Based on the astronomers definition of stars, All stars are part of some constellation. Only the statement II is correct.
All stars visible from Earth are part of a constellation, which is a region of the sky defined by the International Astronomical Union (IAU). The IAU has divided the sky into 88 official constellations, each with its own boundaries and set of stars. Astronomers use the constellations as a way to locate and identify stars and other celestial objects in the night sky.
The brightness of a star is not necessarily related to its position or designation within a constellation, so the first statement is not correct. Also, the presence or absence of an Arabic name is not an indication of a star's brightness, so the third statement is not correct.
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--The complete question is, Every star is part of a constellation and is assigned a Greek letter within the constellation. Many stars still bear ancient Arabic names. Based on how astronomers refer to stars, select all of the correct statements from the following list.
I The star in a constellation is usually brighter than the B star.
II All stars are part of some constellation.
III A star with an ancient Arabic name is probably relatively bright.--
A particle of charge + is at point A a distance from the center of a sphere of charge +Q.The particle is moved to point B, a distance expressions represents the electric potential difference between the two points due to the charge on the sphere? from the center of the sphere. Which of the following (1-4) % (4) (9) (-)
The expressions which represent the electric potential difference between the two points due to the charge on the sphere qQ/4πε0 (1/rb- 1/ra). (Option C)
The electric eventuality, also known as implicit drop or the electrostatic eventuality refers to the quantum of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. The electric implicit generated from a point charge, Q, at a distance, r, from the position of Q is given by
V = 1/ 4πε0 Q/ r
When the electric eventuality is told by two point charge, Q and q, the formula is
V = 1/ 4πε0 qQ/ r
Electric implicit difference refers to the external work needed to bring a charge from one position to another position in an electric field. It's the change of implicit energy endured by a test charge that has a value of 1. The electric implicit difference is given as ΔV = Vb- Va
Hence, the expressions which represent the electric implicit difference between the two points due to the charge on the sphere qQ/ 4πε0( 1/ rb- 1/ ra).
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Complete question:
a particle of charge q is at point a , a distance ra from the center of a sphere of charge q . the particle is moved to point b , a distance rb from the center of the sphere. which of the following expressions represents the electric potential difference between the two points due to the charge on the sphere?
A) Q/4πε0 (1/rb- 1/ra) B) 1/4πε0 Q/rb C) qQ/4πε0 (1/rb- 1/ra) D) 1/4πε0 qQ/rb
a lot of extra positive charges are on one plate, and negative charges on the other, how do these extra charges affect the batteries ability to push even more charges onto the plates
The battery's ability to push more charges onto the plates is determined by its maximum electromotive force, which is reached when the battery has already pushed as many charges as it can onto the plates.
A battery's electromotive force (EMF), a measurement of the potential difference between its positive and negative terminals, determines its capacity to drive additional charges onto the plates. A battery's EMF reaches its peak when it is completely charged and can no longer push charges onto the plates.
The battery has already pushed as many charges as it can onto the plates if one plate has a lot of excess positive charges, while the other plate has a lot of extra negative charges. The battery can no longer put charges onto the plates since it has achieved its maximum EMF.
As a result, the additional fees on the plates have no bearing.
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A 6N force pushes to gliders along an air track. The 200 g spring between the gliders is compressed. How much force does the spring exert on (a) glider A and (b) glider B? Mass of Glider A = 400 g Mass of Glider B = 600 g
The spring exerts the same force on both glider A and glider B, equal to (Fa - (-6N))/2 = (Fb - (6N))/2.
How much force does the spring exert on each other?The force of the spring can be calculated using Hooke's law, which states that the force exerted by a spring is proportional to the displacement of the spring from its rest position.
Let's call the spring force "Fspring". Since the spring is compressed, it is exerting a force on both gliders, which we can call "Fa" for glider A and "Fb" for glider B.
The total force on each glider is equal to the spring force plus the force of the 6N external force:
Fa = Fspring + (-6N)
Fb = Fspring + (6N)
Since the forces on both gliders are equal in magnitude and opposite in direction, we can write the following equation:
Fspring = (Fa - (-6N))/2 = (Fb - (6N))/2
To calculate the spring force, we need to find the displacement of the spring from its rest position. This can be done using the following equation:
Fspring = kx
Where;
k is the spring constant
Since the spring is compressed, the displacement "x" is negative. We can find the spring constant "k" using the equation:
k = Fspring / x
We can substitute the equation for Fspring from above into this equation:
k = (Fa - (-6N))/2x / x
We can rearrange this equation to solve for "x":
x = (Fa - (-6N))/(2k)
We can then substitute this value of "x" back into the equation for Fspring to find the force exerted by the spring:
Fspring = (Fa - (-6N))/2 = (Fb - (6N))/2
Since the masses of glider A and glider B are known, we can use the following equation to find the acceleration of each glider:
Fa = ma
Fb = mb
Where "m" is the mass of each glider and "a" is the acceleration of each glider.
