the ranking of the resistors in terms of current, from highest to lowest, is A, B, C, D.
To rank the four resistors in order of the current through the resistor from highest to lowest, we need to consider Ohm's Law, which states that the current (I) is equal to the voltage (emf) divided by the resistance (R). Mathematically, this is represented as I = emf / R.
Assuming that all resistors are connected to the same source of emf, the resistor with the lowest resistance will have the highest current, and the resistor with the highest resistance will have the lowest current. Therefore, we can rank the resistors based on their resistance values:
1. A. the 6.00-Ω resistor
2. B. the 8.00-Ω resistor
3. C. the 20.0-Ω resistor
4. D. the 25.0-Ω resistor
So the ranking of the resistors in terms of current, from highest to lowest, is A, B, C, D.
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a wire 35 cm long is parallel to a 0.53- t uni- form magnetic field. the current through the wire is 4.5 a. what force acts on the wire?
Answer: The force acting on the wire is 0 N
The formula for the force exerted by a magnetic field on a current-carrying wire is F = BIL sin(theta), Where, F = force B = magnetic field strength, I = current, L = length of the wire, Theta = angle between the wire and the magnetic field direction Given that Length of the wire (L) = 35 cm = 0.35 m. Magnetic field strength (B) = 0.53 T
Current through the wire (I) = 4.5 A, We need to find the force acting on the wire (F).The angle between the wire and the magnetic field is 0° as the wire is parallel to the field. Therefore, sin(theta) = sin(0°) = 0° Using the formula, F = BIL sin(theta) F = 0.53 T × 4.5 A × 0.35 m × sin(0°) = 0 N
Therefore, the force acting on the wire is 0 N, as the wire is parallel to the magnetic field direction. It means that the magnetic field does not exert any force on the wire. Note that the force will be non zero if the wire is not parallel to the magnetic field direction.
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two balls with masses of 2 kg and 6,3 kg travel toward each other at speeds of f14 and 3 respectively if the balls have a head on inelastic collision and the 2 kg ball recoils with a speed of 3.2 how much kinetic energy
If the balls have a head-on, inelastic collision and the 2.0-kg ball recoils with a speed of 3.2 m/s, the kinetic energy lost in the collision is 364.6 J.
Using conservation of momentum, we can find the final velocity:
m1 * v1 + m2 * v2 = (m1 + m2) * vf
Solving for vf, we get:
vf = (m1 * v1 + m2 * v2) / (m1 + m2)
= (2.0 kg * 14 m/s + 6.3 kg * 4.0 m/s) / (2.0 kg + 6.3 kg)
= 6.0 m/s
The final total kinetic energy of the system is:
KEf = (1/2) * (m1 + m2) * vf^2
= (1/2) * 8.3 kg * (6.0 m/s)^2
= 112.2 J
The kinetic energy lost in the collision is the difference between the initial and final kinetic energies:
KE lost = KEi - KEf
= 476.8 J - 112.2 J
= 364.6 J
An inelastic collision is a type of collision between two or more objects in which the total kinetic energy of the system is not conserved. In an inelastic collision, some or all of the kinetic energy of the colliding objects is converted into other forms of energy such as heat, sound, or deformation of the objects.
In an inelastic collision, the colliding objects stick together after the collision and move with a common velocity. This is in contrast to an elastic collision, in which the colliding objects bounce off each other and the total kinetic energy of the system is conserved. Inelastic collisions can occur in many different situations, such as in car crashes, when two objects collide and stick together, or when a ball hits a wall and loses some of its kinetic energy due to deformation.
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Complete Question: -
Two balls with masses of 2.0 kg and 6.3 kg travel toward each other at speeds of 14 m/s and 4.0 m/s, respectively. If the balls have a head-on, inelastic collision and the 2.0-kg ball recoils with a speed of 3.2 m/s, how much kinetic energy is lost in the collision?
star f is known to have an apparent magnitude of -26.7 and an absolute magnitude of 4.8. where might this star be located? what is the name of this star? explain your reasoning.
