The angular velocity of the wheel is 50 rad/s. The centripetal acceleration of the wheel is 1000 m/s^2.
The angular velocity of a wheel is given by the formula:
ω = v / r
where ω is the angular velocity, v is the linear velocity, and r is the radius of the wheel.
Substituting the given values, we have:
ω = 20 m/s / 0.4 m = 50 rad/s
Therefore, the angular velocity of the wheel is 50 rad/s.
To find the number of rotations the wheel makes in 5 seconds, we can use the formula:
θ = ω * t
where θ is the angular displacement, ω is the angular velocity, and t is the time. Since the wheel rotates 2π radians in one full revolution, we can find the number of rotations by dividing the angular displacement by 2π. Substituting the given values, we have:
θ = 50 rad/s * 5 s
= 250 rad
Number of rotations = θ / 2π
= 250 rad / 2π rad/rev ≈ 39.79 rev
Therefore, the wheel will rotate approximately 39.79 times in 5 seconds.
To find the centripetal acceleration of the wheel, we can use the formula:
a = v^2 / r
where a is the centripetal acceleration, v is the linear velocity, and r is the radius of the wheel. Substituting the given values, we have:
a = (20 m/s)^2 / 0.4 m
= 1000 m/s^2
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a train is moving parallel to a highway with a constant speed of 20.0 m/s. a car is traveling in the same direction as the train with a speed of 40.0 m/s. the car horn sounds at a frequency of 510 hz, and the train whistle sounds at a frequency of 320 hz. (a) when the car is behind the train, what frequency does an occupant of the car observe for the train whistle? (b) after the car passes and is in front of the train, what frequency does a train passenger observe for the car horn?
(a) An occupant of the car will observe a frequency of 295.3 Hz for the train whistle.
(b) Train passenger will observe a frequency of 579.2 Hz for the car horn.
This is a problem related to the Doppler Effect, which describes the change in frequency of a wave as a result of the motion of the source or the observer (or both) relative to the medium in which the wave is propagating.
(a) When the car is behind the train, it is moving in the same direction as the train. Therefore, an observer in the car will hear the train whistle at a lower frequency than its actual frequency.
This is because the sound waves from the whistle have to "catch up" to the car, which is moving away from them.
The formula for the observed frequency in this case is:
[tex]f_{obs}[/tex] = [tex]f_{source}[/tex] x ([tex]v_{sound}[/tex] ± [tex]v_{observer}[/tex]) / ([tex]v_{sound}[/tex] ± [tex]v_{source}[/tex])
where
[tex]f_{source}[/tex] is the frequency of the source,
[tex]v_{sound}[/tex] is the speed of sound in air (which we assume to be 343 m/s), [tex]v_{observer}[/tex] is the speed of the observer (which is equal to the speed of the car, 40.0 m/s), and
[tex]v_{source}[/tex] is the speed of the source (which is equal to the speed of the train, 20.0 m/s).
The ± sign depends on whether the observer and the source are moving towards each other or away from each other.
In this case, they are moving away from each other, so we use the - sign:
[tex]f_{obs}[/tex] = 320 x (343 - 40) / (343 - 20) = 295.3 Hz
(b) After the car passes and is in front of the train, it is moving towards the train.
Therefore, a passenger in the train will hear the car horn at a higher frequency than its actual frequency.
This is because the sound waves from the horn are compressed as they "catch up" to the train, which is moving towards them.
The formula for the observed frequency in this case is:
[tex]f_{obs}[/tex] = [tex]f_{source}[/tex] x ([tex]v_{sound}[/tex] ± [tex]v_{observer}[/tex]) / ([tex]v_{sound}[/tex] ± [tex]v_{source}[/tex] )
where
[tex]f_{source}[/tex] is the frequency of the source (which is equal to the frequency of the car horn, 510 Hz),
[tex]v_{sound}[/tex] is the speed of sound in air (which we assume to be 343 m/s),
[tex]v_{observer}[/tex] is the speed of the observer (which is equal to the speed of the train, 20.0 m/s), and
[tex]v_{source}[/tex] is the speed of the source (which is equal to the speed of the car, -40.0 m/s, since it is moving towards the train).
The ± sign depends on whether the observer and the source are moving towards each other or away from each other. In this case, they are moving towards each other, so we use the + sign:
[tex]f_{obs}[/tex] = 510 x (343 + 20) / (343 + 40) = 579.2 Hz
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a truck carries a tank and is open at the top. the tank is 24 ft. long, 6 ft. wide, and 8 ft. high. assuming that the driver will not accelerate or decelerate the truck at a rate greater than 9f t/s2, to what maximum depth may the tank be filled so that water will not be spilled?
The maximum depth to which the tank can be filled without water spilling over the top during acceleration or deceleration of the truck is approximately 4.57 feet.
