the clockwise moment due to the 50 N weight at the 30 cm mark is:
(50 N)(20 cm) = 1000 Ncm
And the counterclockwise moment due to the 20 N weight at the x cm mark is:
(20 N)(x - 50 cm)
To balance the system, these two moments must be equal:
1000 Ncm = (20 N)(x - 50 cm)
Solving for x, we get:
x - 50 cm = 50 cm
x = 100 cm
Therefore, the 20 N weight must be hung at the 70 cm mark (100 cm - 30 cm) to balance the system.
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Which of these best describes convection currents?Aextremely fastBunpredictableCslow-movingDeasily visible
The correct option is D, These best convection currents are easily visible.
Convection is a mode of heat transfer that occurs through the motion of fluids, such as gases or liquids. When there is a temperature difference between two fluid regions, hotter regions tend to expand and become less dense, causing them to rise and cooler regions to sink. This process creates a flow of fluid, which carries heat away from the hotter region and toward the cooler one.
Convection plays a vital role in many natural phenomena, such as weather patterns, ocean currents, and the movement of magma in the Earth's mantle. It is also used in various engineering applications, such as cooling, heating, and fluidized bed reactors.
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Complete Question:
Which of these best describes convection currents?
A. extremely fast
B. unpredictable
C. slow-moving
D. easily visible
determine whether each geologic feature is being caused by tensional, compressional, or shear stresses by analyzing the directions of the forces being applied.
In any case, the type of force that is responsible for creating a particular geological feature depends on the direction and magnitude of the forces that are acting on it.
Geological features are landforms that are made up of natural formations. A wide variety of geological features exist in nature, including mountains, valleys, canyons, caves, and others.
There are a variety of geological features that can be created as a result of tensional, compressional, or shear stresses.
Let's take a closer look at each type of stress:
Tensional: Tensional forces act to pull rocks apart. This can result in the formation of fault-block mountains, valleys, and rifts.
Compressional: Compressional forces act to push rocks together. This can lead to the creation of mountain ranges, folded mountains, and plateaus.
Shear Stresses: Shear stresses act to twist or bend rocks. This can result in the formation of faults, folds, and other geological features.
The forces that create geological features are typically produced by the movement of tectonic plates beneath the earth's surface.
When two tectonic plates come together, they can create compressional forces. When they move apart, they can create tensional forces.
When they slide past each other, they can create shear stresses. Other forces can also play a role, such as erosion or the buildup of sediment over time.
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amanda weighs about 600 n on earth, but would only weigh about 100 n on the moon. which best explains why amanda would weigh less on the moon than on earth? A. the mass of the moon is less than that of earth, therefore it has a weaker gravitational force. B. the circumference of the moon is smaller than earth, therefore it has less gravity. C. the pull from the gravity from earth decreases the pull of gravity from the moon. D. the lack of air pressure on the moon weakens the gravitational force of the moon.
Option A is the correct answer. The mass of the moon is less than that of earth, therefore it has a weaker gravitational force.
The correct option that explains why Amanda would weigh less on the moon than on earth is "A. the mass of the moon is less than that of the earth, therefore it has a weaker gravitational force." This is because weight is the result of the gravitational force that acts on an object, which is determined by both the mass of the object and the gravitational force acting on it. Therefore, the weight of an object varies depending on the mass and gravity.
The gravity of an object is the force that attracts it towards the center of the earth or the celestial object. The amount of gravity an object has depends on its mass and the mass of the object that it is attracting. The moon has a smaller mass than the earth, which means that it has a weaker gravitational force.
Consequently, the pull of gravity on the moon is weaker than on earth. The weight of Amanda is less because pull of gravity on the moon is weaker than on earth. Therefore, option A is the correct answer.
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If you stand on one foot while holding your other leg up behind you, your muscles apply a force to hold your leg in this raised position. We can model this situation as in Figure 1). The leg pivots at the knee joint, and the force that holds the leg up is provided by a tendon attached to the lower leg as shown Assume that the lower leg and the foot have a combined mass of 3.6kg, and that the combined center of gravity is at the center of Figure he knot What is the magnitude of this force? The london provides you hold your leg in this position the upper legeerts a force Express your answer with the appropriate units the lower le TARO? Value Units Sube
To keep the leg in the raised position, the tendon should provide 160N force.
The rotating force or moment of a force around a particular axis or pivot point is measured by torque. The tendency of a force to cause an object to spin along an axis is described as a vector quantity, torque.
Given: combined mass of the lower leg and the foot, m = 3.6kg
position of the center of gravity, r1 = 25cm
r = 0.25m
distance between tendon and lower leg, r2 = 5cm = 0.05m
torque applied will be τ = 3.6 × 10 × 0.25
τ = 8 N-m
the force applied by tendon
F = τ/ r2
F = 8/ 0.05
F = 160N
Therefore, To keep the leg in the raised position, the tendon should provide 160N force.
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Zach, whose mass is 80 kg, is in an elevator descending at 10 m/s.The elevator takes 3.0s to brake to a stop at the first floor.a. What is Zach's apparent weight before the elevator startsbraking ?b. What is Zach's apparent wight while the elevator is braking?a precise and step by step solution will help me understandthe problem.. thank you..