The star that has an apparent magnitude of -26.7 and an absolute magnitude of 4.8 would be located in the milky way galaxy, in the local group. The name of this star would be considered a supergiant
The apparent magnitude of a star is the brightness of the star as seen from Earth, while the absolute magnitude is the brightness of a star if it were located at a distance of 10 parsecs (32.6 light-years) from the Earth. The brightness or luminosity of a star determines where it is located. Stars that are more massive and luminous than the sun would be found in star clusters or in the spiral arms of the galaxy. On the other hand, stars that are less massive than the sun are typically found in the galaxy's central bulge or halo.
The distance to the star can be determined from the apparent and absolute magnitudes, using the formula:d = 10^((m-M+5)/5),where d is the distance in parsecs, m is the apparent magnitude, and M is the absolute magnitude. Substituting the values given in the question:d = 10^((-26.7-4.8+5)/5) = 10^(-26/5) = 2.51 x 10^(-6) parsecsThe distance calculated is extremely small (less than a thousandth of a light-year), so the star would be located within our Milky Way galaxy. As the star has a high luminosity, it would be considered a supergiant, so it would be visible from Earth.
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if there is no change in the charge distributions, what is the direction of the net electrostatic force on an electron located at the center of the circle?
If there is no change in the charge distributions, the direction of the net electrostatic force on an electron located at the center of the circle would be zero.
The electric field is a force that acts on a charged particle in an electric field. The electric field exerts a force on a charged particle that is proportional to the charge on the particle and the strength of the electric field.The force is exerted in the direction of the electric field. If an electron is placed in the electric field, it will experience a force in the opposite direction to the electric field.
When a charged particle is placed in a uniform electric field, the net electrostatic force on the particle is zero, as the direction of the force is opposite to the direction of the electric field.This can be understood through the principle of superposition. Since there is no change in the charge distribution, the electric field at the center of the circle will be zero. Therefore, the net electrostatic force on an electron located at the center of the circle will be zero.
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what is the approimate electrostatic force between two protons seperated bvy a distance of 1.0x10^-6
The electrostatic force between two protons separated by a distance of 1.0 × 10^-6 m is 2.3 × 10^-8 N.
Electrostatic force is the force between two electrically charged objects. This force can either be attractive or repulsive.
It is proportional to the product of the two charges and inversely proportional to the square of the distance between them.
The force is defined by Coulomb's law which states that:
The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is given as:F = kq1q2/r2
WhereF is the electrostatic forcek is Coulomb's constantq1 and q2 are the charges of the two particles is the distance between the two particles in meters.
The value of Coulomb's constant is 9.0 × 10^9 Nm^2/C^2.Let's consider two protons separated by a distance of 1.0 × 10^-6 m. The charge on each proton is +1.6 × 10^-19 C.
F = kq1q2/r2F = (9.0 × 10^9 Nm^2/C^2)(+1.6 × 10^-19 C)(+1.6 × 10^-19 C)/(1.0 × 10^-6 m)^2F = 2.3 × 10^-8 N
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mars has a mass of about 0.1075 times the mass of earth and a diameter of about 0.533 times the diameter of earth. the acceleration of a body falling near the surface of mars is about:group of answer choices0.30 m/s21.4 m/s22.0 m/s23.7 m/s226 m/s2
Mars has a mass of about 0.1075 times the mass of earth and a diameter of about 0.533 times the diameter of earth.The acceleration of a body falling near the surface of Mars is about 3.7 m/s².
Given, Mars has a mass of about 0.1075 times the mass of Earth and a diameter of about 0.533 times the diameter of Earth.
To find the acceleration of a body falling near the surface of Mars, we can use the formula:
g = GM/r²
where: g = acceleration due to gravity on Mars
M = mass of Mars
r = radius of Mars
We know that mass is proportional to the cube of the radius. So we can say:
Mars/Mass of Earth = (radius of Mars / radius of Earth)³0.1075M/ME
= (0.533RE/RE)³0.1075
= 0.01514ME
Simplifying this equation, we can say:
M = 0.1075 × MEr = 0.533 × RE
Now, let's calculate the acceleration due to gravity on Mars:
g = GM/r²g
= (6.674 × 10⁻¹¹) × (0.1075 × ME) / (0.533 × RE)²g
= 3.7 m/s².
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if a parcel of air rises without any external forcing, what can be said about the temperature of the parcel?
When a parcel of air rises without any external forces ,the temperature decreases so it expands because there's less pressure higher up in the atmosphere.
