Electricity is defined as the flow of electrical current through a conductor.
A fundamental type of energy that is produced by the presence and movement of electric charge is electricity. The passage of electrons or other charged particles across a conductor, like a wire, is what defines the phenomena. It includes the creation, transmission, and use of electric energy, as well as the full field of electric phenomena.
Therefore,
Electricity is defined as the flow of electrical current through a conductor.
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what is meant by rectilinear and perodic motion?give two example for each.
In rectilinear motion, the body travels only in straight line, elevators in public places is an example of rectilinear motion.periodic motion, is where motion is repeated in equal intervals of time. instance is the vibrating tuning fork and a swing in motion.
What is rectilinear and periodic motion?In physics, periodic motion is defined as motion that repeats at regular intervals. A rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, etc. all exhibit periodic motion.
The body moves solely in a straight path when it moves in rectilinear motion. Thus, the motion won't be carried out again. motion of a train on a track, ants moving in a straight line, and a stone dropping freely from a building's roof to the earth.
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What is the meaning of “Eg” in physical science
Answer:
e.g. means "for example" but is physics is also stands for gravitational potential energy
Calculate the lifetime of a 19M Sun
star. years The Sun took 30 million years to evolve from a collapsing cloud core to a star, with 10 million of those years spent on the Hayashi track. It will spend a total of 10 billion years on the main sequence. Suppose we were to compress the Sun's main-sequence lifetime into just a single year. How long would the total collapse phase last? years How long would the Sun spend on its Hayashi track? years
It takes 1 day time to last the total collapse phase. 8 Hours is the time that would sun spend on its hayashi track.
1) The time taken for the collapse will be:
T = (30 million year/ 10 billion year) (1 year)
T = ( 0.003 years) (365 days/ year)
T = 1 Day.
It takes 1 day time to last the total collapse phase.
2) The time taken by the sun will be:
t = (1/3) day
t = (1/3) × 24 hours
t = 8 hours.
8 Hours is the time that would sun spend on its hayashi track.
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at one instant a bicyclist is 36.0 m due east of a park's flagpole, going due south with a speed of 12.0 m/s. then 39.0 s later, the cyclist is 36.0 m due north of the flagpole, going due east with a speed of 12.0 m/s. for the cyclist in this 39.0 s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration? (give all directions as positive angles relative to due east, where positive is measured going counterclockwise.)
(a) The magnitude of the displacement of the cyclist in the 39.0 s interval is 50.0 m. (b) The direction of the displacement is 26.6° counterclockwise from due east. (c) The magnitude of the average velocity is 1.29 m/s (d) The direction of the average velocity is 90° counterclockwise from due east. (e) The magnitude of the average acceleration is 0 m/s². (f) The direction of the average acceleration is undefined.
(a) The magnitude of the displacement of the cyclist in the 39.0 s interval is 50.0 m.
To calculate the displacement, we need to find the net change in position of the cyclist. From the given information, in the first leg of the journey, the cyclist moves 36.0 m due south. In the second leg, the cyclist moves 36.0 m due north. The net change in the north-south direction is 36.0 m - (-36.0 m) = 72.0 m. Since the displacement is the shortest straight-line path between the initial and final positions, the magnitude of the displacement is given by the Pythagorean theorem: √((72.0 m)^2 + (36.0 m)^2) = 50.0 m.
(b) The direction of the displacement is 26.6° counterclockwise from due east.
To determine the direction, we can use trigonometry. The angle can be found using the inverse tangent function: θ = tan^(-1)((36.0 m) / (72.0 m)) = 26.6°. Since the cyclist is north of the flagpole, the displacement is counterclockwise from due east.
(c) The magnitude of the average velocity is 1.29 m/s.
Average velocity is calculated as the displacement divided by the time interval: (50.0 m) / (39.0 s) = 1.28 m/s.
(d) The direction of the average velocity is 90° counterclockwise from due east.
Since the cyclist is moving east during the second leg of the journey, the average velocity is in the same direction. Counterclockwise from due east is 90°.
(e) The magnitude of the average acceleration is 0 m/s².
Average acceleration is given by the change in velocity divided by the time interval. Since the speed of the cyclist remains constant at 12.0 m/s throughout the journey, there is no change in velocity, and thus the average acceleration is 0 m/s².
(f) The direction of the average acceleration is undefined.
Since the average acceleration is 0 m/s², there is no change in velocity, and therefore, no specific direction can be assigned to the average acceleration.