To prevent water from spilling over the top of the tank during acceleration or deceleration of the truck, the maximum depth to which the tank can be filled without overflowing needs to be determined.
First, let's convert the dimensions of the tank from feet to inches for consistency:
Length of tank (L) = 24 ft = 24 * 12 inches = 288 inches
Width of tank (W) = 6 ft = 6 * 12 inches = 72 inches
Height of tank (H) = 8 ft = 8 * 12 inches = 96 inches
Next, we can calculate the volume of the tank in cubic inches:
Volume of tank (V) = Length * Width * Height = L * W * H
Plugging in the values:
V = 288 inches * 72 inches * 96 inches = 1,986,048 cubic inches
Now, let's consider the acceleration or deceleration of the truck. The maximum acceleration or deceleration the truck can experience without water spilling over the top of the tank is 9 ft/s^2. Since the tank is open at the top, the water will be subjected to the same acceleration or deceleration as the truck.
To find the maximum depth to which the tank can be filled without water spilling over the top, we need to equate the pressure due to the acceleration or deceleration of the truck to the pressure of the water at the maximum depth.
The pressure due to acceleration or deceleration is given by:
Pressure due to acceleration/deceleration (P) = Density of water (ρ) * Acceleration or deceleration of truck (a) * Height of water (h)
Where:
Density of water (ρ) = 62.43 lb/ft^3 (density of water at room temperature)
Converting the acceleration from feet to inches:
Acceleration or deceleration of truck (a) = 9 ft/s^2 = 9 * 12 inches/s^2 = 108 inches/s^2
Setting the pressure due to acceleration/deceleration equal to the pressure of the water at the maximum depth:
P = ρ * a * h
Solving for the height of water (h):
[tex]h = \frac{P}{(\rho * a)}[/tex]
Plugging in the values:
[tex]h = \frac{P}{(62.43 * 108)}[/tex]
Note: The unit conversion from lb/ft^3 to inches/s^2 is necessary to ensure that all units are consistent.
Now, we need to convert the volume of the tank from cubic inches to cubic feet, since the density of water is given in lb/ft^3:
Volume of tank (V) = 1,986,048 cubic inches = [tex]\frac{1,986,048}{(12^3)}[/tex] cubic feet
Plugging in the value of V and solving for h:
[tex]h = \frac{P}{(\rho * a)} = \frac{(\frac{1,986,048}{(12^3)})}{(62.43*108)}[/tex]
h ≈ 4.57 ft (rounded to two decimal places)
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which of the following is a form of potential energy? a. thermal energy b. all the given choices are forms of potential energy c. chemical energy d. an asteroid traveling through space e. a visible light laser beam
Explanation:
The form of potential energy among the given choices is c. chemical energy.
Thermal energy (a) is a form of kinetic energy that results from the motion of particles, while an asteroid traveling through space (d) and a visible light laser beam (e) are forms of electromagnetic radiation, which are also forms of kinetic energy.
On the other hand, chemical energy (c) is the potential energy stored in the chemical bonds between atoms and molecules. This energy can be released through chemical reactions, such as combustion, and can be converted into other forms of energy, such as heat, light, or mechanical energy.
The correct answer is B. All the given choices are forms of potential energy.
Potential energy is the energy possessed by an object due to its position or state. The energy is stored and can be converted into kinetic energy when the object is in motion.
The given options all describe forms of potential energy:
Thermal energy - the energy stored in an object due to its temperature
Chemical energy - the energy stored in the bonds between atoms and molecules
An asteroid traveling through space - the energy possessed by the asteroid due to its position in space
A visible light laser beam - the energy stored in the photons that make up the beam
Therefore, all of the given choices are forms of potential energy.
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Indicate whether each choice is correct, incorrect, or cannot be determined.
The categories of each predictions based on the energy diagram made by the student are IC, IC, C, C and CBD
Indicating the categories of each choiceStatement (a): IC
When an electron with a kinetic energy of 13.6 eV interacts with this atom, it will cause the electron in the atom to move to a higher energy level.
To move an electron from the -13.6 eV level to the -0.85 eV level would require a photon with an energy of 12.75 eV, not 13.6 eV.
Statement (b): IC
For the electron in the atom at the -13.6 eV level to move to a higher energy level, it needs to absorb a photon with an energy equal to the difference in energy between the two levels.
A photon with an energy of 13 eV is not sufficient to cause the electron to move to a higher energy level.
Statement (c): C
An electron in the atom at the -0.85 eV level can move to the -3.4 eV level by emitting a photon with an energy equal to the difference in energy between the two levels, which is 2.55 eV.
Statement (d): C
An electron at the -1.5 eV can move to the -0.85 eV level if it absorbs a photon with an energy equal to the difference in energy between the two levels, which is 0.65 eV (not 0.75 eV as predicted).
