Option b is correct. If enough material from the accretion disk makes it onto the white dwarf then it will cause the white dwarf to collapse and produce a Type Ia supernova.
A white dwarf is a remnant of low or medium-mass stars that have stopped producing energy through nuclear fusion. White dwarfs are not planets; instead, they are the small cores of stars that have collapsed under the weight of their own gravity. They are the densest form of matter apart from black holes and neutron stars.
If a lot of material from the accretion disk makes it onto the white dwarf (before it can all fuse away), the material will accumulate on the surface of the white dwarf, increasing its mass.
The white dwarf will have a mass greater than the Chandrasekhar limit (1.4 solar masses). This can lead to a catastrophic event known as a Type Ia supernova, in which the white dwarf collapses and explodes. hence correct option is b.
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describe the single touch method of making magnet. Also describe the process of making electromagnet.
Answer:
The single touch method of making a magnet is a simple way to magnetize a ferromagnetic material, such as a piece of iron. The process involves taking a magnet and bringing it into contact with the material to be magnetized, while moving it along the length of the material in a single direction. This process aligns the magnetic domains within the material, causing it to become magnetized. The strength of the resulting magnet depends on the strength of the magnet used and the length of time it is in contact with the material.
The process of making an electromagnet involves using an electric current to create a magnetic field. To make an electromagnet, a wire is wound into a coil around a magnetic core, such as an iron rod. When an electric current flows through the wire, it creates a magnetic field around the coil, which in turn magnetizes the magnetic core. The strength of the resulting magnet depends on the strength of the current flowing through the wire, the number of coils in the wire, and the type of magnetic core used. Electromagnets are commonly used in a variety of applications, including electric motors, speakers, and MRI machines.
Explanation:
You plunge a basketball beneath the surface of a swimming pool until half the volume of the basketball is submerged. If the basketball has a radius of 12cm what is that buoyancy force on the ball due to the water
If the basketball has a radius of 12cm,the buoyant force on the basketball due to the water is 35.5 N.
When an object is submerged in a fluid, it experiences an upward force called the buoyant force. The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.
In this case, the basketball is submerged in water until half its volume is submerged. The volume of a sphere of radius r is given by:
V = (4/3) * pi * r^3
So, the volume of the basketball is:
V_basketball = (4/3) * pi * (0.12 m)^3 = 7.24 x 10^-3 m^3
Half of the volume of the basketball that is submerged is:
V_submerged = (1/2) * V_basketball = 3.62 x 10^-3 m^3
The weight of the water displaced by the basketball is equal to the weight of the volume of water that is submerged. The density of water is 1000 kg/m^3, so the mass of the displaced water is:
m_water = rho * V_submerged = 1000 kg/m^3 * 3.62 x 10^-3 m^3 = 3.62 kg
The buoyant force on the basketball due to the water is equal to the weight of the displaced water, which is:
F_buoyant = m_water * g
= 3.62 kg * 9.81 m/s^2
= 35.5 N
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Exercise 2: Julie, the clown, carries two balloons that rub against a circus
elephant, causing the balloons to separate. Each balloon acquires 2. 0 x107 C of
in charge. How large is the electric force between them when they are separated by
a distance of 0. 50 m?
When the spheres are separated by 0.5 m, the electric force between them is 1.44 x [tex]10^{20}[/tex] N.
To determine the electric force between two balloons, we can use Coulomb's law:
F = k * (q₁ * q₂) / r²
where F is the electric force, k is Coulomb's constant (9 × [tex]10^{9}[/tex] N·m²/C²), q1 and q2 are the charges on the balloons (2.0 x [tex]10^{7}[/tex] C), and r is the distance between the balloons (0.5 m).
Plugging in these values, we get:
F = 9 × [tex]10^{9}[/tex] * (2.0 x [tex]10^{7}[/tex])² / (0.5)²
F = 1.44 x [tex]10^{20}[/tex] N
Therefore, the electric force between the balloons is 1.44 x [tex]10^{20}[/tex] N when they are separated by a distance of 0.5 m.
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At which location is contact metamorphism least likely to occur?