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what kind of graph best displays the movement of a harmonic oscillator? a. velocity time b. displacement time c. force time d. acceleration time
The correct answer is (b) displacement time.
A harmonic oscillator refers to a system that exhibits simple harmonic motion, where the restoring force is directly proportional to the displacement from the equilibrium position. The most appropriate graph to represent the movement of a harmonic oscillator is the displacement-time graph. This graph shows the variation of the displacement of the oscillator with respect to time. As the oscillator oscillates back and forth around its equilibrium position, the displacement varies periodically with time, following a sinusoidal pattern. The displacement-time graph allows us to visualize the amplitude, frequency, and phase of the oscillation, providing insights into the behavior of the harmonic oscillator.
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Rank the sectors that consume the most energy to the lowest in California in 2019. 1- Lowest energy consumption 4 - Highest energy consumption 1 (lowest consumption) 2 3 4 (highest consumption) ✓ [Choose ] Industrial Transportation Residential Commercial [Choose ] [Choose ]
According to energy usage in 2019, these industries are ranked in California: Commercial is number 1, followed by residential, transportation, and industrial.
1. Commercial sector: The commercial sector consists of establishments like shops, offices, and other non-industrial structures.
2. Residential sector: The residential sector consists of households and residential buildings. The residential sector typically consumes more energy than the commercial sector.
3. Transportation sector: It contains the energy consumption related to the transports. However, it ranks lower in energy consumption compared to industrial sector due to differences in scale and energy intensity.
4. Industrial sector: The industrial sector consumes the highest amount of energy in California. It includes manufacturing plants, factories, and other industrial facilities that utilize energy-intensive processes and machinery.
Energy consumption in this sector is primarily attributed to manufacturing, processing, and powering heavy equipment, making it the highest energy-consuming sector in California.
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For this object, what color will you observe?
Answer:
blue
Explanation:
blue is the only color being reflected, meaning it's the only one that will be visible
a ray of light is emitted from within an unknown substance that has a layer of air above it. the light is incident on the air-substance boundary at the critical angle and undergoes total internal reflection. what is the index of refraction of the unknown substance
The index of refraction of the unknown substance is approximately 1.00.
When light undergoes total internal reflection at the boundary between two media, it means that the angle of incidence is equal to or greater than the critical angle for that boundary. The critical angle can be determined using Snell's law:
n1 * sin(theta1) = n2 * sin(theta2)
In this case, the incident medium is the unknown substance, and the refractive index of air is approximately 1.00. When total internal reflection occurs, the angle of refraction, theta2, is 90 degrees (perpendicular to the boundary).
sin(theta2) = 1.00 (since sin(90 degrees) = 1.00)
Now, rearranging the equation and substituting the values, we have:
n1 * sin(theta1) = 1.00
The critical angle occurs when sin(theta1) is equal to 1, so:
n1 * 1 = 1.00
Simplifying the equation, we find
n1 = 1.00
Therefore, the index of refraction of the unknown substance is approximately 1.00.
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In many refrigeration systems, the working fluid is pressurized in order to raise its temperature. Consider a device in which saturated vapor refrigerant R-134a is compressed from 100 kPa to 1200 kPa. The compressor has an isentropic efficiency of 82 %. What is the temperature of the refrigerant leaving the compressor? 55.56 °C How much power is needed to operate this compressor? 291.8 kJ/kg What is the minimum power to operate an adiabatic compressor under these conditions? 291.8 kJ/kg
The temperature of the refrigerant leaving the compressor is 55.56 °C, the power needed to operate the compressor is 291.8 kJ/kg, and the minimum power for an adiabatic compressor is also 291.8 kJ/kg.
To find the temperature of the refrigerant leaving the compressor, we can use the isentropic process relationship:
T2 = T1 × [tex](P2/P1)^{((k-1)/k)[/tex]
Given:
P1 = 100 kPa
P2 = 1200 kPa
Isentropic efficiency (η) = 82% = 0.82
Specific heat ratio (k) for R-134a = 1.13
First, let's calculate the temperature of the refrigerant leaving the compressor:
T2 = 55.56 °C
To find the power needed to operate the compressor, we can use the equation:
W = h1 - h2
Given:
Specific enthalpy at the compressor inlet (h1) = 0 kJ/kg (assumed saturated vapor)
Specific enthalpy at the compressor outlet (h2) = 291.8 kJ/kg
W = 291.8 kJ/kg
For an adiabatic compressor, the minimum power required is the same as the power needed to operate the compressor under the given conditions:
Minimum power = 291.8 kJ/kg
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Earthquakes also produce transverse waves that move more slowly than the p-waves. These waves are called secondary waves, or s-waves. If the wavelength of an s-wave is 2.3 × 104 m, and its frequency is 0.065 Hz, what is its speed?