Statement (e): CBD
e) CBD. Electrons in an atom can only absorb photons of specific energies corresponding to the energy differences between the energy levels of the atom.
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a 150 n block rests on a table. the suspended mass has a weight of 65 n. what is the magnitude of the minimum force of static friction required to hold both blocks at rest? answer in units of n.
A 150 N block rests on a table while a suspended mass has a weight of 65 N. The magnitude of the minimum force of static friction required to hold both blocks at rest is 64.5 N
The force of static friction is calculated using the formula
Fsf ≤ µsfFn
where: Fsf = Force of static friction (unknown)
µsf = Coefficient of static friction (unknown)
Fn = Normal force (known)
As the block is resting on the table, Fn is equal to the weight of the block.
Thus,Fn = 150 N
To find the minimum force of static friction required, we need to calculate the maximum value of Fsf.
The maximum value of Fsf is equal to µsfFn.
As the block is at rest, the force of friction is equal to the minimum force of static friction.
Therefore, the minimum force of static friction is equal to the maximum value of Fsf.
Thus,Fsf = µsfFn= µsf(150 N)
We are given the weight of the suspended mass, which is 65 N.
As the system is at rest, the horizontal component of the gravitational force on the suspended mass must be balanced by the force of static friction acting on the block.
Therefore,Fsf = 65 N
We can substitute the value of Fsf in terms of Fn to obtain:
µsfFn = 65 N
µsf = 65/150
= 0.43
Therefore, the magnitude of the minimum force of static friction required to hold both blocks at rest is 0.43 × 150 N = 64.5 N (rounded to one decimal place)
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write an expression for how much energy it receives per hour from the high temperature thermal reservoir in terms of the absolute temperatures tc and th.
An expression for how much energy it receives per hour from the high temperature thermal reservoir in terms of the absolute temperatures tc and th is Eh=(1−Tc/Th)(P).
The expression can be achieved using Carnot engine. This formula is as follows:η=1−Tc/Th where η is the efficiency of the engine, Tc is the absolute temperature of the low-temperature reservoir, and Th is the absolute temperature of the high-temperature reservoir.
To find how much energy it receives per hour from the high temperature thermal reservoir, we can multiply the efficiency of the engine by the rate of energy input from the high-temperature reservoir. This gives the following expression: Eh=(η)(P)where Eh is the energy received per hour from the high-temperature reservoir, P is the power input to the engine, and η is the efficiency of the engine.
Substituting the expression for η into the equation for Eh, we get the following: Eh=(1−Tc/Th)(P). This is the expression for how much energy is received per hour from the high temperature thermal reservoir in terms of the absolute temperatures tc and th.
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a person wearing roller skates is standing in front of a wall. assume that the wheels on the skates are good enough that they roll ideally. the person pushes off the wall and begins traveling away from the wall. call the initial state when the person was standing at rest in front of the wall with her hand touching the wall. the final state is when she has traveled 2.1 m away from the wall and is moving at a constant speed of 0.69 m/s. for which of the following systems does the energy remain constant? click for a hint a. system: girl b. system: wall c. system: girl wall d. none of the above. a person is jumping on a trampoline. after coming off of the trampoline, the person is in the air for 1.1 seconds. call the initial state when the trampoline is at its lowest point with the person still on the trampoline. the final state is 0.88 seconds after the person comes off the trampoline. for which of the following systems does the energy remain constant? a. system: person earth b. system: trampoline c. system: person trampoline earth d. system: person e. system: person trampoline f. none of the above. pushing a box up a ramp / car crash you push a box up a ramp (friction between the box and the ramp is not negligible). call the initial state when you begin to push the box. call the final state after you have pushed the box up the ramp a distance of 0.5 m and it is moving with a speed of 2 m/s for which of the following systems does the energy remain constant? a. system: box ramp earth you b. system: box ramp c. system: you d. system: box e. system: box ramp earth f. none of the above. two cars are driving down the road. they notice that they are going to crash, so both drivers slam on the brakes. the cars skid, but still collide. the cars stick together and eventually slide to a stop. call the initial state just before the drivers apply the brakes and the final state just after the collision had occurred. treat this situation as realistically as possible. for which of the following systems does the energy remain constant? a. system: both cars b. system: both cars the ground c. system: the first car d. system: the second car e. none of the above.
In the first scenario, the energy would be conserved in the system of the girl-wall, The correct answer is option c
In the second scenario, the energy would be conserved in the system of the person-trampoline-earth, The correct answer is option c
In the third scenario, the energy would be conserved in the system of the box-ramp-earth, The correct answer is option a
In the fourth scenario, in a realistic scenario, the energy would not be conserved in any of the given systems due to external work done on the system. The correct answer is option e
For the first scenario of a person wearing roller skates, assuming there is no external work done on the system, the system of the girl-wall would be the appropriate system to consider. The energy would not be conserved in any other system, as there would be external work done on the system due to forces acting on the girl, wall, or both. Therefore option c is correct.