Contact metamorphism is least likely to occur in low-temperature areas with sedimentary rocks, such as the Earth's oceanic crust.
Contact metamorphism occurs due to heating, with or without burial, of rocks that lie close to a magma intrusion. It is characterized by low P/T gradients, as strong thermal gradients between an intruding magma and adjacent country rock are best established at shallow crustal level. It is a type of metamorphism where rock minerals and texture are changed, mainly by heat, due to contact with magma. Regional metamorphism is a type of metamorphism where rock minerals and texture are changed by heat and pressure over a wide area or region.
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Heeelp please…
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.
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The reason that the melting point of table salt is higher than that of sugar is A, the particles of salt are attracted more strongly than those of sugar.
How is melting point measured?Melting point is measured by heating a substance until it melts and then recording the temperature at which melting occurs. This is typically done using a device called a melting point apparatus or a melting point apparatus.
The sample is placed in a small capillary tube and inserted into the melting point apparatus, which heats the sample gradually and detects the temperature at which the first signs of melting are observed. The temperature is recorded as the melting point of the substance.
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Image transcribed and translated:
What is the reason that the melting point of table salt is higher than that of sugar?
A. The particles of salt are attracted more strongly than those of sugar.
B. The particles of sugar are attracted more strongly than those of salt.
C. The particles of salt are attracted more weakly than those of sugar.
D. The particles of sugar repel while those of salt attract.
A pipe, 30 cm long, is open at both ends. Which harmonic mode of the pipe resonates with a 1. 1 kHz source? Speed of sound in air = 330 m/s
The harmonic mode of the pipe that resonates with a 1. 1 kHz source is second harmonic.
The length of the pipe is given as 30 cm = 30/100 m = 0.3 m
Speed of light is known to be 330 m/s
Frequency of the source is given as 1.1 kHz = 1100 Hz
We know the relation between frequency, harmonic mode and length as,
vn = n v / 2 l
where,
n is number of harmonic mode
v is speed of light
l is length
vn is frequency of nth harmonic in an open pipe
Putting in the values, by making n as subject we have,
n = vn × 2 l/v = 1100 × 2 × 0.3/330 = 2
Thus, the harmonic mode of the pipe is calculated to be 2nd harmonic mode.
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A 80 kg passenger is seated 12 m from the center of the loop of a roller coaster. What centripetal force does the passenger experience when the roller coaster reaches an angular speed of 3.14 rad/s? Fc = (m)(r)(ω2)
The centripetal force the passenger experience is 30167.52 N
Centripetal force calculation'The centripetal force experienced by the passenger can be calculated using the formula:
Fc = (m)(r)(ω^2)
where Fc is the centripetal force, m is the mass of the passenger, r is the radius of the loop, and ω is the angular speed of the roller coaster.
Given:
m = 80 kg
r = 12 m
ω = 3.14 rad/s
Substituting these values into the formula, we get:
Fc = (80 kg)(12 m)(3.14 rad/s)^2
Fc = 30167.52 N
Therefore, the passenger experiences a centripetal force of approximately 30167.52 N when the roller coaster reaches an angular speed of 3.14 rad/s.
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In a physics lab experiment, a 6kg box is pushed across a flat table by a horizontal force F . What the magnitude of force F if the box is speeding up with a constant acceleration of 0.18ms^-2?
Answer:
The F's accelerated
where m is the box's mass, an is its acceleration, and F net is the net force.
The net force in this situation is equal to the applied force less any potential frictional forces. The net force is just the force applied to the box, assuming the table is smooth and there is no friction:
F net = F
Thus, the equation can be rewritten as follows:
F = m * a
Inputting the values provided yields:
F = 6 kg times 0.18 m/s2 to get 1.08 N
As a result, 1.08 N of force F is needed to accelerate a 6 kilogramme box at a constant rate of 0.18 m/s2.
Determine the power consumed by
The power consumed by 18 watts.
What is power ?
Work is done at a certain rate according to power. P=Wt. The watt (W), which is equal to 1 joule per second (1J/s), is the SI unit of power. Power is the rate at which energy is used up since power is the transfer of energy. For instance, a 60-W light bulb uses about 60 J of energy every second.