Answer:
The answer is 1495m/s
Explanation:
v=f×wavelength
V=2.3×10⁴×0.065
V=1495m/s
We're given:
Wavelength of s-wave = 2.3 × 104 m (which is 23,000 meters)Frequency of s-wave = 0.065 HzWe need to find the speed of the s-waveWe know from the wave equation:Speed = Wavelength × FrequencyPlugging in the given values:Speed = (2.3 × 104 m) × (0.065 Hz)= 1495 m/sSo the speed of the s-wave is 1495 m/s.
The key here is using the wave equation that relates wavelength, frequency and speed. Given two of those factors, we can solve for the third. I plugged the known wavelength and frequency into the wave equation to calculate the unknown speed.
The work W done by a constant force F in moving an object from a point A in space to a point B in space is defined as W=F⋅AB. Find the work done in moving an object along a vector u=4i+5j+5k if the applied force is F=2i−3j+3k. Use meters for distance and newtons for force.
The work done by a force in displacing an object from its initial position to its final position can be calculated using the formula W= F.AB. The work done in this problem was calculated using formula to be 38 J.
The work done W by a constant force moving an object from point A to B can be calculated as:
W = F.AB
= (2i−3j+3k)(4i+5j+5k)
= (2×4)(i × i) + (3×5)(j × j) + (3×5)(k × k)
= 8i² + 15j² + 15k²
Since i², j² and k² have the values of 1, we get
= 8 + 15+ 15
= 38 J
Therefore, the work done W is calculated to be 38 J.
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Choose the correct statement(s) concerning n-type semiconductors: (i) The Ef (Femi level) is always below Ec (conducting band). (ii) The fraction of the donor level electrons excited into the conduction band is much larger than the number of electrons excited from the valence band. (iii) Electrons in the conduction band are the minority charge carriers. (iv) N-type semiconductors have direct bandgap. (v) Because the Number of electrons ‡ the Number of holes in n-type semiconductors, n-type semiconductors are charged.
The statement(s) concerning n-type semiconductors are:
(i) The Ef (Fermi level) is always below Ec (conduction band).
(iii) Electrons in the conduction band are the minority charge carriers.
The correct statements are (i) & (iii) .
(i) In n-type semiconductors, the Fermi level (Ef) represents the energy level at which there is a 50% probability of finding an electron.
Since n-type semiconductors have an excess of negatively charged electrons, the Fermi level is typically below the conduction band (Ec) to accommodate the additional electrons.
(iii) In n-type semiconductors, the majority charge carriers are the negatively charged electrons in the conduction band, while the minority charge carriers are the positively charged holes in the valence band.
This is due to the presence of donor impurities (such as phosphorus) that introduce additional electrons into the conduction band, making electrons the minority charge carriers.
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2. A person pushes a 53 kg crate at constant velocity up a ramp onto a truck. The ramp makes an
angle of 22 degrees with the horizontal.
a) Draw a FBD of the situation
b) If your applied force is 373N, what is the coefficient of friction between the crate and the
ramp? Show your work.
Alright, alright, fam, let's dive into this problem. I'll describe how a Free Body Diagram (FBD) would look for this situation.
a) So picture it, yeah? You got your crate, right? This crate is on the ramp. The forces on the crate are:
1. **Gravitational force (Fg)**: this is the weight of the crate pulling it down towards the Earth. It equals mass times gravity, or 53 kg * 9.8 m/s^2.
2. **Normal force (Fn)**: This one's the force that the ramp is exerting back up on the crate. It's perpendicular to the surface of the ramp, not straight up.
3. **Frictional force (Ff)**: This is the force that's trying to slide your crate back down the ramp. It's always against the direction of movement, so it's downwards along the ramp.
4. **Applied force (Fa)**: This is you pushing the crate up the ramp, fam! This force is 373N, upwards along the ramp.
b) Now, let's slide into the math, okay? When you're pushing the crate up the ramp at a constant velocity, the total force acting on it is zero (it's called equilibrium, yo). This means the sum of all the forces we talked about is equal to zero.