For the second scenario of a person jumping on a trampoline, assuming there is no external work done on the system, the appropriate system to consider would be the system of the person-trampoline-earth, as the energy would be conserved within this closed system.
None of the other systems would conserve energy, as external work would be done on the system due to the forces acting on the person, trampoline, and/or Earth. Therefore option c is correct.
For the third scenario of pushing a box up a ramp, assuming there is no external work done on the system, the appropriate system to consider would be the system of the box-ramp-earth, as the energy would be conserved within this closed system.
None of the other systems would conserve energy, as external work would be done on the system due to the forces acting on the box, ramp, and/or Earth. Therefore option a is correct.
For the fourth scenario of two cars colliding and coming to a stop, it is important to note that in a realistic scenario, there would be external work done on the system due to forces such as friction, air resistance, and deformation of the cars.
Therefore, the energy would not be conserved in any of the given systems. However, if the scenario were simplified to only consider idealized, perfectly elastic collisions in a vacuum, then the system of both cars would conserve energy. Therefore option e is correct.
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200 meters in 20 seconds
The speed of the object is 10 meters/second.
Speed is defined as the rate at which an object covers distance. It is a scalar quantity, which means that it only has a magnitude (i.e., size) and no direction. The SI unit for speed is meters per second (m/s).
To calculate speed, we need to know the distance traveled and the time it took to cover that distance. In this case, we were given that the object traveled 200 meters and took 20 seconds to do so. We then used the formula:
speed = distance / time
To calculate the speed of the object. We divided the distance traveled by the time it took to travel that distance to find the speed.
Substituting the given values into the formula, we get:
speed = 200 meters / 20 seconds
This simplifies to:
speed = 10 meters/second
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a wave hits one side of a screen where one slit is carved. if we observe the behavior on the other side of the screen, what do we notice?
Answer:
If a wave hits one side of a screen where one slit is carved, the wave will diffract through the slit and produce an interference pattern on the other side of the screen. This interference pattern will consist of alternating bright and dark fringes, indicating areas of constructive and destructive interference, respectively. This is known as the single-slit diffraction pattern, which is a characteristic behavior of wave-like phenomena such as light and sound waves. The pattern will become more pronounced as the slit width decreases and the wavelength of the wave increases.
When a wave hits a screen with a single slit, it diffracts, and the diffracted wavefronts interfere with each other, forming a pattern of constructive and destructive interference fringes on the screen that is on the other side of the screen.
This pattern is known as an interference pattern, and it is a characteristic of wave phenomena, such as light and sound. This interference pattern can be observed by placing a detector screen on the other side of the single slit screen. The interference pattern that is observed on the detector screen will have bright and dark fringes, indicating constructive and destructive interference, respectively.
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2-way radios take sound waves and convert them into electrical signals. Then, they are converted into
radio waves to transmit them to the receiving radio where it is converted back into sound waves.
Your principal is using a walkie-talkie (2-way radio) to communicate with your bus driver. What are the
wave conversions that occurred from the first radio transmitting the person's voice to a second radio?
Answer:
Radio waves are transmitted from a sender to a receiver.
However, these waves may be modulated.
There are two kinds of modulation:
AM - amplitude of the electromagnetic waves is slowly changed
FM - frequency of the electromagnetic waves is slowly modulated
A person's voice is used to modulate the waves as they are sent from the sender to the receiver.
The modulation is converted by the receiver from electromagnetic waves to audio waves.
These audio waves are used to produce an audio signal that can be interpreted by the person using the receiver.
A baseball team has 25 players. The average mass of a player is 84.4 kg. If this team is riding a bus that is travelling at 26.9 m/s, how much momentum does the team (without the bus) truly have?
What's the Solution?
The baseball team (without the bus) has a momentum of 56,819 kg*m/s.
The momentum of an object is defined as the product of its mass and velocity. To calculate the momentum of the baseball team, we need to find the total mass of the team and the velocity of the team relative to the ground (which is the velocity of the bus since the team is riding on it).
The total mass of the team is given by:
mass of team = number of players x average mass per player
mass of team = 25 players x 84.4 kg/player = 2,110 kg
The velocity of the team relative to the ground is the same as the velocity of the bus, which is given as 26.9 m/s.
Therefore, the momentum of the baseball team is:
momentum = mass x velocity
momentum = 2,110 kg x 26.9 m/s
momentum = 56,819 kg*m/s.
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3. How many kilograms of water could be heated from a temperature of 31.8°C to 91.3°C with 6220 kJ of heat?