What is current ?
A flow of electrical charge carriers, most frequently electrons or atoms lacking in electrons, is referred to as current. I in capital letters is a typical symbol for current. Ampere, which is denoted by the letter A, is the common unit.I, which is derived from the French term intensité du courant, is the commonly used symbol for current (current intensity).
Therefore, power consumed by 18 watts.
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a fidget spinner rotates at about 4000 rpm (revolutions per minute). calculate the speed of the edge of a fidget spinner that has a diameter of 6.0 cm.
The speed of the edge of a fidget spinner that has a diameter of 6.0 cm is 12.58 m/s.
To calculate the speed of the edge of the fidget spinner, we need to convert the rotational speed from revolutions per minute to radians per second, and then multiply by the radius of the spinner.
First, we convert 4000 rpm to radians per second:
4000 rpm = 4000 revolutions / 60 seconds = 66.67 revolutions per second
Each revolution is 2π radians, so:
66.67 revolutions per second = 66.67 * 2π radians per second = 419.47 radians per second
Now that we have the rotational speed in radians per second, we can calculate the speed of the edge of the fidget spinner:
speed = rotational speed x radius
The radius of the fidget spinner is half of its diameter, which is 6.0 cm / 2 = 3.0 cm = 0.03 m.
So, the speed of the edge of the fidget spinner is:
speed = 419.47 radians per second x 0.03 m = 12.58 m/s
Therefore, the speed of the edge of the fidget spinner is approximately 12.58 m/s.
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What is the dewpoint when the air temperature is 28°C and the relative humidity is 47%?
The dewpoint when the air temperature is 28°C and the relative humidity is 47% is 15.9°C.
What is temperature?Temperature is a physical property of matter that quantitatively expresses hot and cold. It is measured with a thermometer, which may work through the bulk behavior of a thermometric material, detection of thermal radiation, or particle kinetic energy. Temperature is one of the principal parameters of thermodynamics. Temperature is also important in all fields of natural science, including physics, geology, chemistry, atmospheric sciences, medicine and biology.
The dewpoint is the temperature at which the air has reached its saturation point. At this point, water vapor will begin to condense into liquid droplets, forming dew.
To calculate the dewpoint when the air temperature is 28°C and the relative humidity is 47%, we use the formula:
Dewpoint (°C) = (Air Temperature (°C) * Relative Humidity) / 100 + (112 - Relative Humidity) / (x + 0.444)
Where x = (17.27 * Air Temperature (°C)) / ( Air Temperature (°C) + 237.3 ).
In this example, x = (17.27 * 28) / (28 + 237.3) = 0.39.
Plugging the values into the formula gives us:
Dewpoint (°C) = (28 * 47) / 100 + (112 - 47) / (0.39 + 0.444) = 15.9°C.
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A hungry 165 kg lion running northward at 76.4 km/hr attacks and holds onto a 35.9kg Thomson’s gazelle running eastward at 64.6 km/hr. Find the final speed of the lion-gazelle system immediately after the attack.
Final speed: _____ m/s
a 0.61 m copper rod with a mass of 0.043 kg carries a current of 15 a in the positive x direction. what are the magnitude and direction of the minimum magnetic field needed to levitate the rod?
The magnitude and direction of the minimum magnetic field needed to levitate the rod is 0.152 T.
The magnetic force on a current-carrying wire in a magnetic field is given by:
F = BIL sin(θ)
where B is the magnetic field,
I is the current in the wire,
L is the length of the wire in the magnetic field, and
θ is the angle between the direction of the magnetic field and the direction of the current.
Therefore, we want to find the minimum magnetic field required to levitate the copper rod the rod is levitating, the magnetic force acting upwards must be equal and opposite to the gravitational force acting downwards.
The gravitational force on the rod is given by:
F_{gravity }= mg
where m is the mass of the rod and
g is the acceleration due to gravity.