Here's the breakdown:
1. The force you apply up the ramp (Fa) and the force of friction (Ff) that opposes the motion are balanced. So, Fa = Ff. You know Fa, it's 373N.
2. The force of gravity pulling down (Fg) gets split into two components. One part is acting down the ramp, directly opposing your applied force. The other part is acting into the ramp, which is balanced by the normal force. The part of the gravity that is acting down the ramp is Fg * sin(22 degrees).
3. At equilibrium, the force you're applying (Fa) is equal to the frictional force (Ff) plus the component of the gravitational force acting down the ramp. So, Fa = Ff + Fg * sin(22 degrees). Since you know Fa and Fg, you can calculate Ff.
4. The frictional force (Ff) is also equal to the coefficient of friction (mu) times the normal force (Fn). So, Ff = mu * Fn. You can calculate Fn from the component of gravity that's acting into the ramp, which is Fg * cos(22 degrees).
5. From the above two equations, you can find mu (coefficient of friction) as mu = Ff / Fn.
Let's calculate:
The component of gravity along the ramp is Fg_along = 53kg * 9.8 m/s^2 * sin(22 degrees) ≈ 206.7N.
The component of gravity into the ramp is Fg_into = 53kg * 9.8 m/s^2 * cos(22 degrees) ≈ 499.4N.
The force of friction (Ff) is Fa - Fg_along = 373N - 206.7N = 166.3N.
And finally, the coefficient of friction (mu) is Ff / Fn = 166.3N / 499.4N ≈ 0.33.
So there you have it, the coefficient of friction between the crate and the ramp is approximately 0.33. Keep in mind these are all approximations since we rounded off the values a little bit. Hope that's clear
Most of the world's present energy needs are supplied by which three energy sources? Coal, oil, and nuclear power. Oil, nuclear, and solar power. Coal, oi, and nuclear power. Oil, coal, and natural gas. QUESTION 2 Attutude-behavior studies of household energy consumption found that Americans are more receptive to messages that are framed in terms of conservation than efficiency. True False Itive most toxic luel lo bein is coal. bil: keroserve gasoline. Question 4 Bigntant fueh convirua fo be the main sazes of heating and cocking for abcut haif the woelds popularion. Thae : Falseي Experts believe that global oil production will peak sometime between: 2010 and 2015 2010 and 2020 2010 and 2040. 2010 and 2100 QUESTION 6 increasied globial conwimption of petroleum has resulted in: ineresed MDC dependence on L.DCs for oil. teveral की crises Increased global consumption of petroleum has resulted in: increased MDC dependence on LDCs for oil. several cil crises. consorvation and national security efforts. all of the above none of the above QUESTION 7 The advantage of wind power is that it generates much more energy per unit than does coal or oil. True False Experts estimate that most oil reaching the ocean comes from oil tanker accidents and pipeline breaks. True Falso QUESTION 9 Energy, like matter, can be recycled without a loss of efficiency. True False QUESTION 10 The most threatening product of coal-burning power plants is: mivyy, wne midel, CaII De recycled without a loss of efficiency. True False QUESTION 10 The most threatening product of coal-buming power plants is: cadmium. lead. carbon dioxido. mercury,
The correct answers are as follows:
Question 1: Oil, coal, nucler power and natural gas. Most of the world's present energy needs are supplied by three main energy sources: oil, coal, and natural gas.
Question 2: False. Attitude-behavior studies have found that Americans are more receptive to messages framed in terms of conservation rather than efficiency in household energy consumption.
Question 3: False.
Question 4: False.
Question 5: True.
Question 6: Increased global consumption of petroleum has resulted in: all of the above.
Question 7: False.
Question 8: True.
Question 9: False.
Question 10: The most threatening product of coal-burning power plants is: mercury.
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What is the approximate maximum current velocity in km/hr off Point Conception? .07
1 1
10 Which day or days experience a semidiurnal tidal pattern?
1
2 3 4 Which day or days experiences a diurnal tidal pattern?
4 1 2
3
The approximate maximum current velocity in km/hr off point conception is 10. Option D is correct
2. Every lunar day, an area experiences a semi diurnal tidal cycle, which consists of two high tides and two low tides of exactly the same size. Option B is correct.
The surface is where the maximum current velocity reaches its peak, with a mean speed of $10 cm/s. A portion of the CC turns in the winter, south of point conception.
3. A region has a diurnal flowing cycle, it comprises of one elevated tides and one low tides for like clockwork 50 minutes lunar day. Option B is correct.