We can rearrange these equations to solve for the acceleration of each glider:
a = Fa / m
b = Fb / m
Now that we have the acceleration of each glider, we can use Newton's second law to find the force exerted by the spring on each glider:
Fa = ma = 400 g * a
Fb = mb = 600 g * b
Finally, we can substitute the value of Fspring from above into these equations to find the force exerted by the spring on each glider:
Fa = Fspring + (-6N) = (Fa - (-6N))/2 + (-6N)
Fb = Fspring + (6N) = (Fb - (6N))/2 + (6N)
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(a) suppose that the center of the outer electron cloud of a carbon atom shifts a distance when the atom is polarized by the pen. calculate algebraically in terms of the charge on the pen. (use the following as necessary: , for acceleration due to gravity, for the distance from the carbon to the pen, for the charge of the electron, , , and
The center of the outer electron cloud of a carbon atom shifts a distance when the atom is polarized by the pen is, [tex]s = \dfrac{\pi \epsilon_0 m g d^3}{2eQ}[/tex].
When a negatively charged object is brought close to a small piece of paper, the paper charges cannot be distributed as they would if it were a conductor.
Outer electron Cloud Charge = −4e
Carbon Atom shifts Distance = s=?
The Pen Charge = Q
Force on Pen Charge is, [tex]F = Q\dfrac{1}{4\pi \epsilon_0} \dfrac{8es}{d^3}[/tex]
The sum of the forces, both that of gravity and that exerted by the charge Q, must be equal to zero to maintain equilibrium at a distance s.
[tex]mg - Q\dfrac{1}{4\pi \epsilon_0} \dfrac{8es}{d^3} = 0[/tex]
This gives the distance s, to be
[tex]s = \dfrac{\pi \epsilon_0 m g d^3}{2eQ}[/tex]
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--The complete question is, Try rubbing a plastic pen through your hair, and you'll find that you can pick up a tiny scrap of paper when the pen is about one centimeter above the paper. From this simple experiment you can estimate how much an atom in the paper is polarized by the pen! You will need to make several assumptions and approximations.
(a) Suppose that the center of the outer electron cloud (q = -4e) of a carbon atom shifts a distance s when the atom is polarized by the pen. Calculate s algebraically in terms of the charge Q on the pen. (Use the following as necessary: m, g for acceleration due to gravity, h for the distance from the carbon to the pen, e for the charge of the electron, Q,ε0, and π. )--
A positive test charge is brought near to, but not touching a conducting sphere that is connected to the ground. Both objects remain at rest in the positions shown above. Charge begins to flow from the ground to the sphere. Which of the following statements best describes when charge stops flowing and provides justification for the claim? A. Charge will flow until the electric field at the surface of the sphere is equivalent to the electric field of the test charge, because then the excess charge on the surface of the sphere will be equal to the charge of the test charge B. Charge will flow until the electric field at the surface of the sphere is equivalent to the electric field of the test charge, because then the net force on all charges on the surface of the sphere will be zero C. Charge will flow until the potential at the surface of the sphere is the same as the potential of the test charge, because then the net force on all charges on the surface of the sphere will be zero. D. Charge will flow from the ground to the sphere until the potential is the same everywhere within the sphere, because then the excess charge on the surface of the sphere will be equal to the charge of the test charge.
E. Charge will flow from the ground to the sphere until the potential is the same everywhere within the sphere, because then the net force on all charges on the surface of the sphere will be zero
A charge will flow until the electric field at the surface of the sphere is equivalent to the electric field of the test charge, because then the net force on all charges on the surface of the sphere will be zero. Here option B is the correct answer.
When the positive test charge is brought near the conducting sphere, the excess negative charge from the ground flows onto the surface of the sphere until equilibrium is reached. At this point, the electric field at the surface of the sphere becomes equal and opposite to the electric field of the test charge, thus canceling out the electric field of the test charge within the sphere.
Since the electric field inside the conductor is zero, the excess charges on the surface of the conductor are uniformly distributed and the net force on them is zero. Therefore, option B is the correct answer. Option A is incorrect because the excess charge on the surface of the sphere will be opposite in sign and not equal in magnitude to the test charge.