The temperature decreases because
As the parcel expands, it cools down, and this cooling happens because the air is doing work against the atmospheric pressure. This cooling is called adiabatic cooling, (The rate of cooling is known as the dry adiabatic lapse rate when the air is dry, while the rate of cooling is known as the moist adiabatic lapse rate when the air is saturated with water vapor) and it causes the temperature of the parcel to decrease by around 10 degrees Celsius for every 1000 meters it rises, assuming the parcel is saturated with water vapor.So, in simpler terms, if a parcel of air rises on its own, it gets cooler as it goes up because it's expanding and doing work against the air around it.(The decrease in pressure causes the parcel to cool, leading to a drop in temperature.)To learn more about the parcel of air:
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which is the proper order of structures through which light must pass in order to perceive and image?
The proper order of structures through which light must pass in order to perceive and image is cornea, aqueous humor, lens, vitreous humor, retina.
These are the five main structures of the human eye that enable vision, including light perception and imaging. Let's delve into each of these structures.
Cornea: The clear, protective outer layer of the eye is the cornea. The cornea has two purposes: to shield the inner eye from harm and to help focus light on the retina at the back of the eye.
The cornea's curved shape bends light waves as they enter the eye, assisting in their concentration.
Aqueous humor: This is a liquid that flows throughout the front of the eye, nourishing and removing waste from its surrounding tissues.
It aids in the maintenance of normal eye pressure, and if this pressure becomes too high, it can lead to glaucoma.
Lens: The lens' job is to concentrate light onto the retina. It's a transparent structure with a biconvex (lens-like) shape that varies in thickness.
It is supported by ciliary muscles that allow it to alter shape when we focus on things at different distances.
Vitreous humor: This gel-like substance fills the eye's posterior (rear) cavity, providing it with structural stability and helping it to maintain its form. It also assists in light refraction.
Retina: This is a thin layer of tissue lining the rear of the eye. The retina's photoreceptor cells, or rods and cones, are sensitive to light.
The retina converts light energy into neural signals that are transmitted to the brain via the optic nerve, which is located behind the eye. The brain translates these signals into images, allowing us to see.
What we see when we open our eyes is formed by light. In order to perceive an image, light must pass through a series of structures in the eye.
The cornea, aqueous humor, lens, vitreous humor, and retina are the five main structures of the human eye that enable vision, including light perception and imaging.
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The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit the magnetic field at 2.4 GHz, and as they do so they emit 2.4 GHz electromagnetic waves.
a. What is the magnetic field strength?
b. If the maximum diameter of the electron orbit before the electron hits the wall of the tube is 2.5 cm, what is the maximum electron kinetic energy?
a. The magnetic field strength is B = 86mT
b. The maximum electron kinetic energy K = 1.6X 10-14J
The magnetic field strength is 86mT and if the maximum diameter of the electron orbit before the electron hits the wall of the tube is 2.5 cm, then the maximum electron kinetic energy is 1.6 x 10^(-14) J.
a. To determine the magnetic field strength, we can use the equation for the electron cyclotron frequency:
f = (eB)/(2πm),
where e is the charge of the electron (1.6 x 10^-19 C),
m is the mass of the electron (9.11 x 10^-31 kg), and
f is the frequency (2.4 GHz).
First, convert the frequency to Hz:
2.4 GHz = 2.4 x 10^9 Hz.
Next, rearrange the equation to solve for B:
B = (2πmf)/e
Finally, plug in the values and calculate B:
B = (2π * (9.11 x 10^-31 kg) * (2.4 x 10^9 Hz))/(1.6 x 10^-19 C)
B = 86mT
So, the magnetic field strength is 86mT.
b. To find the maximum electron kinetic energy, we first need to determine the maximum electron velocity, using the maximum diameter of the electron orbit (2.5 cm) and the cyclotron frequency.
The circumference of the orbit is
C = πd
= π * (2.5 x 10^-2 m).
The time for one orbit is
T = 1/f
= 1/(2.4 x 10^9 Hz).
Now, find the electron's maximum velocity:
v = C/T
= (π * (2.5 x 10^-2 m))/(1/(2.4 x 10^9 Hz))
= 1.884 x 10^8 m/s.
Finally, use the equation for kinetic energy to find the maximum electron kinetic energy:
K = 0.5mv^2
= 0.5 * (9.11 x 10^-31 kg) * (1.884 x 10^8 m/s)^2
= 1.6 x 10^-14 J.