Substituting the given values, we get:
F_{gravity} = (0.043 kg) * (9.81 m/s²) = 0.42283 N
For the rod to levitate, the magnetic force must be equal in magnitude and opposite in direction to the gravitational force:
F_{magnetic} = F_{gravity}
Substituting the expression for the magnetic force, we get:
BIL sin(θ) = mg
Solving for B, we get:
B = mg / IL sin(θ)
Substituting the given values, we get:
B = (0.043 kg * 9.81 m/s²) / (15 A * 0.61 m * sin(90°))
B = 0.152 T
Therefore, the magnitude of the minimum magnetic field needed to levitate the copper rod is 0.152 T. The direction of the magnetic field should be perpendicular to the direction of the current flow in the rod, which is in the positive x direction in this case.
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A man sees a magnified image of a closely kept object by holding a simple 'clear glass' device in his hand. this device is likely to be a ____________
a) Concave mirror
b) Concave lens
c) Convex lens
d) Convex mirror
A man sees a magnified image of a closely kept object by holding a simple 'clear glass' device in his hand. this device is likely to be a convex lens .
Option C.
What is the device described?The device described in the scenario is a clear glass, which suggests that it is likely a lens rather than a mirror. Additionally, since the image is magnified, it is likely a converging lens that is being used to view the object.
Therefore, the device in question is most likely a convex lens (option c) that is being used as a simple magnifying glass.
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Which of the following will increase the strength of the magnetic field? Select the TWO best answers.
A
Decrease the number of turns of the wire on the core.
B
When you decrease the current.
C
The composition of the core material.
D
The shape and size of the core.
E
The use of non-alloy materials such as glass or plastic.
Answer: Option : C and D are correct .
Explanation: Because core increases the strength of the magnetic field since core is having iron and nickel metals that helps to increase the current.
Why is it difficult to walk on wet marble floor?
Answer:
On a wet marble floor, the friction is very less because marble floor has irregularities that get filled with water. We can walk only when there is more friction. Due to the less friction, we may slip. So, it is very difficult to walk on a wet marble floor.
Answer:
because marbles are always sliding and water makes them more slimy
A 2. 72-nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and is then isolated. The dielectric material between the plates is Pyrex glass, with a dielectric constant of 5. 6
The work done by the capacitor is obtained as 1.36 * 10^-5 J
What is the capacitor?A capacitor is an electrical component that stores electrical energy in an electric field. It consists of two conductive plates separated by a non-conductive material, known as the dielectric.
When a voltage is applied across the plates, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. The dielectric material prevents the charges from flowing between the plates and instead stores the energy in the form of an electric field.
Work done by the Capacitor = 1/2CV^2
= 0.5 * 2.72 * 10^-9 * (100)^2
= 1.36 * 10^-5 J
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Missing parts;
A 2. 72-nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and is then isolated. The dielectric material between the plates is Pyrex glass, with a dielectric constant of 5. 6. What is the work done by the capacitor?
If the sun somehow became twice as massive, your weight as normally measured here on earth would A) double. B) quadruple. C) not change
Even if the sun doubled in size, a person's weight on earth wouldn't alter.
If the sun somehow became twice as massive, your weight as normally measured here on earth would not change.
This is because weight is the force of gravity acting on an object, which depends on the mass of the object and the distance between the centers of mass of the two objects. While the mass of the sun has doubled, the distance between the centers of mass of the earth and the sun remains the same. Therefore, the gravitational force acting on the earth and on a person on the earth remains the same.
As a result, the weight of a person on the earth would not change if the sun became twice as massive.
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Stopping Distance Activity
Background: This activity involves observing the effect of original car speed upon the
skid distance. The simulation involves a toy car rolling down a hill, hitting a box, and
skidding to a stop. The height from which the car is released can be modified. The
speed of the car at the bottom of the hill (prior to contact with the box) is reported. The
distance that the car (and box) skid before stopping can be measured. Before answering
the questions, it would be useful to first run the simulation for each of the provided
heights and to record the corresponding pre-collision car speed and the stopping
distance. The background grid on the simulation screen can be used to determine the
stopping distance. Each square on the grid is 10 cm in length along its edge.