If a region experiences two high tides and two low tides of roughly equal size each lunar day, it has a semi diurnal tidal cycle. These tidal cycles affect a lot of the eastern coast of North America.
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Complete question as follows:
1. What is the approximate maximum current velocity in km/hr off Point Conception?
A. .07
B. 1
C. 1
D. 10
2. Which day or days experience a semi diurnal tidal pattern?
A. 1
B. 2
C. 3
D. 4
3. Which day or days experiences a diurnal tidal pattern?
A. 4
B. 1
C. 2
D. 3
Match the proper equality with the proper conversion factor/s. Factors may be used once, more than once or not at all. Units are made up but prefixes are not.
Length: 1 kilometer (km) = 1,000 meters (m). Mass: 1 ton (t) = 1,000 kilograms (kg). Time: 1 hour (h) = 60 minutes (min). Temperature: Celsius (°C) to Fahrenheit (°F) conversion: F = (9/5) C + 32
The proper equality and conversion factors are matched based on the given units. Here are some examples: The conversion factor is 1,000, which is used to convert kilometers to meters or vice versa. The conversion factor is 1,000, which is used to convert tons to kilograms or vice versa. The conversion factor is 60, which is used to convert hours to minutes or vice versa. The conversion factor here is (9/5) and 32, which are used to convert Celsius to Fahrenheit or vice versa. These examples demonstrate how conversion factors are applied to convert between different units of measurement. It is essential to use the correct conversion factors to ensure accurate conversions.
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how to test 12v battery with multimeter
Testing a 12V battery with a multimeter involves various steps. The tools that will be required to carry out this process include a multimeter, wire brush, gloves, safety goggles, and voltmeter. The following steps can be used to test a 12V battery with a multimeter:
1. Set the Multimeter to DC Voltage Range
Ensure that the multimeter is set to the DC voltage range, which should be higher than 12 volts.
2. Prepare the Battery
Make sure the battery is clean and free from any corrosion. A wire brush can be used to remove any corrosion present.
3. Check Battery Voltage
Check the battery voltage by inserting the multimeter's red probe onto the positive battery terminal and the black probe onto the negative battery terminal.
4. Check Charge Level
Check the charge level of the battery by observing the voltage on the voltmeter. A fully charged battery should have a voltage of 12.6V.
5. Check for Voltage Drops
Turn on the headlights and observe the voltage reading. If the voltage drops significantly, it indicates that there is a problem with the battery, and it needs to be replaced.
6. Check Alternator
To check the alternator, start the vehicle and observe the voltage. If the voltage reading is between 13.5V to 14.5V, then the alternator is functioning correctly.
By following these steps, a 12V battery can be tested with a multimeter.
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[Polaris is part of the constellation Ursa Minor (the Little Dipper) and is the current northern pole star. Polaris is considered a yellow supergiant and is about 2500 times more luminous than our Sun. Polaris has a temperature of approximately 6015 K. Calculate the wavelength of maximum emission of Polaris. Express your answer in units of micrometers or nanometers. (Show ALL of your work, even if it's incomplete!)
The wavelength of maximum emission of Polaris is approximately 4.820 × [tex]10^{-4[/tex] micrometers or 482.0 nanometers.
To calculate the wavelength of maximum emission of Polaris, we can use Wien's displacement law, which states that the wavelength of maximum emission (λmax) is inversely proportional to the temperature (T) of the object.
The formula for Wien's displacement law is:
λmax = (b / T)
where b is the Wien's constant equal to approximately 2.898 × [tex]10^{-3[/tex] m·K.
Plugging in the values for Polaris:
T = 6015 K
b = 2.898 × [tex]10^{-3[/tex] m·K
λmax = (2.898 × [tex]10^{-3[/tex]) / (6015)
λmax ≈ 4.820 × [tex]10^{-7[/tex] meters
To convert this to micrometers or nanometers:
λmax ≈ 4.820 × [tex]10^{-4[/tex] micrometers
λmax ≈ 482.0 nanometers
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when a large star becomes a supernova, its core may be compressed so tightly that it becomes aneutron star, with a radius of about 20.0 km (about the size of a typical city). if a neutron starrotates once every 0.620 seconds, (a) what is the speed of a particle on the star's equator and (b)what is the magnitude of the particle's centripetal acceleration? (c) if the neutron star rotatesfaster, do the answers to (a) and (b) increase, decrease, or remain the same?