Option C is incorrect because the potential at the surface of the sphere is not necessarily the same as that of the test charge. Option D is incorrect because the excess charge on the surface of the sphere will be opposite in sign and not equal in magnitude to the test charge. Option E is incorrect because the net force on charges on the surface of the sphere is zero when the electric field at the surface of the sphere is equivalent to the electric field of the test charge.
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A cat is being chased by a dog. both are running in a straight line at constant speeds. the cat has a head start of 4.2 m. the dog is running with a speed of 9 m/s and catches the cat after 5.6 s. How fast did the cat run?
The required speed of the cat when the relative speed and the speed of the dog are given is calculated to be 8.25 m/s.
Distance by which the dog lags behind the cat is given as d = 4.2 m
Speed of the dog v₂ = 9 m/s
The dog catches the cat in time t = 5.6 s
Relative speed of the dog with respect to the cat is given by,
vr = v₂ - v₁ = 9 - v₁
where,
v₁ is the speed of the cat
Therefore, we can write,
t = d/(9 - v₁)
Making v₁ as subject, we have,
v₁ = v₂ - d/t = 9 - 4.2/5.6 = 9 - 0.75 = 8.25 m/s
Thus, the speed of the cat is calculated to be 8.25 m/s.
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suppose that the force applied to the object were twice as large. sketch with dashed lines on the same axes above the force, acceleration, and velocity vs. time
Force is a physical quantity that describes the interaction between two objects, which can cause a change in the motion of an object. When a force is applied to an object, it can either cause the object to start moving, speed up, slow down, or change direction.
Force is typically measured in units of Newtons (N) and is represented by a vector, which has both magnitude (strength) and direction.
For example, pushing a book across a table requires a force that is applied in the direction of the push. If the force is strong enough to overcome the book's friction with the table, the book will begin to move in the direction of the force.
The amount of force required to produce a given effect on an object depends on several factors, including the mass of the object, the nature of the force, and the length of time the force is applied. In general, the greater the force applied to an object, the greater the acceleration (or deceleration) of the object.
Assuming that the force applied to an object is directly proportional to the acceleration it experiences, if the force were doubled, the acceleration would also double. This would result in a steeper slope on the acceleration vs. time graph.
As velocity is the integral of acceleration over time, if the acceleration were doubled, the velocity would increase at a faster rate than before. This would result in a steeper slope on the velocity vs. time graph.
Lastly, if the force applied to an object were doubled, the object would reach a higher velocity than before in the same amount of time. This would result in a higher maximum velocity and a longer time to come to a stop (assuming a constant force is applied). The velocity vs. time graph would shift upward and to the right, indicating a higher maximum velocity and a longer time to reach that maximum velocity.
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Two identical arrows, one with twice the kinetic energy of the other, are fired into a hay bale. The faster arrow will penetrate______
a. the same distance as the slower arrow b. twice as far as the slower arrow. c. four times as far as the slower arrow. d. more than four times as far as the slower arrow. e. none of these
The correct answer is b. twice distance as far as the slower arrow.
The kinetic energy of an object is given by the equation:
KE = 1/2mv^2
where m is the mass and v is the velocity.
Since the two arrows are identical, they have the same mass. Therefore, the kinetic energy is directly proportional to the square of the velocity.
If one arrow has twice the kinetic energy of the other, this means that its velocity is √2 times greater. The distance that an object travels is directly proportional to its velocity, so the faster arrow will travel √2 times farther than the slower arrow.
However, since the kinetic energy is proportional to the square of the velocity, the distance that the faster arrow travels will be (√2)^2 = 2 times greater than the distance traveled by the slower arrow.
Thus, the faster arrow will penetrate twice as far as the slower arrow.
Therefore, the correct answer is option b)twice as far as the slower arrow.
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Two children, who are 224 meters apart, start walking toward each other at the same instant at rates of 1.5 m/sec and 2 m/sec, respectively when will they meet? How far will each have walked?
Both the children meet at t = 64 s and the distance travelled by first child is 96 m, distance travelled by second child is 128 m.
Two children are said to be 224 m apart.
First child walks towards the second with 1.5 m/s.
Second child walks towards the first with 2 m/s.
Let V = V₁ + V₂ = 1.5 + 2 = 3.5 m/s
We know the expression for, distance, velocity and time as,
s = v t
where,
v is velocity
t is time
s is distance
Let us find out the time at which they meet.
t = s/v = 224/3.5 = 64 s.
They meet at t = 64 s.
The distance travelled by first child s₁ = V₁ t = 1.5 × 64 = 96 m.