Data:
Initial Height Pre-Collision Speed Skid Distance
(m/s)
(cm)
(cm)
10. 0
20. 0
30. 0
40. 0
50. 0
60. 0
70. 0
80. 0
By using the formula:- [tex]mgh= \frac{1}{2} mv^{2} , v = \sqrt{2gh}[/tex], as the release height of the car increase the speed of the car increases.
As the release height of the car increases, the skid distance of the car increaseincrease due to increased velocity.
The directional speed of an object in motion, as measured by a specific unit of time and witnessed from a specific point of reference, is what is referred to as velocity.
Velocity can be defined as the rate at which something travels in a specific direction. Like the speed of a vehicle travelling north on a highway or the speed of a rocket after launch, for instance.
Less momentum with increased height. Less height, more speed. The rate at which it is exchanging height for velocity at any given moment in the fall is therefore equal to the velocity divided by the gravitational constant. As a result, at the top when its velocity is low, it loses only a tiny amount of height with each increase in velocity.
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complete question:-
two out-of-tune flutes play the same note. one produces a tone that has a frequency of 262 hz, while the other produces 266 hz. when a tuning fork is sounded together with the 262 hz tone, a beat frequency of 1 hz is produced. when the same tuning fork is sounded together with the 266-hz tone, a beat frequency of 3 hz is produced. what is the frequency of the tuning fork?
When the same tuning fork is sounded together with the 266-hz tone, a beat frequency of 3 hz is produced, the frequency of the tuning fork is: 264 Hz.
When two out-of-tune flutes play the same note, and one produces a tone that has a frequency of 262 Hz, and the other produces 266 Hz. The beat frequency of 1 Hz is produced when a tuning fork is sounded together with the 262 Hz tone, and the same tuning fork is sounded together with the 266-Hz tone.
In this case, the frequency of the tuning fork is required. The formula for beat frequency is expressed as follows: Beat frequency = frequency 2 − frequency 1
The frequency of the tuning fork can be determined by subtracting the frequency of the 266 Hz tone from that of the 262 Hz tone (i.e. 266 - 262 = 4 Hz).
This implies that the tuning fork is 4 Hz higher than the 266 Hz tone. The beat frequency of 3 Hz when the same tuning fork is sounded together with the 266 Hz tone implies that the tuning fork's frequency is 3 Hz lower than the 269 Hz tone. The average of these two frequencies is 264 Hz (i.e. 266 Hz - 2 Hz). Therefore, the frequency of the tuning fork is 264 Hz.
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a football is kicked from the ground with a speed of 28.65 m/s at an angle of 37.18 degrees. what is the horizontal displacement of the football when it lands back on the ground?
49.87 m. is the horizontal displacement of the football when it lands back on the ground.
The horizontal displacement of the football can be determined using the formula
x = vxt,
where vx is the initial x-component of the velocity, and t is the time the football is in the air.
The initial x-component of the velocity (vx) can be calculated using the formula
vx = vcos(θ),
where v is the initial velocity, and θ is the angle of projection.
Therefore, vx = 28.65cos(37.18) = 25.36 m/s
The time the football is in the air (t) can be calculated using the formula t = 2vsin(θ)/g, where g is the acceleration due to gravity (9.81 m/s2).
Therefore, t = 2(28.65sin(37.18))/9.81 = 1.97 seconds
Finally, the horizontal displacement (x) can be calculated using the formula x = vxt.
Therefore, x = 25.36*1.97 = 49.87 m.
Therefore, the horizontal displacement of the football when it lands back on the ground is 49.87 m.
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If the water vapor comprises 0. 2% of an air parcel and the pressure exerted by it is 1mb, what is the total atmospheric pressure of the parcel?
If the water vapor comprises 0.2% of an air parcel and the pressure exerted by it is 1 mb, we can use the ideal gas law to determine the total atmospheric pressure of the parcel.
The ideal gas law states that:
PV = nRT
where P is the pressure of the gas, V is its volume, n is the number of moles of gas, R is the ideal gas constant, and T is the absolute temperature.
Assuming that the air in the parcel behaves as an ideal gas, we can use the ideal gas law to relate the pressure of the water vapor to the total atmospheric pressure of the parcel.