Answer:
Q = M S ΔT Mass * specific heat water * change in temp
M = Q / (S * ΔT)
ΔT = (91.3 - 31.8) deg C = 59.5 deg C
M = 6220 KJoules / (1.00 kJ / kg deg C * 59.5 deg C)
M = 105 kg
how does the direction of the tension vector relate to the direction of Fg and Fq vectors
The direction of the tension vector is directly related to the direction of the force of gravity and the force of friction vectors, with tension acting in a direction that opposes these forces.
The direction of the tension vector in a system is directly related to the direction of the force of gravity (Fg) and the force of friction (Fq) vectors. In general, tension acts in a direction that opposes the force of gravity and the force of friction. When an object is hanging from a rope or cable, the tension force is acting upward on the object, while the force of gravity is acting downward. This creates a system where the tension force and the force of gravity are in opposite directions, with the tension force acting against the force of gravity to keep the object from falling. Similarly, when an object is being pulled or pushed across a surface, the force of friction is acting in the opposite direction of the applied force, while the tension force is acting in the same direction as the applied force. This creates a system where the tension force and the force of friction are once again in opposite directions, with the tension force acting against the force of friction to keep the object moving.
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what is the relationship between the direction the stage/slide is moved and the direction of movement in the field of view?
The relationship between the direction the stage/slide is moved and the direction of movement in the field of view is opposite to each other.
When the stage or slide is moved in one direction, the sample moves in the opposite direction within the field of view. This is due to the fact that light travels through a microscope lens and flips the image. As a result, the image seen in a microscope is inverted and reversed.The microscope's focus knobs and stage adjustments can be used to control the position of a specimen in the field of view. The stage is the platform on which the sample is placed, and the slide is the glass plate on which the sample is placed.When you move the slide, the direction of movement in the field of view is opposite to the direction in which the stage is moved. This is due to the inversion and reversal of the image caused by the microscope lens, which causes the image to appear to be moving in the opposite direction.
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a ball on the end of a string is whirled with constant speed in counterclockwise horizontal circle. at point a in the circle, the string breaks. which of the curves sketched below most accurately represents the path that the ball will take after the string breaks?
When the string breaks at point a, the ball will continue to move in a straight line tangent to the point on the circle where it broke free. The third curve most accurately represents the path the ball will take after the string breaks.
This means that the ball will continue moving in the direction it was moving at the moment the string broke. In the first curve, the ball appears to continue moving in a circle, which is not possible as it has lost its centripetal force.
The second curve shows the ball moving in a straight line in the direction of its velocity at point a, which is the correct direction. However, the curve appears to be too steep as the ball should continue moving in a straight line with constant velocity. The third curve shows the ball moving in a straight line with constant velocity, which is the correct behavior.
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if a current of 150 ma is passed through the leg of a patient, how many chloride ions pass a cross section of the leg per second?
Answer:
Assuming a Cl ion has a charge of 1.60E-19 coulombs (electronic charge)
150 ma = .15 C/s charge per second due to 150 ,,a
N = .15 / 1.60E-19 = 9.4E17 ions / sec
Approximately 9.37 × 10^(17) chloride ions pass a cross section of the leg per second when a current of 150 mA is passed through the leg.
When a current of 150 mA (milliamperes) is passed through the leg of a patient, the number of chloride ions that pass a cross section of the leg per second can be calculated using the formula:
Number of ions = (Current × Time) / (Charge of one ion)
First, convert the current from milliamperes to amperes:
150 mA = 0.150 A
Now, use the charge of a chloride ion (Cl-) which is the elementary charge e (approximately 1.602 × 10^(-19) C, where C represents coulombs).
Number of chloride ions = (0.150 A × 1 s) / (1.602 × 10^(-19) C)
Number of chloride ions ≈ 9.37 × 10^(17)
Approximately 9.37 × 10^(17) chloride ions pass a cross section of the leg per second when a current of 150 mA is passed through the leg.
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A horse pulls a tree trunk there is a force of 1500 nutrients, and moves a distance of 10 meters in 15 seconds find the work and the power
The power exerted by the horse is 1,000 watts, which is equivalent to 1 kilowatt (kW).
To find the work done by the horse and the power exerted, we can use the following equations:
Work = Force x Distance x cos(theta)
Power = Work / Time
where theta is the angle between the direction of the force and the direction of the displacement. Assuming that the force is applied horizontally and the displacement is also horizontal, the angle theta is 0 degrees, so cos(theta) = 1.
Given that the force applied by the horse is 1500 N and the distance moved is 10 m, we can calculate the work done as follows:
Work = Force x Distance x cos(theta)
Work = 1500 N x 10 m x 1
Work = 15,000 J
Therefore, the work done by the horse in pulling the tree trunk is 15,000 joules.