The distance travelled by first child s₂ = V₂ t = 2 × 64 = 128 m.
Thus, the first child walked 96 m and the second child walked 128 m.
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The total current density in a semiconductor is constant and equal to J10 A/cm2. The total current is composed of a hole drift current and an electron diffusion current. Assume that the hole concentration is constant and equal to 1016 cm3 and that the electron concentration is given by n(x) = 2x 1015c"1cm" where L = 15 μm. The electron diffusion coefficient is D,-27 cm2/s and the hole mobility is μ,-420 cm2/Vs. Calculate (a) the electron diffusion current density for x > 0, (b) the hole drift current density for >0, and (c) the required electric field for x >0.
The required electric field for x > 0 is 2.0 x 10^3 V/cm.
Given:
Total current density, J = 10 A/cm²
Hole concentration, p = 10¹⁶ cm⁻³
Electron concentration, n(x) = 2x10¹⁵x cm⁻³, where L = 15 μm, and x > 0
Electron diffusion coefficient, Dₙ = 27 cm²/s
Hole mobility, μₚ = 420 cm²/Vs
To calculate:
(a) Electron diffusion current density for x > 0
(b) Hole drift current density for x > 0
(c) Electric field for x > 0
Solution:
(a) The electron diffusion current density can be calculated using the equation:
Jn = -qDn(dn/dx)
where q is the electronic charge, and dn/dx is the gradient of the electron concentration profile.
q = 1.6x10⁻¹⁹ C (electronic charge) (electronic charge)
Dn/Dx = 2 x 1015 cm4 (gradient of electron concentration)
If we apply these values, we obtain:
Jn = -(1.6x1019 C) (27 cm2/s) (2x1015 cm4) = -8.64 A/cm2.
The electron diffusion current is said to be flowing in the opposite direction from the direction of the total current density when it has a negative sign.
(b) The equation: Jp = qppE, where E is the electric field and pp is the hole concentration, can be used to calculate the hole drift current density.
pp = 10¹⁶ cm⁻³ (constant hole concentration) (constant hole concentration)
The values of Jn and pp, along with the formula J = Jn + Jp, allow us to calculate Jp:
Jp = Jn = 10 A/cm2 - (-8.64 A/cm2) = 18.64 A/cm2, where
This value, q, and pp are all substituted into the equation for Jp to produce the following result:
Jp equals (1.6x1019 C)(420 cm2/Vs)(1016 cm3).
E = 18.64 A/cm²
Solving for E, we get:
E = Jp/(qμpp) = (18.64 A/cm²)/[(1.6x10⁻¹⁹ C)(420 cm²/Vs)(10¹⁶ cm⁻³)] = 0.28 V/cm
(c) The electric field required for x > 0 is 0.28 V/cm.
Therefore, the electron diffusion current density is -8.64 A/cm², the hole drift current density is 18.64 A/cm², and the required electric field is 0.28 V/cm for x > 0.
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A student fires a 0.07 kg arrow at an object with mass m that is initially at rest on a frictionless surface. The speed of the arrow before the collision is 90 m/s. The speed when the arrow emerges from the object is v. What is the resulting velocity of the object? (4 pts)
b) Is the collision between the arrow and the object elastic or inelastic? Include evidence to support your answer
To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a system is conserved in the absence of external forces. In this case, the system consists of the arrow and the object.
Before the collision, the arrow has a momentum of:
p1 = m1v1 = (0.07 kg)(90 m/s) = 6.3 kg m/s; where m1 is the mass of the arrow and v1 is its velocity.
Since the object is initially at rest, its momentum is zero:
p2 = m2v2 = (m)(0) = 0; where m2 is the mass of the object and v2 is its velocity.
After the collision, the arrow and the object move together with a common velocity v. The total momentum of the system after the collision is:
p3 = (m1 + m2)v
Using the law of conservation of momentum, we can equate the initial and final momenta:
p1 + p2 = p3
Solving for v, we get:
v = p1/(m1 + m2) = (6.3 kg m/s)/(0.07 kg + m)
To determine the resulting velocity of the object, we need to solve for v. One way to do this is to use the law of conservation of energy, which states that the total energy of a system is conserved in the absence of external forces.
The initial kinetic energy of the system is: KE1 = (1/2)m1v1^2 = (1/2)(0.07 kg)(90 m/s)^2 = 283.5 J; where KE1 is the initial kinetic energy of the arrow.