So, the maximum electron kinetic energy is 1.6 x 10^-14 J.
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A student holds a 0.06 kg egg out a window. Just before the student releases the egg, the egg has a 8.0 J of gravitational potential energy with respect to the ground. How far is the students arm from the ground? a.) 133m b.) 13.3m c.) 0.8m d.) 0.08m
a student used the setup below to investigate electric current and fields. which action will increase the current in the wire
The final answer are current is directly proportional to the potential difference and inversely proportional to the wire's resistance. Therefore, decreasing the resistance of the wire increases the current in the wire.
To increase the current in the wire of an electric current and field investigating setup, the action to be taken is to decrease the resistance of the wire. What is an electric current? The flow of electrons in a conductor is known as an electric current. To complete an electric circuit, the electrons must flow continuously in a circular pattern.
The electron movement is generated by a power supply, such as a battery. Electrons are pushed out of one end of the battery by a voltage differential between the battery terminals (the potential difference). Electrons enter the other end of the battery and complete the circuit.
The potential difference between the battery terminals drives the electrons around the circuit. This generates an electric current. The formula for current is: I = Q/t Where I is the current, Q is the amount of charge transferred, and t is the time taken.
What is the relationship between electric current and fields? When a charged particle moves through a magnetic field, a force is exerted on it. This force is proportional to the particle's velocity, as well as the magnetic field strength and the charge's magnitude.
The mathematical equation that describes this relationship is: F = qvB sinθ Where F is the force on the charged particle, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In the wire, the current is directly proportional to the potential difference and inversely proportional to the wire's resistance. Therefore, decreasing the resistance of the wire increases the current in the wire.
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magnus has reached the finals of a strength competition. in the first round, he has to pull a city bus as far as he can. one end of a rope is attached to the bus and the other is tied around magnus's waist. if a force gauge placed halfway down the rope reads out a constant 1400 newtons while magnus pulls the bus a distance of 1.55 meters, how much work does the tension force do on magnus? the rope is perfectly horizontal during the pull.
The work done by the tension force on Magnus is 2170 J.
What is work?
Work is the product of the force acting on an object and the distance through which the object moves. In other words, work is accomplished when a force is used to transfer energy to an object, causing the object to move some distance as a result.
The force of 1400 N, Distance of 1.55 meters, and a rope tied around Magnus's waist.
The work done by the tension force on Magnus is the product of the force exerted by the tension force and the distance through which Magnus is moved.
W = Fd
where W = Work done by the tension force on Magnus
F = Force of tension force
= 1400 Nd
= Distance moved by Magnus
= 1.55 m
Substituting these values:
W = 1400 N x 1.55 mW
= 2170 J
Hence, the work done by the tension force on Magnus is 2170 J.
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a fan is rotating at a constant 339.0 rev/min. what is the magnitude of the acceleration (in m/s2) of a point on one of its blades 13.0 cm from the axis of rotation? m/s2
Magnitude of the acceleration of fan blades will be approximately 165.9 m/s2
The first step is to convert the rotational speed from revolutions per minute (RPM) to radians per second (rad/s), since acceleration is measured in meters per second squared (m/s^2), which is a linear unit of measurement. We can use the conversion factor:
1 revolution = 2π radians
Therefore, 339.0 rev/min is equal to:
339.0 rev/min * (2π rad/1 rev) * (1 min/60 s) = 35.6 rad/s
The acceleration of a point on one of the fan blades can be found using the formula for centripetal acceleration:
a = rω^2
where:
a is the centripetal acceleration
r is the radius of the fan blade (0.13 m)
ω is the angular velocity in radians per second (35.6 rad/s)
Plugging in the values:
a = (0.13 m)(35.6 rad/s)^2
a = 165.9 m/s^2 (rounded to three significant figures)
Therefore, the magnitude of the acceleration of a point on one of the fan blades is 165.9 m/s^2.
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A common technique used to measure the
force constant k of a spring is the following:
Hang the spring vertically, then allow a
mass m to stretch it a distance d from the
equilibrium position under the action of the
“load” m g.
Find the spring constant k if the spring
is stretched a distance 87 m by a suspended
weight of 69 N. The acceleration of gravity is
9.8 m/s^2.
Answer in units of N/m.