Let's assume that the total number of moles of gas in the parcel is N. Since the water vapor comprises 0.2% of the parcel, the number of moles of water vapor in the parcel is 0.002N. Therefore, the number of moles of dry air in the parcel is (1 - 0.002)N = 0.998N.
Using the ideal gas law, we can write:
(P_water vapor)(V) = (0.002N)(R)(T)
(P_dry air)(V) = (0.998N)(R)(T)
Adding these two equations together, we get:
P total can be solved as follows:
N(R)(T) / V Equals P total
(P_total)(V) = N(R)(T)
where P_total is the total atmospheric pressure of the parcel.
Therefore, we can solve for P_total as:
P_total = (N(R)(T)) / V
We can substitute the value of P_water vapor, which is 1 mb, into the first equation to find the volume of the water vapor in the parcel. Then, we can substitute the given values of N, R, and T, and the calculated volume of the water vapor into the equation for P_total to find the total atmospheric pressure of the parcel.
In summary, by using the ideal gas law and assuming that the air in the parcel behaves as an ideal gas, we can determine the total atmospheric pressure of the parcel given the pressure and volume of the water vapor, and the temperature and total number of moles of gas in the parcel.
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a luminous object moves along the optical axis of a concave spherical mirror. if the object approaches the mirror and its focal point, where does the image of the object move?
When a luminous object moves along the optical axis of a concave spherical mirror, if the object approaches the mirror and its focal point, the image of the object moves to infinity.
A concave mirror is a mirror with a reflecting surface that curves inward. A concave mirror, often known as a converging mirror, causes light to concentrate at a point known as the focal point when light rays hit it. The distance between the mirror's center and its focal point is known as the focal length. A concave mirror, for example, is used in searchlights to direct the beam of light, in telescopes to gather more light from faraway objects, and in headlights to concentrate the beam of light to illuminate the road ahead.
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Compare resistance of electrical flow through wires to the resistance of water flow in a pipe. How are they similar?
Answer:
Resistance to the flow of electricity in wires and resistance to the flow of water in pipes are similar in several ways:
Both are affected by the size of the conductor: In both cases, the resistance to flow is affected by the size of the conductor. A larger diameter wire or pipe will offer less resistance to the flow of electricity or water than a smaller diameter one.
Both are affected by the length of the conductor: The resistance to flow also increases with the length of the conductor. A longer wire or pipe will offer more resistance to the flow of electricity or water than a shorter one.
Both are affected by the material of the conductor: Different materials have different resistances to flow. In electrical wires, metals are commonly used as they have low resistance. In pipes, materials such as copper, plastic, or steel may be used depending on the specific application.
Both can be reduced by minimizing obstructions: Resistance to flow can be reduced in both cases by minimizing obstructions in the path of the flow. In pipes, for example, the use of smooth inner walls and proper pipe bends can help reduce resistance. In electrical wires, minimizing the number of connections and using high-quality connections can help reduce resistance.
Overall, resistance to flow is a fundamental concept that applies to the flow of electricity and water, and the factors affecting resistance are similar in both cases.
If an object is suspended from a ceiling by a cord of length 16
ft, then the period of oscillation will be
4.43 seconds
The length of the pendulum and the acceleration caused by gravity affect how long an object will oscillate when it is suspended from the ceiling by a 16-foot string.
The following formula determines a pendulum's period T:
T = 2π * sqrt(L/g)
where g is the acceleration caused by gravity and L is the pendulum's length.
The pendulum in this scenario is 16 feet long. The acceleration brought on by gravity is roughly 32 feet per second. As a result, the oscillation period T is:
T = 2π * sqrt(16/32) = 2π * sqrt(1/2) = 2π * 0.707 = 4.43 seconds (approximately)
T = 2π * sqrt(16/32)
= 2π * sqrt(1/2)
= 2π * 0.707
= 4.43 seconds (approximately)
Thus, the object's period of oscillation is approximately 4.43 seconds.