The final kinetic energy of the system is:
KE2 = (1/2)(m1 + m2)v^2
Using the law of conservation of energy, we can equate the initial and final kinetic energies:
KE1 = KE2
Solving for v, we get:
v = sqrt(2KE1/(m1 + m2)) = sqrt(2283.5 J/(0.07 kg + m))
To determine whether the collision is elastic or inelastic, we can compare the system's initial and final kinetic energies. If the kinetic energy is conserved, then the collision is elastic. If the kinetic energy is not conserved, then the collision is inelastic.
The initial kinetic energy of the system is 283.5 J. The final kinetic energy of the system is:
KE2 = (1/2)(m1 + m2)v^2
Substituting the value of v we obtained earlier, we get:
KE2 = (1/2)(0.07 kg + m)(6.3 kg m/s)^2/(0.07 kg + m)
To determine whether the collision is elastic or inelastic, we must compare KE1 and KE2. Since the object's mass is unknown, we cannot make a definitive conclusion about the nature of the collision. However, we can provide some evidence to support our answer.
If the collision is elastic, then KE2 should be equal to KE1. However, if we substitute v = 90 m/s into the expression for KE2, we get: KE2 = (1/2)(0.07 kg + m)(90 m/s)^2/(0.07 kg + m) = 4050 J. This value is greater than KE1, which indicates that the collision is not elastic. Therefore, we can conclude that the collision is inelastic.
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According to the 1998 Guinness Book of World Records stuntman Dan Koko fell a distance of 365 feet into an airbag after jumping from the Vegas World Hotel and Casino. The distance d in feet traveled by a freefalling object in t seconds is given by the formula d = 16t 2. To the nearest tenth of a second, how long did the stuntman's freefall last?
Solution of the question with the information which is given is:
Ignoring wind resistance and using 32ft/sec/sec as acceleration due to gravity.
d= 16 t2
t2= 365/ 15
t= 4.8 seconds
Therefore the hit impact was about 4.8 seconds
In Newtonian physics, free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it.
According to Newton's second law of motion, force (F) = mass (m) times acceleration, an item in free fall will move with an acceleration (a). We can use a little mathematics to determine the object's acceleration in terms of the net external force and the object's mass (a = F / m).
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We are given 5 (independent) equations and 6 unknowns. Currents (the I variables) are given in amps and voltage (the V variables) in volts. I = I_1 + I_2 V_1 = 2l_1 V_2 = 5l_2 V_1 - V_2 = 0 V - 5 = V_1 Solve (using multiple substitutions) to find I in terms of V and no other variables. I = V+ (within three significant digits) If we are giving one more fact (a 6^th equation) V = 3 volts what will be the value of 1? I = amps (within three significant digits)
An independent equation is an equation in a system of simultaneous equations which cannot be derived algebraically from the other equations. The concept typically arises in the context of linear equations.
I = [tex]I_{1} + I_{2}[/tex]
[tex]V_{1} = 2I_{1}[/tex]
[tex]V_{2} = 5I_{2}\\V_{1} - V_{2} = 0 \\V_{1} = V_{2}\\V-5= V_{1}\\V = V_{1} + 5[/tex]
put equation 2 and 3 in 1
[tex]I =V_{1}/2 + V_{2} /5\\= V_{1} [1/2 + 1/5]\\= 7V_{1}/ 10\\I = 7V_{1} / 10[/tex]
put 5 in 6
[tex]I = 7/10 (v-5) [ I = 0.7V- 3.5][/tex]
given V= 3
I = 0.7 V - 3.5
I = 0.7 (3) - 3.5
[ I = -1.4 A]
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Why there are positive and negative magnetic anomalies on the seafloor?
Positive anomalies result when the crust is magnetised in a normal polarity parallel to the ambient field of Earth, and negative anomalies result when the crust is reversely magnetised in an opposite sense.
A positive magnetic anomaly is a reading that exceeds the average magnetic field strength and is usually related to more strongly magnetic rocks, such as mafic rocks or magnetite‐bearing rocks, underneath the magnetometer. A value that is lower than the typical magnetic field is referred to as a negative magnetic anomaly.
These anomalies are classified as negative or magnetic lows if their strength is below normal and positive or magnetic highs if it is above normal. These anomalies are typically caused by variations in magnetic susceptibility, which in turn mostly reflect the rocks' magnetite content.
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What is the average force of gravitation between Venus and the sun?
The average force of gravitation between Venus and the Sun is3.24 x 10²² N.
What is average gravitational force?The force of gravitation between two objects can be calculated using the equation;
F = G (m1m2) /d²
where;
F is the force of gravitation, G is the gravitational constantm1 and m2 are the masses of the two objects, and d is the distance between them.To find the average force of gravitation between Venus and the Sun, we need to know their masses and the average distance between them.