Answer:
The answer to your problem is, the spring constant K should be 6 N/m.
Explanation:
When the force F should be applied at the spring so it generated the stretching distance
i.e.
F = k.x
F means the force
K means the constant
x means the stretching distance
Here the weight should be 96 N, the stretching distance is 16 km
Now the spring constant should be:
K = f/x
= 96/16
= 6 N/m
Thus the answer to your problem is, the spring constant K should be 6 N/m.
a sphere of radius r has charge q. the electric field strength at distance . what is the ratio of the final to initial electric field strengths if (a) q is halved, (b) r is halved, and (c) r is halved (but is )? each part changes only one quantity; the other quantities have their initial values.
A decrease in charge, q, will result in a decrease in the electric field strength. The ratio of the final to initial field strengths can be expressed as qf/qi. If q is halved, the ratio would be 0.5. If the radius, r, is halved, the ratio would be 1/2r2i/r2i, which is equal to 0.25. If r is halved, but the distance remains the same, the ratio would be 1/2r2i/r2i, which is equal to 0.25.
The electric field strength is inversely proportional to the distance from the charge, and directly proportional to the charge and the radius of the sphere. Therefore, halving the charge or radius will result in a decrease in the electric field strength. Halving the radius, with the distance remaining the same, will result in the same ratio as halving the charge because the distance will be the same in both cases.
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if a current of 5.5 a is used, what is the force generated per unit field strength on the 20.0 cm wide section of the loop? use units of newtons per tesla.
The force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is: 0.001 newtons per tesla
The force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is given by the formula F = (μI) / 2πr,
where μ is the permeability of free space, (4π x 10-7 N/A²)
I is current, and r is the radius of the loop.
In this case, the force is (4π x 10-7 x 5.5) / (2π x 0.1) = 0.001 N/T.
In other words, the force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is 0.001 newtons per tesla.
The formula for the force generated per unit field strength on a loop is derived from the fact that the force is a result of the magnetic field generated by the current flowing in the loop.
The magnitude of the magnetic field generated is proportional to the current and inversely proportional to the radius of the loop. Since the force is a product of the current and the magnetic field, it is proportional to the square of the current and inversely proportional to the square of the radius of the loop.
In summary, the force generated per unit field strength on a 20.0 cm wide section of the loop with a current of 5.5 A is 0.001 newtons per tesla, given by the formula F = (μI) / 2πr, where μ is the permeability of free space (4π x 10-7 N/A²), I is current, and r is the radius of the loop.
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which planets are you more likely to see transit, planets closer to or further from their host stars? explain your reasoning.
The transit technique used to identify planets involves looking for small dips in a star's brightness as a planet crosses in front of it. This causes a slight decrease in the amount of light received by the Earth from that star, which is then detected by astronomers.
Transit method is a technique that uses the detection of planetary transits to identify exoplanets. By detecting dips in the brightness of a star, caused by a planet crossing in front of it, this method allows for the detection of planets orbiting other stars beyond our own solar system.
To find exoplanets, astronomers look for periodic dips in the brightness of stars that are caused by a planet passing in front of them. The amount of light that a planet blocks depends on its size, so larger planets create deeper dips in the star's brightness.
The timing and duration of the dips also provide information about the planet's orbit, size, and composition.
Transiting planets are therefore more likely to be detected if they have a large radius compared to their host stars, or if their orbital periods are short.
The transit method is also more effective when the host star is relatively small and bright, as this makes the planet's transit easier to detect.
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ball and a magnet are released simultaneously from the same altitude. they both fall vertically, but the magnet passes through a coil on its way down. which one reaches the ground first? please make a couple of statements to support your answer.
Both the ball and the magnet will reach the ground at the same time because the presence of the coil does not affect the rate of free-fall acceleration of the magnet.
This is because, according to the principle of equivalence, objects with different masses fall at the same rate in a vacuum. In this case, the effect of the coil on the magnet is negligible since the magnet's mass is much smaller than that of the Earth. Therefore, both the ball and the magnet will experience the same acceleration due to gravity and reach the ground at the same time, regardless of the presence of the coil.
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tom has a 4-inch refracting telescope and steve has a 3-inch reflecting telescope. whose telescope has a higher resolving power?
Tom's 4-inch refracting telescope has a higher resolving power.