Explanation:
Let's use Zach's weight F = ma
F = 80 (9.81) = 785 N
at constant speed descending a = 9.81 and Zach's weight is still 785 N
As the elevator is slowing to a stop, Zach's apparent weight will increase
a = change in velocity / change in time = 10 m/s / 3s = 3.33 m/s^2
Zach's APPARENT weight will be F = ma where a = 9.81 + 3.33 m/s^2
F = 80 ( 9.81+3.33) = 1051 N
(a) how many kilometers does light traverse in 1 ly? km (b) what is the speed of light c in terms of ly per year. ly/y (c) express your answer from (b) in terms of feet per nanosecond. ft/ns
a) Light traverses approximately 9.461 × 10^12 kilometers in 1 light-year
.b) The speed of light in terms of ly per year is 1 ly/y.c) 1 light-year equals 5.8785 × 10^12 miles. 1 mile is equal to 5,280 feet.
Therefore, 1 light-year is equal to 31.0688 × 10^12 feet. A nanosecond is equal to one billionth of a second (1/1,000,000,000 second). Therefore, 1 second is equal to 1 × 10^9 nanoseconds. Speed is equal to distance divided by time.
Therefore, Speed of light in feet per nanosecond = (31.0688 × 10^12 feet) / (1 year × 365 days/year × 24 hours/day × 3600 seconds/hour × 1 × 10^9 nanoseconds/second) = 1.005 × 10^5 feet per nanosecond (approximately).
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matches have the potential to light on fire, but they will not do so without sufficient activation energy. explain what activation energy means and what type of activation energy the matches need.
Activation energy is the minimum energy that reactants must summon in other to become products during reactions.
What is activation energy?Activation energy is the minimum amount of energy required for a chemical reaction to occur. It is the energy barrier that must be overcome to start a chemical reaction. In order for a match to light on fire, it needs to be exposed to enough activation energy to initiate the chemical reaction between the match head and the striking surface.
When the match head is struck against the rough surface of the matchbox, friction generates heat, which provides the activation energy necessary to ignite the match.
The heat generated by friction between the match head and the striking surface provides enough activation energy to initiate a chemical reaction between the chemicals in the match head and the oxygen in the air. This chemical reaction produces heat and a flame, which can then be used to light a candle, stove, or other combustible material.
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why is europa considered a good candidate for the possible existence of life?
Europa is considered a good candidate for the possible existence of life because it has a subsurface ocean of liquid water that is believed to be in contact with a rocky seafloor, which provides a source of chemical energy and nutrients that could support life.
Additionally, observations by the Galileo spacecraft and ground-based telescopes suggest the presence of plumes of water vapor and other materials erupting from Europa's surface, providing a possible means for organic molecules to reach the surface and potentially support life.
Furthermore, Europa is believed to have a relatively stable environment and is shielded from harmful solar and cosmic radiation by its thick ice crust, which could protect any potential life forms from harmful radiation.
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Storm clouds build up large negative charges, as described in the chapter. The charges dwell in charge centers, regions of concentrated charge. Suppose a cloud has -25 C in a 1.0-km-diameter spherical charge center located 10 km above the ground, as sketched in (Figure 1) . The negative charge center attracts a similar amount of positive charge that is spread on the ground below the cloud.
The charge center and the ground function as a charged capacitor, with a potential difference of approximately 4.1×108 V . The large electric field between these two "electrodes" may ionize the air, leading to a conducting path between the cloud and the ground. Charges will flow along this conducting path, causing a discharge of the capacitor−a lightning strike.
What is the approximate magnitude of the electric field between the charge center and the ground??
What is the approximate capacitance of the charge center + ground system?
If 12.5 C of charge is transferred from the cloud to the ground in a lightning strike, what fraction of the stored energy is dissipated?
If the cloud transfers all of its charge to the ground via several rapid lightning flashes lasting a total of 1 s, what is the average power?
The electric field between the charge center and the ground can be calculated using the formula:
E = V/d
where E is the electric field, V is the potential difference, and d is the distance between the two electrodes. In this case, the potential difference is 4.1×10^8 V and the distance is 10 km (which we need to convert to meters):
d = 10 km = 10,000 m
So, the electric field is:
E = 4.1×10^8 V / 10,000 m = 4.1×10^4 V/m
The capacitance of the charge center + ground system can be calculated using the formula:
C = Q/V
where C is the capacitance, Q is the charge stored, and V is the potential difference. In this case, the charge stored is -25 C (since it's a negative charge) and the potential difference is 4.1×10^8 V:
C = -25 C / 4.1×10^8 V = -6.1×10^-8 F
Note that capacitance is always positive, but in this case, it came out negative because the charge is negative.
The energy stored in a capacitor is given by the formula:
U = 1/2 CV^2
where U is the energy stored, C is the capacitance, and V is the potential difference. In this case, the energy stored before the lightning strike is:
U = 1/2 (-6.1×10^-8 F) (4.1×10^8 V)^2 = 5.1×10^14 J
If 12.5 C of charge is transferred from the cloud to the ground in a lightning strike, the energy dissipated is:
U' = 1/2 (-6.1×10^-8 F) (4.1×10^8 V - 12.5 C/(-6.1×10^-8 F))^2 = 3.3×10^14 J
So, the fraction of the stored energy that is dissipated is:
(U - U') / U = (5.1×10^14 J - 3.3×10^14 J) / 5.1×10^14 J = 0.35 or 35%
The average power of the lightning flashes can be calculated using the formula:
P = U/t
where P is the power, U is the energy transferred, and t is the time taken. In this case, the energy transferred is 25 C × 4.1×10^8 V = 1.03×10^10 J (since the potential difference is the same as before the lightning strike), and the time taken is 1 s (since the flashes last a total of 1 s):
P = 1.03×10^10 J / 1 s = 1.03×10^10 W or 10.3 GW (since 1 GW = 10^9 W)
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