Venus has a mass of approximately 4.87 x 10²⁴ kg, and the average distance between Jupiter and the Sun is about 108 million kilometers.
Plugging these values into the equation, we get:
F = 6.67 x 10⁻¹¹ (4.87 x 10²⁴) x (1.989 x 10³⁰ kg) / (108 x 10⁶)²
F = 3.24 x 10²² N.
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Car A travels with speed v around curve number one, which has a radius r. Car B travels with speed 2v around curve number two, which has a radius 2r. The acceleration will be ?
1-greater for car A.
2-greater for car B.
3-zero for both cars.
4-the same for both cars.
chose one
The acceleration will be greater for car B.
What is Acceleration ?Acceleration is a physical quantity that describes the rate at which an object's velocity changes over time. It is defined as the change in velocity per unit of time, and it is a vector quantity, meaning it has both magnitude and direction.
When an object experiences acceleration, its velocity changes either in speed, direction, or both. If the acceleration is in the same direction as the velocity, the object's speed increases. If the acceleration is in the opposite direction as the velocity, the object's speed decreases. If the acceleration is perpendicular to the velocity, the object's direction changes.
The standard unit of acceleration is meters per second squared (m/s^2). A positive acceleration indicates that an object is speeding up, while a negative acceleration indicates that an object is slowing down. If the acceleration is zero, then the object's velocity is constant.
According to the given information:To calculate the acceleration of each car around its respective curve, we can use the following formula:
a = v^2 / r
where a is the centripetal acceleration of the car, v is its speed, and r is the radius of the curve.
For car A, traveling at speed v around curve number one with radius r, the acceleration is:
a_A = v^2 / r
For car B, traveling at speed 2v around curve number two with radius 2r, the acceleration is:
a_B = (2v)^2 / (2r) = 4v^2 / (2r) = 2v^2 / r
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13) With which force did Jupiter, and the other large planets, attract gases to its surface?
Answer: The force of gravity
Jupiter and other gas Giants use gravitational force to attract gas to the top, while they still have a solid core,the force in the center keeps the planet together.
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You have a horizontal grindstone (a disk) that is 93 kg, has a 0.33 m radius, is turning at 86 rpm (in the positive direction), and you press a steel axe against the edge with a force of 21 N in the radial direction.
a) assuming that the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone in rad/s2.
b) what is the number of turns, N, that the stone will make before coming to rest?
a) The angular acceleration of the grindstone 0.27 rad per square second. b) The number of turns stone will make before coming to rest is 23.9 revolutions.
a) The tangential force slowing the grindstone is fF.
Its torque is rfF.
By Newton's Second Law
rfF = Iα
Therefore, α = 2fF/mr
Substituting actual numbers
α = 2×0.2×21/(93×0.33) = 0.27 rad per square second
b) Let t be the time it takes the grindstone to come to rest, θ be the total angle by which the grindstone will turn before coming to rest.
ω = αt
θ = αt²/2
From the first and second,
θ = ω²/2α
Substitute actual numbers
θ = 9²/2×0.27 = 150
The above result is in radians.
Since the result is supposed to be given in revolutions, 150/2π = 23.9 revolutions.
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the large positive charge inside the shell causes equal in magnitude charges distributed on the inner and outer surfaces of the spherical shell. which of the pictures best represents the charge distribution on the inner and outer walls of the shell? (figure 1)
The picture that best represents the charge distribution on the inner and outer walls of the shell is picture 4
How does charge distribution affect a conducting shell?Charge distribution on a conducting shell can affect the electric field both inside and outside the shell. If the charge is distributed uniformly on the surface of the shell, the electric field inside the shell will be zero, while the electric field outside the shell will be the same as that of a point charge located at the center of the shell.
If the charge is distributed non-uniformly, then the electric field inside and outside the shell will be affected accordingly. In general, the charge distribution will determine the potential difference between the inside and outside of the shell, which can have important implications for the behavior of charged particles both inside and outside the shell.
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Three ice skaters, numbered 1, 2, and 3, stand in a line, each with her hands on the shoulders of the skater in front. Skater 3, at the rear, pushes forward on skater 2. Assume the ice is frictionless. Draw a free body diagram for skater 2 in the middle.Â
Since skater 2 is in the middle, force acting on her from both skater 1 in front and skater 3 behind. The free body diagram for skater 2 would include two arrows pointing towards her, representing the forces from skater 1 and skater 3. Since there is no friction, there are no other forces to consider.
The diagram would look something like this:
∧
|
F3 ← I → F1
|
v
where F1 is the force from skater 1 and F3 is the force from skater 3.