Refracting telescopes have higher resolving power than reflecting telescopes, as the size of the objective lens in a refractor can be larger than the size of the mirror in a reflector.
Resolving power is the ability of a telescope to distinguish between two closely spaced objects. It is determined by the diameter of the telescope's objective lens or mirror. The resolving power is proportional to the diameter of the objective, so a larger objective will have a higher resolving power.
Therefore, Tom's 4-inch refracting telescope has a higher resolving power than Steve's 3-inch reflecting telescope.
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what is the relative permittivity of the dielectric medium in which the electromagnetic wave traveling g
The relative permittivity of a dielectric medium is a measure of the electric field created within it when exposed to an electromagnetic wave. The relative permittivity is calculated by comparing the electric field in the dielectric medium to the electric field in a vacuum.
In a dielectric medium, the electric field is weakened due to the presence of molecules that absorb some of the energy from the electromagnetic wave. The greater the number of molecules, the higher the relative permittivity of the dielectric medium. The lower the relative permittivity of a dielectric medium, the faster the speed of the electromagnetic wave traveling through it.
In conclusion, the relative permittivity of the dielectric medium in which the electromagnetic wave is traveling can be used to measure the electric field within it. The greater the number of molecules present in the dielectric medium, the higher the relative permittivity and the slower the speed of the electromagnetic wave.
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Two wires cross, one carrying current to the east and the other to the north. The force between the two wires is_____ O repulsive. Ozero. O attractive.
Attractive
Explanation:
Like poles repel
And
Unlike poles attract
jacob asks imad to explain to jacob how the number of field lines and the magnitude of the charge are related. which response is correct?
To explain to Jacob how the number of field lines and the magnitude of the charge are related, Imad should mention that the number of field lines is proportional to the magnitude of the charge.
There is a relationship between the number of field lines and the magnitude of the charge. The magnitude of the charge is directly proportional to the number of field lines that pass through the surface that is perpendicular to the field lines. The number of field lines created by a charge or charges is proportional to the charge or charges' magnitude.
In the absence of any other charges or objects, the field lines emanating from a charge with magnitude q will terminate on another charge with magnitude q of opposite polarity, according to Coulomb's law. Therefore, Jacob should be told that the number of field lines is proportional to the magnitude of the charge, meaning that if the charge's magnitude increases, the number of field lines will increase as well.
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suppose that a 50-kilogram cart and a 70-kilogram cart, both traveling at 5 meters per second in opposite directions, collide and stick together. in meters per second with one significant figure, what is the speed of the final composite object?
The final speed of the composite object is 0.8 m/s.
We can use the law of conservation of momentum, which states that the total momentum of a closed system remains constant. In this case, the initial momentum of the system is,
initial momentum = (50 kg) x (-5 m/s) + (70 kg) x (5 m/s)
= -250 kg m/s + 350 kg m/s
= 100 kg m/s
Since the carts stick together after the collision, their masses add up to give the mass of the final composite object,
mass of final object = 50 kg + 70 kg
= 120 kg
Using the conservation of momentum, we can solve for the final velocity of the composite object,
initial momentum = final momentum
100 kg m/s = (120 kg) x (v) m/s
Solving for v,
v = 0.83 m/s
Rounding off to one significant figure, velocity is, 0.8 m/s.
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a 4.0 kg body has two times the kinetic energy of an 8.5 kg body. calculate the ratio of the speeds of these bodies.
The ratio of the speeds of these bodies is 2.06
The kinetic energy of an object is equal to 1/2mv^2.
For the 4.0 kg body, the kinetic energy is 1/2 (4.0 kg)v^2
For the 8.5 kg body, the kinetic energy is 1/2 (8.5 kg)u^2
Given that the kinetic energy of the 4.0 kg body is twice the kinetic energy of the 8.5 kg body, we can set up the following equation:
1/2 (4.0 kg)v^2 = 2 * (1/2 (8.5 kg)u^2)
Simplifying the equation, we have:
2 (4.0 kg)v^2 = (8.5 kg)u^2
Solving for the ratio of the speeds, we get:
v^2/u^2 = (8.5 kg)/(2 (4.0 kg)) = 4.25
Therefore, the ratio of the speeds of the two bodies is equal to the square root of 4.25, which is approximately equal to 2.06.