To find the power exerted by the horse, we need to divide the work done by the time taken. Given that the time taken is 15 seconds, we have:
Power = Work / Time
Power = 15,000 J / 15 s
Power = 1,000 W
Therefore, the power exerted by the horse is 1,000 watts, which is equivalent to 1 kilowatt (kW).
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a 70g bullet moving east at 40 m/s strikes a 1.2kg block suspended and becomes embedded in the block. how high will the bullet-block system rise above its original point?
A 70g bullet moving east at 40 m/s strikes a 1.2kg block suspended and becomes embedded in the block. the height to which the bullet-block system rises above its original point is 4.75 m.
The height to which the bullet-block system rises above its original point can be calculated using the law of conservation of energy.
The initial kinetic energy of the bullet is converted into gravitational potential energy of the system when it reaches its highest point.
The initial kinetic energy of the bullet-block system is given by the expression:
(1/2)mv²
where m = 70 g = 0.07 kg (mass of bullet) and v = 40 m/s (velocity of bullet)
Hence, initial kinetic energy of the bullet-block system is:
(1/2)(0.07)(40)² = 56 J
At the highest point of the system, all of the initial kinetic energy is converted into gravitational potential energy. Therefore, the change in potential energy is equal to the initial kinetic energy.
Change in potential energy = initial kinetic energy= 56 J
At the highest point of the system, the potential energy of the system is given by the expression:
mgh
where m = 1.2 kg (mass of block), g = 9.8 m/s² (acceleration due to gravity) and h is the height to which the system rises above its original point.
Substituting the given values in the above expression, we get:
56 = (1.2)(9.8)h
Solving for h, we get:
h = 56 / (1.2 × 9.8)
h = 4.75 m
Therefore, the height to which the bullet-block system rises above its original point is 4.75 m.
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A body of mass 10kg is moving with a velocity of 5m/s in a circle of radius 5m ,what is the centrpetal acceleration of the body
Answer:ac = v^2/r
= (5)^2/5=5
Explanation:
Can someone help me
ASAP pleazs
The horizontal component of the velocity is 10.6 m/s and the vertical component of the velocity is 7.5 m/s.
What are the values of the horizontal and vertical components of velocity?The horizontal and vertical components of the velocity can be found using the following equations:
Vx = Vcos(θ)
Vy = Vsin(θ)
where V is the initial velocity, θ is the angle of projection, Vx is the horizontal component of the velocity, and Vy is the vertical component of the velocity.
Substituting the given values, we get:
Vx = 13 m/s × cos(35°) = 10.6 m/s
Vy = 13 m/s × sin(35°) = 7.5 m/s
The time of flight can be found using the following equation:
t = 2Vsin(θ) / g
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Substituting the given values, we get:
t = 2 × 13 m/s × sin(35°) / 9.8 m/s^2 ≈ 1.9 s
Therefore, the giant snowball is in the air for approximately 1.9 seconds.
The horizontal distance traveled can be found using the following equation:
d = Vx × t
Substituting the values we found earlier, we get:
d = 10.6 m/s × 1.9 s
d ≈ 20.1 m
Therefore, the giant snowball travels approximately 20.1 meters horizontally before hitting the ground.
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1. a 1 m diameter wheel rotates clockwise 10 times per second. what is its angular velocity? what is the tangential velocity of the outside of the wheel? what is the angular displacement after 3 seconds? how far linearly will the wheel roll after 3 seconds?
Answer: Use a search engine so you can learn and stuff it’s a move I don’t study in this stuff I’m a history major
The angular velocity of the wheel is 62.83 radians per second. The tangential velocity of the outside of the wheel is 31.42 meters per second. The angular displacement after 3 seconds is 188.5 radians. The wheel will roll 94.25 meters linearly after 3 seconds.
The angular velocity of a rotating object is the rate at which it rotates about its axis, measured in radians per second. Since the wheel rotates 10 times per second, the angular velocity can be calculated as:
angular velocity = 10 rotations/second * 2π radians/rotation
angular velocity = 62.83 radians/second
The tangential velocity of the outside of the wheel can be calculated using the formula:
tangential velocity = radius * angular velocity
The radius of the wheel is half its diameter, or 0.5 meters. Thus, the tangential velocity is:
tangential velocity = 0.5 meters * 62.83 radians/second
tangential velocity = 31.42 meters/second
The angular displacement after 3 seconds can be calculated using the formula:
angular displacement = angular velocity * time
angular displacement = 62.83 radians/second * 3 seconds
angular displacement = 188.5 radians
Finally, to find the distance that the wheel rolls linearly, we can use the formula:
distance = circumference * number of rotations
The circumference of the wheel is π times its diameter, or π meters. After 3 seconds, the wheel has completed 30 rotations (10 rotations/second * 3 seconds). Thus, the distance rolled is:
distance = π meters * 30 rotations
distance = 94.25 meters
Therefore, the angular velocity of the wheel is 62.83 radians per second, the tangential velocity of the outside of the wheel is 31.42 meters per second, the angular displacement after 3 seconds is 188.5 radians, and the distance that the wheel rolls linearly after 3 seconds is 94.25 meters.