What is friction?A force called friction works against the relative velocity of two surfaces coming into contact. The minute imperfections on the surfaces that interlock with one another when they come into touch are what cause it. We experience friction as a result of this interlocking as resistance to motion. The type of surfaces in contact, the force forcing the surfaces together, and the rate of motion can all have an impact on the frictional force. As some of the kinetic energy is transformed into heat energy during friction, this can also result in energy loss. Designing machinery, vehicles, and structures to reduce wear and tear and maximize performance requires an understanding of friction.
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a ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. ignore air resistance. the initial vertical component (y axis) of velocity is group of answer choices 34.3 m/s upward 0, since the ball is thrown horizontally none of the above 28 m/s downward 34.3 m/s downward 28 m/s upward
Because the ball is thrown horizontally and lacks a vertical component at the beginning, the correct response is "0."
How does air resistance function? What is it?Air exerts a force known as air resistance. The force acts in the opposite direction when an object is flying through the air. Although a sports car with a sleek design will experience less drag and less air resistance, the car will be able to drive more quickly than a truck with a flat front.
The reason why air resistance is a forceA moving object experiences air resistance force when the air is pressing up against it. An example of a frictional force is air resistance. A force is always applied to stop an item in motion. Initially, air The force of resistance is not particularly strong.
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In a mechanical compressional wave, the wave energy causes the matter in the medium to move up and down or back and forth at right angles to the direction the wave travels.
answer choices
True
False
In a mechanical compressional wave, the wave energy causes the matter in the medium to move up and down or back and forth at right angles to the direction the wave travels. [FALSE]
About Compressional waveP-waves, or primary waves, are one of two types of seismic waves and are often called surface waves (so called because they propagate through the earth). A wave produced by an earthquake and recorded by a seismograph. It gets its name because it is faster than other seismic waves, reaches each seismic station first, and the next wave is called the S wave or secondary wave. Like pressure waves and longitudinal waves, sound is also a type of P wave. This means that particles in the ground (body of the earth) vibrate along or parallel to the direction of propagation of the energy of the propagating wave.
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A parallel-plate capacitor has plate area A. A battery is used to charge the capacitor so that the magnitude of charge on each plate is Q, and then is disconnected. Initially, the capacitor has a plate separation of d. At this separation the capacitor contains energy U. The plates are then moved to a separation of 2d without disturbing the charge. What is the energy of the capacitor at this larger plate separation? Show work and steps.
(a) U
(b) U/2
(c) 2U
(d) U/4
(e) 4U
The energy of the capacitor at this larger plate separation will be= D) U/4
The energy stored in a parallel-plate capacitor with plate area A, plate separation d, and charge Q is given by the formula:
U = (1/2) * (Q^2 / (ε_0 * A * d))
where ε_0 is the permittivity of free space.
If the separation between the plates is increased from d to 2d without changing the charge on the plates, the capacitance of the capacitor will be reduced by a factor of 2. This is because the capacitance of a parallel-plate capacitor is given by the formula:
C = ε_0 * A / d
So, when the separation is doubled, the capacitance is halved.
Since the charge on the plates remains constant, the energy stored in the capacitor is proportional to the square of the charge and inversely proportional to the capacitance. Thus, the new energy of the capacitor when the plates are separated by 2d is given by:
U' = (1/2) * (Q^2 / (ε_0 * A * (2d))) * (1/2)
where the factor of 1/2 is included because the capacitance is halved.
Simplifying this expression, we get:
U' = U / 4
Therefore, the energy of the capacitor at the larger plate separation is (d) U/4.
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(6%) problem 12: a stone is tossed horizontally from the highest point of a 85 m building and lands 105 m from the base of the building. ignore air resistance, and use a coordinate system whose origin is at the highest point of the building, with positive y upwards and positive x in the direction of the throw.A. How long is the stone in the air in s?
B. What must have been the initial horizontal component of the velocity, in m/s?
C. What is the vertical component of the velocity just before the stone hits the ground, in m/s?
D. What is the magnitude of the velocity of the stone just before it hits the ground, in m/s?
A ball is thrown horizontally from a 95 m-high building's highest point and landing 125 m from the building's base. Use a coordinate and disregard air resistance.
What does the word "resistance" mean?
Describe resistance. Electrical | January 12, 2021 The obstruction to part of the electrical circuit is measured by resistance. The Greek letter beta () represents the unit of measurement for resistance, known as ohms. Georg Simon Ohm (1784–1854), a German physicist that investigated the connection between voltage, current, and resistance, is the name given to the unit of resistance.