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A metal cube with a negative charge -Q_1−Q
1
minus, Q, start subscript, 1, end subscript is brought into contact with a cylinder of a negative charge -Q_2−Q
2
minus, Q, start subscript, 2, end subscript, where |Q_1| > |Q_2|∣Q
1
∣>∣Q
2
∣vertical bar, Q, start subscript, 1, end subscript, vertical bar, is greater than, vertical bar, Q, start subscript, 2, end subscript, vertical bar
The metal cube will transfer some of its charge to the cylinder, causing both objects to have a net positive charge. This is because when two objects with opposite charges are brought into contact, they will exchange charges until they have the same magnitude and the same sign.
In this case, the cube with a greater charge will transfer charge until both objects have a positive charge.When an object has more electrons than protons is called a negative charge. A positive charge occurs when an atom has more protons than electrons. Electrons have a negative charge and protons have a positive charge. The unit of charge is the coulomb(C). Like charges repel each other.
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Question:
A metal cube with a negative charge -Q_1−Q 1 minus, Q, start subscript, 1, end subscript is brought into contact with a cylinder of a negative charge -Q_2−Q 2 minus, Q, start subscript, 2, end subscript, where
|Q_1| > |Q_2|Q1∣>∣Q 2∣vertical bar, Q, start subscript, 1, end subscript, vertical bar, is greater than, vertical bar, Q, start subscript, 2, end subscript, vertical bar.
a front-loading washing machine is mounted on a thick rubber pad that acts like a spring; the weight (with m/s ) of the machine depresses the pad exactly 0.89 cm. when its rotor spins at radians per second, the rotor exerts a vertical force of newtons on the machine. neglecting friction, at what speed (in revolutions per minute) will resonance vibrations will occur? resonance occurs at rpms
The speed (in revolutions per minute) at which resonance vibrations will occur is 273.91 RPM.
The front-loading washing machine is mounted on a thick rubber pad that acts as a spring. When the rotor spins at radians per second, the rotor exerts a vertical force of newtons on the machine.
Resonance vibrations occur when the natural frequency of the machine matches the frequency of the external force applied to it. This can cause the machine to vibrate excessively and even damage it.
In this scenario, the natural frequency of the washing machine is related to its spring constant and the mass of the machine. Neglecting friction, resonance vibrations occur at RPMs.
Given,
The weight (with m/s) of the machine depresses the pad exactly 0.89 cm.
The rotor exerts a vertical force of N on the machine.
The natural frequency of a machine is given by:
f = 1/2π x √(k/m)
where f is the frequency of oscillation,
k is the spring constant of the spring,
m is the mass of the machine.
In this case, the weight of the machine depresses the spring by 0.89 cm.
So, the effective spring constant is given by,
k = F/x = (mg)/x
where F is the weight of the machine,
m is its mass,
g is the acceleration due to gravity,
and x is the amount by which the spring is compressed.
Substituting the given values,
k = (m x g) / x = (m x 9.8) / 0.0089 = 1101.12m N/m
Substituting the given values,
f = 1/2π x √(k/m) = 1/2π x √(1101.12m/m) = 4.9448 √m Hz
The frequency of oscillation is given in Hz.
To convert it to RPMs, we need to divide it by 60 (the number of seconds in a minute) and multiply it by 2π (the number of radians in a circle).
f = 4.9448 √m Hz = (4.9448 √m x 2π)/60 RPM
At resonance, the frequency of oscillation should match the frequency of the rotor.
So, the frequency of the rotor is given by,
f = ω/2π
where ω is the angular velocity of the rotor in radians per second.
Substituting the given values,
N = mg = F,
ω = N/(m x r) = 94.4912/r rads/s
where r is the radius of the rotor in meters.
Substituting these values in the equation for frequency,
f = ω/2π = (94.4912/r) / 2π = 15.0415 / r Hz
At resonance, the frequency of oscillation should match the frequency of the rotor.
Therefore,4.9448 √m = 15.0415 / r
Squaring both sides,19.4542 m = (15.0415/r)^2
Solving for r, r = 0.0275 m = 2.75 cm
Therefore, the radius of the rotor is 2.75 cm.
The speed (in revolutions per minute) at which resonance vibrations will occur is 273.91 RPM.