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what is the resistance of the 0.800 mm -diameter, 4.70 cm -long lead from a mechanical pencil? studysmarter
The resistance of the 0.800 mm-diameter, 4.70 cm-long lead from a mechanical pencil is approximately 20.668 Ω.
To find the resistance of the 0.800 mm-diameter, 4.70 cm-long lead from a mechanical pencil, we'll need to use the formula:
Resistance (R) = Resistivity (ρ) * Length (L) / Area (A)
The resistivity (ρ) of lead is approximately 2.2 x [tex]10^{-7}[/tex]Ωm. The length (L) is given as 4.70 cm, which is equal to 0.047 m. The diameter of the lead is 0.800 mm or 0.0008 m. We can find the area (A) using the formula for the area of a circle:
A = π * (Diameter / 2[tex])^{2}[/tex]
A = π * (0.0008 / 2[tex])^{2}[/tex]
A ≈ 5.027 x 10^-10 [tex]m^{2}[/tex]
Now, we can find the resistance:
R = (2.2 x 10^-7 Ωm) * 0.047 m / (5.027 x 10^-10 m^2)
R ≈ 20.668 Ω
So, the resistance of the 0.800 mm-diameter, 4.70 cm-long lead from a mechanical pencil is approximately 20.668 Ω.
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a thin copper wire in the same circuit is 8 mm long and has a constant cross section of 0.2 mm. the conductivitiy is. calculate the resistance r of the copper wire and the potential v atend at the other end of the wire.
Potential V at the other end of the copper wire is 4.82 V.
The resistance of a wire is given by the formula: R = ρL/A
Using the given values:
R = (1.68 x 10^-8 Ωm) x (8 x 10^-3 m) / (0.2 x 10^-6 m^2) = 0.672 Ω
The potential difference (voltage) at the end of the wire can be calculated using Ohm's law: V = IR
I = V/R = 12 V / 1.672 Ω = 7.18 A
Substituting this value of current and the calculated resistance into Ohm's law gives:
V = IR = (7.18 A) x (0.672 Ω) = 4.82 V
Therefore, the potential at the other end of the copper wire is 4.82 V.
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a body is moving with constant speed over frictionless horizontal surface. What is the work done by the weight?
Answer:
Zero.
Explanation:
Work done by the weight is Zero, since the force and displacement are at right angles to each other.
A light, rigid rod of length
ℓ = 1.00 m
joins two particles, with masses
m1 = 4.00 kg
and
m2 = 3.00 kg,
at its ends. The combination rotates in the xy-plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 6.40 m/s. (Enter the magnitude to at least two decimal places in kg · m2/s.)
What If? What would be the new angular momentum of the system (in kg · m2/s) if each of the masses were instead a solid sphere 14.5 cm in diameter? (Round your answer to at least two decimal places.)
The new angular momentum of the system, if each of the masses were instead a solid sphere 14.5 cm in diameter, is 0.008275 M kg · m² + 111 kg · m²/s. The mass M is unknown, so we cannot provide an exact value. However, we have shown the method to calculate the new angular momentum.
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The angular momentum of the system about the origin is given by:L = L1 + L2, whereL1 = I1ω1L2 = I2ω2I1 and I2 are moments of inertia and ω1 and ω2 are angular velocities.Let's assume that the rod is massless, and the masses m1 and m2 are concentrated at their ends, so we can write:I1 = m1r1², I2 = m2r2²where r1 = l/2 and r2 = l/2.I1 = (1/3)m1(l/2)² and I2 = (1/3)m2(l/2)²
The total moment of inertia of the system about the pivot point is:I = I1 + I2 = (1/3)(m1 + m2)(l/2)²The angular momentum of the system is:L = (1/3)(m1 + m2)(l/2)²(ω1 + ω2)When the speed of each particle is 6.40 m/s, we haveω1 = v1/r1 = 6.40/(l/2)ω2 = v2/r2 = 6.40/(l/2)So,L = (1/3)(m1 + m2)(l/2)²(2ω1) = (1/3)(m1 + m2)(l/2)²(2ω2)Therefore, L = (1/3)(m1 + m2)(l/2)²(2ω), whereω = ω1 = ω2 = 6.40/(l/2)The angular momentum of the system is L = (1/3)(m1 + m2)(l/2)²(2ω) = (1/3)(4.6 kg)((2.00 m)/2)²(2(6.40 m/s)) = 111 kg · m²/s.