What is a circuit's electrical resistance?
The relationship between the induced voltage and the current that flows through a circuit determines its electrical resistance. Ohms is the symbol for electrical resistance. Some materials permit the flow of electric charge more easily than others. The electrical resistance gauges how much the circuit restricts the flow of the this electric charge.
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What are the magnitude and direction of the electric field at P 1 when t=3.00s and r 1 =0.02m.
a. 0,3 v/m
b. 0,6 v/m
c. 0,9 v/m
d. 0,12 v/m
The magnitude and direction of the electric field at P 1 when t = 3.00 s and r 1 = 0.02m is 0.3 v/m (option A). The direction is anti clockwise.
The force per charge on the test charge is a straightforward way to define the size of the electric field. From its definition, the common metric units for electric field strength are derived. Electric field units would be force units divided by charge units since the definition of an electric field is a force per charge.
The magnitude of the electric field at P1 is as follows:
E1 = dΦ/dt
E (2πr1) = S dB/dt
= r1/2 (dB/dt)
= r1/2 ║6t² + 8t║
= 0.2 / 2 ║6 (3²) + 8 (3)
= 0.3 v/m
The electric field is oriented counter clockwise to the direction of the induced current in a hypothetical circular conducting loop that passes through P 1.
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to determine the effect of the temperature dependence of the thermal conductivity on the temperature distribution in a solid, consider a material for which this dependence may be represented as k
To determine the effect of the temperature dependence of the thermal conductivity on the temperature distribution in a solid, consider a material for which this dependence may be represented as K=K1+aT.
Temperature and conductivity is inversely related as temeprature increases ,thermal diffusivity and thermal conductivity decrease. Thermal conductivity is related linearly so it decrease with increasing temperature. At high temperature, thermal diffusivity and conductivity approach constant.
Given that
k = ko + aT
From Fourier law
Heat transfer per unit volume given as
x measured from left hand side of wall.
By integrating
qdx=-(ko + aT)dT
When a = 0 :
qx=-koT+C
This is become straight line.
When a > 0 :
dT/dX= increase
K decrease when x is increases.
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A man, with his arms at his sides, is spinning on a light frictionless turntable. When he extends his arms
a)his moment of inertia decreases.
b)his angular momentum decreases.
c)his angular momentum remains the same.
d)his angular velocity increases.
e)his angular velocity remains the same.
A man, with his arms at his sides, is spinning on a light frictionless turntable. When he extends his arms his angular velocity remains the same. Option e is the correct choice.
When the arms are folded the rotational inertia is lower. When the man extends his arm he is basically increasing his rotational inertia thereby loosing angular velocity using the principle of conservation of angular momentum. This implies less angular distance is covered in the same time giving us negative work.
After pulling arms, the angular velocity increases but moment of inertia of decreases in, such a way that angular momentum remains constant.
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in terms of known quantities, write an expression for the time the arrow is in the air until it returns to launch height.
Assuming that the arrow is launched vertically upward with an initial velocity v and air resistance is negligible, the time the arrow is in the air until it returns to launch height can be expressed as:
What is air resistance ?When air resistance is minimal and an arrow is shot vertically upward with an initial velocity of v, the time it spends in the air before returning to launch height can be represented as:t = 2v/g
where g is the acceleration brought on by gravity (about 9.81 m/s2), t is the passage of time, v is the arrow's starting speed.To know more about air resistance , check out :
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compared to the electrical potential energy of the system at a separation of 12 meters, the electrical potential energy of the system at a separation of 6 meters is
The electrical potential energy of the system at a separation of 6 meters is greater than the electrical potential energy of the system at a separation of 12 meters.
The electrical potential energy of the system at a separation of 6 meters is greater than the electrical potential energy of the system at a separation of 12 meters. This is because electrical potential energy is inversely proportional to the separation distance between the charges. As the separation distance decreases, the electrical potential energy increases. This relationship can be represented by the equation:
U = kq1q2/r
where U is the electrical potential energy, k is a constant, q1 and q2 are the charges, and r is the separation distance. As r decreases, U increases.
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Sally was removing her nail polish to get ready for a dance. She was also drinking a bottle of water. The doorbell rang and she rushed to answer it, leaving the open bottle of nail polish remover (acetone) and the bottle of water open. Sally forgot about them until the next morning and when she checked, the nail polish remover (acetone) was empty and the water looked untouched.
Sally believes the particles in water stick together more. In terms of your knowledge of the strength of electrical forces and the structure of substances, explain Sally's statement.
Answer:
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