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An electron is accelerated through a distance of 15mm the work done on the electron is 1.2x10^-13J calculate the force on the electron
help pleaseee
The satellite orbits at a distance from the center of the moon. Which of the following is a correct expression for the time it takes the satellite to make one complete revolution around the moon?
A) T = 2π[tex]\sqrt \frac{R^3}{Gm}[/tex]
B) T = 2π[tex]\sqrt\frac{R^3}{GM}[/tex]
C) T = 2π[tex]\sqrt\frac{Gm}{R^3}[/tex]
D) T = 2π[tex]\sqrt\frac{GM}{R^3}[/tex]
Answer:
[tex]\displaystyle 2\, \pi\, \sqrt{\frac{R^{3}}{G\m M}}[/tex], where [tex]R[/tex] is the orbital radius, [tex]M[/tex] is the mass of the Moon, and [tex]G[/tex] is the gravitational constant.
Explanation:
Let [tex]m[/tex] denote the mass of the satellite. Let [tex]R[/tex] denote the orbital radius, let [tex]M[/tex] denote the mass of the Moon, and let [tex]G[/tex] denote the gravitational constant.
The Moon would exert the following gravitational attraction on the satellite:
[tex]\displaystyle \frac{G\, M\, m}{R^{2}}[/tex].
Let [tex]\omega[/tex] denote the angular velocity of the satellite. For the satellite to stay in this orbit of radius [tex]R[/tex], the net force on the satellite needs to be:
[tex]m\, \omega^{2}\, R[/tex].
Since the gravitational force is the only force on this satellite, the net force on the satellite would be equal to the gravitational force:
[tex]\displaystyle m\, \omega^{2}\, R = \frac{G\, M\, m}{R^{2}}[/tex].
Rearrange this equation to find the angular velocity:
[tex]\displaystyle \omega^{2} = \frac{G\, M}{R^{3}}[/tex].
[tex]\displaystyle \omega = \sqrt{\frac{G\, M}{R^{3}}}[/tex].
Note that with the Moon as the center, a full revolution around the Moon would take an angular distance of [tex]2\, \pi[/tex]. Divide the angular distance by the angular velocity to find the time required for this revolution:
[tex]\begin{aligned}T &= \frac{2\, \pi }{\omega} && \genfrac{}{}{0}{}{\text{angular displacement}}{\text{angular velocity}} \\ &= 2\, \pi \, \sqrt{\frac{R^{3}}{G\, M}}\end{aligned}[/tex].
Starting from rest, a car accelerates at 2.2 m/s2 up a hill that is inclined 5.3 ∘
above the horizontal.
Part A: How far horizontally has the car traveled in 15 s?
Part B: How far vertically has the car traveled in 15 ?
Express your answer using two significant figures.
The car travels approximately 245 m horizontally and 14.1 m vertically in 15 seconds up an inclined hill.
Section A: To find how far evenly the vehicle has gone in 15 s, we really want to utilize the equation:distance = starting speed × time + 1/2 × speed increase × time^2.Since the vehicle is beginning from rest, the underlying speed is 0. We are searching for the even distance, which is the distance along the ground. Hence, we can disregard the grade of the slope. Hence, we can utilize the level part of the speed increase, which is given by:a_horizontal = a × cosθ.where θ is the point of grade and an is the speed increase. Subbing the qualities given, we get:a_horizontal = 2.2 m/s^2 × cos(5.3°) ≈ 2.18 m/s^2.Utilizing this worth, we can find the even distance went in 15 s:distance = 0 + 1/2 × 2.18 m/s^2 × (15 s)^2 ≈ 245 m.Hence, the vehicle has voyaged around 245 m evenly in 15 s.
Part B: To find how far in an upward direction the vehicle has gone in 15 s, we want to utilize the upward part of the speed increase, which is given by:a_vertical = a × sinθ
Subbing the qualities given, we get:
a_vertical = 2.2 m/s^2 × sin(5.3°) ≈ 0.197 m/s^2
Utilizing this worth, we can find the upward distance went in 15 s:
distance = 0 + 1/2 × 0.197 m/s^2 × (15 s)^2 ≈ 14.1 m
Hence, the vehicle has voyaged roughly 14.1 m in an upward direction in 15 s.
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