If each of the masses were instead a solid sphere 14.5 cm in diameter, we can calculate the moment of inertia for each sphere. A solid sphere of uniform density has moment of inertia:I = (2/5)MR²where M is the mass of the sphere and R is the radius. Since the diameter is given as 14.5 cm, we have R = 7.25 cm = 0.0725 m. Thus:I = (2/5)M(0.0725 m)²The new moment of inertia for the system will be:I' = I + I + (1/3)(m1 + m2)(l/2)²where I is the moment of inertia of each sphere about its own center of mass.
So,I' = (2/5)M1(0.0725 m)² + (2/5)M2(0.0725 m)² + (1/3)(m1 + m2)(l/2)²We know that each sphere has a diameter of 14.5 cm, so its radius is 7.25 cm or 0.0725 m. Let the mass of each sphere be M. Therefore, I = (2/5)MR² = (2/5)M(0.0725 m)².Substituting the values,I' = 2(2/5)M(0.0725 m)² + (1/3)(4.6 kg)((2.00 m)/2)²= 0.008275 M kg · m² + 111 kg · m²/s
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Which species has the largest influence on the Earth?
Answer:
The species that has had the largest influence on Earth is undoubtedly humans. Human activity has significantly altered the natural environment, leading to climate change, deforestation, ocean acidification, and many other damaging effects on the planet.
q21. a bar magnet is falling though a loop of wire with constant velocity. the north pole enters first. as the south pole leaves the loop, the induced current (as viewed from above) will be
The south magnetic pole of a bar magnet enters the wire circle first as it falls vertically through it. As the north pole exits the circle, the induced current (viewed from above) flows anticlockwise. (viewed from above).
Faraday's Law states that whenever the magnetic flux in a coil of wire changes, an induced EMF (electromotive force) manifests itself in the wire, resulting in an induced current. Lenz's Law, which is a corollary of Faraday's Law, states that this induced EMF will flow in the direction that is opposite the shift in magnetic flux that caused it.
According to Lenz's Law, the induced current's direction opposes the shift in magnetic flux that caused it. The magnetic field created by the current induced in the wire loop resists the motion of the magnet, slowing it down. As a consequence, as the north magnetic pole exits the loop, the current moves anticlockwise (when viewed from above).
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. what is the resonant frequency of a 0.500 mh inductor connected to a 40.0 μf capacitor?
Therefore, the resonant frequency of the LC circuit is approximately 795.8 Hertz..
The resonant frequency of an LC circuit can be calculated using the formula: f = 1 / (2π √(LC))
The given equation relates the resonant frequency (f) in Hertz with the inductance (L) in Henrys and capacitance (C) in Farads.
In this case, the inductance of the circuit is 0.500 mH, which is equivalent to 0.0005 H, and the capacitance is 40.0 μF, which is equivalent to 0.000040 F. Substituting these values into the formula, we get:
f = 1 / (2π √(0.0005 H × 0.000040 F))
f ≈ 795.8 Hz
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true or false? a record is a homogeneous collection and an array is a heterogeneous collection.
A record is a homogeneous collection and an array is a heterogeneous collection.
False. The statement is incorrect.
An array is a homogeneous collection because it contains elements of the same data type. In contrast, a record is a heterogeneous collection because it can contain elements of different data types.
Arrays are data structures that store a collection of elements of the same data type, such as integers or characters. Each element in an array is identified by an index, which represents its position within the array. Since all elements in an array are of the same data type, arrays are considered homogeneous collections.
On the other hand, a record is a data structure that groups together elements of different data types into a single unit, such as a student's name, age, and grade. The elements in a record are referred to as fields or members, and each field can have a different data type. Records are used to represent complex data structures, and they are considered heterogeneous collections.
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a pitcher accelerates a 150 g baseball from rest to 30 m/s . part a how much work does the pitcher do on the ball?
To accelerate a 150 g baseball from rest to 30 m/s, the pitcher does 67.5 J of work, transferring energy to the ball.
To compute the work done by the pitcher on the baseball, we can utilize the recipe:
Work = Power x Distance x Cosine(theta)
In any case, since the issue doesn't give us the power or distance, we can utilize an alternate recipe:
Work = 1/2 x Mass x Velocity^2
Here, the mass of the baseball is 150 grams or 0.15 kg, and the last speed is 30 m/s. Accordingly, the work done by the pitcher ready is:
Work = 1/2 x 0.15 kg x (30 m/s)^2 = 67.5 Joules
This implies that the pitcher did 67.5 Joules of work on the baseball to speed up it from rest to 30 m/s. Work is a scalar amount that actions how much energy moved to an item, for this situation, the baseball, because of the utilization of a power. The work done by the pitcher made the baseball gain dynamic energy and move with a specific speed.
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