So, the 4.0 kg body is moving at approximately 2.06 times the speed of the 8.5 kg body.
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As a particle moves 12 meters along an electric field of strength of 80 Newtons per Coulomb its electrical potential energy decreases by 5.2 x 10^-18 Joules.
What is the particle charge?
Giving out brainliest please help this is due today.
Answer:
The electric potential energy (EPE) of a particle with charge q moving through an electric field of strength E over a distance d is given by the formula:
EPE = qEd
In this problem, we are given:
EPE = 5.2 x 10^-18 J
E = 80 N/C
d = 12 m
Substituting these values into the formula, we get:
5.2 x 10^-18 J = q(80 N/C)(12 m)
q = 5.2 x 10^-18 J / (80 N/C)(12 m)
q = 6.875 x 10^-21 C
Therefore, the particle charge is 6.875 x 10^-21 Coulombs.
Explanation:
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A new planet is discovered orbiting a star with a mass of [tex]3.5\times 10^{31}[/tex] kg at a distance of [tex]1.2\times10^{11}[/tex] m. Assume that the orbit is circular. What is the orbital speed of the planet? What is the orbital period of the planet?
The orbital speed of the planet is [tex]4.41 * 10^3 m/s[/tex] and the orbital period of the planet is 22.608s when its mass is [tex]3.5 * 10^{31}kg.[/tex]
The mass of the planet (m) = [tex]3.5 * 10^{31}kg[/tex]
The distance of star from the planet (r) = [tex]1.2 * 10^{11}m[/tex]
The planet is moving in circular shape around the star.
The orbital speed of the planet = v
The orbital speed of the planet can be calculated using the equation for the orbital speed, which is given by [tex]v = \sqrt{(GM/r)}[/tex], where G is the gravitational constant = [tex]6.67 * 10^{-11} m^3/(kg*s^2)[/tex]
[tex]v = \sqrt{6.67 * 10^{-11} m^3/(kg*s^2) * 3.5 * 10^{31}kg/1.2 * 10^{11}m}[/tex]
[tex]v = \sqrt{23.345 * 10^{20}/1.2 * 10^{11}} = \sqrt{19.45 * 10^9}[/tex]
[tex]v = 4.41 * 10^3 m/s[/tex]
the orbital speed of the planet is = [tex]4.41 * 10^3 m/s[/tex]
The orbital time period of the planet can be calculated using the equation for the orbital period, which is given by [tex]T = 2 * \pi * \sqrt{(r^3/GM)}[/tex]
[tex]T = 2 * \pi * \sqrt{(1.2 * 10^{11})^3/6.67 * 10^{-11} m^3/(kg*s^2) * 3.5 * 10^{31}kg}[/tex]
[tex]T = 2 * \pi * \sqrt{(1.2 * 10^{11})^3/23.345 * 10^{20}}[/tex]
[tex]T = 2 * \pi * \sqrt{0.07 * 10^{13}}[/tex]
[tex]T = 2 * \pi * 0.26 * 3.9[/tex]
T = 22.608s
Hence the orbital period of the planet is 22.608s
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if the coconut from the taller tree reaches the ground with a speed v , what will be the speed (in terms of v ) of the coconut from the other tree when it reaches the ground?
Since both coconuts fall the same distance, they will reach the ground with the same speed (v). The speed of the coconut from the other tree when it reaches the ground is equal to the speed of the coconut from the taller tree (v).
The speed of the coconut from the other tree when it reaches the ground is equal to the speed of the coconut from the taller tree (v). This is because the force of gravity is the same on both coconuts and they experience the same acceleration. This means that they will reach the ground with the same speed, regardless of the height of the tree they are falling from.
The gravitational acceleration (g) is a constant and is independent of the mass of the coconut. Since both coconuts have the same mass, they will experience the same force of gravity, resulting in the same acceleration. This acceleration is independent of the initial height of the coconut, meaning that the coconuts will reach the ground with the same speed regardless of their initial height.
The speed (v) of the coconuts when they reach the ground is determined by their initial speed at the top of the tree (v0) and the distance they fall (d). If the initial speed is 0 (which is the case when the coconut is released from rest) then the final speed is determined by the distance the coconut has fallen (d). According to the equation v2 = 2gx, v = sqrt(2gd), where g is the gravitational acceleration and d is the distance fallen. Therefore, since both coconuts fall the same distance, they will reach the ground with the same speed (v).
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Like for the Mars Pathfinder mission, the entry and landing on Mars use a combination of aerodynamic drag (during entry, the spacecraft is protected by a heat shield), rockets, parachutes, and inflated airbags. The last phase of the entry & landing sequence is controlled by the on-board computer system. When the altitude reaches a certain critical value, the spacecraft velocity is 40 m/s. At this altitude, the airbags are inflated and a solid rocket engine is turned on to slow down the spacecraft prior to impact on the Martian soil.
Knowing that the thrust generated by the rocket engine is 4810 N, and the propellant burns for 10 seconds before impact, what will be the velocity at impact in m/s? . Assume that the spacecraft drag (due to parachute & inflated airbags) is constant and is 7500 N, and that the spacecraft mass is 2060 kg. Also, the Martian gravitational acceleration is equal to 3.7 m/s2.
Hint: to solve this problem, make sure to include all forces acting on the spacecraft (weight, drag and thrust).
The velocity at impact will be (-10) m/s if all forces acting on the spacecraft (weight, drag, and thrust) are included.
To find the velocity at impact, first, we need to consider all the forces acting on the spacecraft (weight, drag, and thrust). Then, we can use these forces to find the net force and acceleration. Finally, we can calculate the impact velocity.
Step 1: Calculate the weight of the spacecraft:-
Weight = mass × gravitational acceleration
Weight = 2060 kg × 3.7 m/s² = 7618 N
Step 2: Calculate the net force acting on the spacecraft:-
Net force = thrust - drag - weight
Net force = 4810 N - 7500 N - 7618 N = -10308 N
Step 3: Calculate the acceleration of the spacecraft:-
Acceleration = net force/mass
Acceleration = -10308 N / 2060 kg = -5 m/s²
Step 4: Calculate the velocity at impact:-
We know that the initial velocity is 40 m/s, and the propellant burns for 10 seconds. We can use the equation of motion (v = u + at) to find the final velocity:-
Final velocity(v) = initial velocity(u) + acceleration(a) × time(t)
Final velocity = 40 m/s + (-5 m/s²) × 10 s = 40 m/s - 50 m/s = -10 m/s
Therefore, the velocity at impact will be -10 m/s (the negative sign indicates that the velocity is in the opposite direction to the initial velocity).
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what law(s) in physics can be used to explain the behaviors of the carts, in this interactive, whether or not the collision was elastic or inelastic?
The laws of physics that can be used to explain the behaviors of the carts in this interactive, whether or not the collision was elastic or inelastic, are the laws of conservation of momentum and conservation of energy.
The law of conservation of momentum states that the total momentum of the objects before the collision is equal to the total momentum of the objects after the collision.
The law of conservation of energy states that the total energy of the system remains constant, regardless of the type of collision that takes place.
If the collision is elastic, then the total kinetic energy of the objects before and after the collision is equal. If the collision is inelastic, then the total kinetic energy of the objects after the collision is less than before the collision.
This law applies to both elastic and inelastic collisions. Conservation of energy also applies to both elastic and inelastic collisions. In elastic collisions, the kinetic energy is conserved.
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a balloon floats inside a stopped car. when the car starts moving forward, the balloon appears to move backward relative to the car. which statement best explains this observation?(1 point)responses
The balloon is being pushed backward by air pressure. When a car begins to move, air pressure builds up in front of the car, pushing air backwards and creating a wind that affects the balloon inside. As the car accelerates, the wind increases, and the balloon is pushed backwards relative to the car. This phenomenon is known as the 'Venturi Effect'.
When a car moves, it creates a pressure difference in front of and behind the car. This difference in pressure creates a force that moves air around the car. In the case of the balloon, the force of the wind created by the car is pushing the balloon backwards. This is the same effect you feel when a fan is turned on, but in reverse.
The Venturi Effect is a phenomenon in fluid dynamics which explains the decrease in pressure when the velocity of the fluid increases. In the case of the balloon, this decrease in pressure created by the wind of the car causes it to move backwards. This is because air is being pushed away from the balloon and the surrounding area, creating a low-pressure environment.
In summary, the balloon is being pushed backwards by air pressure as the car moves forward. This is known as the Venturi Effect, and it is caused by the decrease in pressure caused by the wind of the car.
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