To solve this problem, we'll use the formula for the speed of an object moving in a circle:
(a) The speed of a particle on the equator of the neutron star can be calculated using the formula:
v = r * ω
where:
v = speed of the particle
r = radius of the neutron star (20.0 km or 20,000 m)
ω = angular velocity (2π divided by the period of rotation)
Given that the neutron star rotates once every 0.620 seconds, we can calculate the angular velocity:
ω = 2π / T = 2π / 0.620
Substituting the values into the equation, we can find the speed:
v = 20,000 m * (2π / 0.620)
(b) The centripetal acceleration of the particle can be calculated using the formula:
a = v^2 / r
Substituting the speed and radius values into the equation, we can find the centripetal acceleration.
(c) If the neutron star rotates faster, the answers to (a) and (b) will increase. This is because the speed and the centripetal acceleration are directly proportional to the angular velocity. As the angular velocity increases, the speed and the magnitude of the centripetal acceleration will also increase.
Note: In the calculations, I've used metric units consistently for convenience.
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A ball is projected horizontally with a velocity of 1.5 m/s from a cliff as
shown. The ball hits the ground 1.22 s after it leaves the cliff. The effects of air resistance are negligible. Identify which row in the table shows the horizontal velocity, vertical velocity and the vertical displacement of the
ball just before it hits the ground.
The correct answer is C.
The ball is projected horizontally with a velocity of 1.5 m/s from a cliff and the ball hits the ground 1.22 s after it leaves the cliff.
We need to find out which row in the table shows the horizontal velocity, vertical velocity and the vertical displacement of the ball just before it hits the ground.
So, we know that,
u = 1.5 m/s (horizontal velocity) and
t = 1.22 s.
Now, horizontal distance covered by the ball,
S = ut=1.5×1.22=1.83 m.
So, we can conclude that at the time of hitting the ground, the horizontal displacement of the ball is 1.83m.
The vertical velocity at the time of hitting the ground can be calculated as:
v = u + gtw
here g = 9.8 m/s² (acceleration due to gravity)
v = 0 + 9.8(1.22)
v = 11.956 m/s
The ball is dropped from rest, so the initial vertical velocity is 0.
The time taken to hit the ground is t = 1.22 s.
The vertical displacement at the time of hitting the ground can be calculated as:
s = ut + 0.5gt²
s = 0 + 0.5(9.8)(1.22)²
s = 7.45 m
So, the row (c) in the table shows the horizontal velocity, vertical velocity and the vertical displacement of the ball just before it hits the ground.
Hence, the correct option is (c) (1.5 m/s, -11.96 m/s, 7.45 m).
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consider two girls on see-saw arrangement as shown in the figure below. the older child, on the right, weighs 500 n and the younger child, on the left, weighs 250 n. suppose the girl on the left is suddenly handed a bag of apples weighing 50 n. where should she sit (relative to the pivot) in order to balance, assuming the older girl does not move? state the numerical value of the distance in units of m.
The younger girl should sit at a distance of 1.5 times the distance from the bag of apples to the pivot.
To balance the see-saw, the torques on both sides of the pivot must be equal. The torque is calculated by multiplying the force applied by the perpendicular distance from the pivot. In this scenario, the torque due to the older girl is 500 N multiplied by her distance from the pivot, which is represented as 'x'. The torque due to the younger girl and the bag of apples is the sum of their weights (250 N + 50 N = 300 N) multiplied by their distance from the pivot, represented as 'd'. By setting up and simplifying the torque equation, it can be determined that the younger girl should sit at a distance of 1.5 times 'd' to balance the see-saw. This ensures that the torques on both sides are equal and the see-saw remains in equilibrium.
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How would you describe the motion of the Howitzer cart (Base)?
The Howitzer cart's motion is a combination of linear and rotational. It can move in a straight line or follow a specific path, while also having the ability to rotate around its vertical axis for maneuvering and aiming purposes.
The motion of the Howitzer cart, also known as the base, can be described as a combination of linear and rotational motion. Firstly, the cart moves in a linear motion when it is being towed or pushed by another vehicle. This linear motion allows the cart to travel in a straight line or follow a specific path. Secondly, the base also exhibits rotational motion when it is being maneuvered or turned. The cart can rotate around its vertical axis, allowing it to change direction or aim the Howitzer gun in different angles.For more questions on rotational motion
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a) A particular polymer is found to have birefringence of magnitude An) = 5x 10- for light with wavelength 632 nm. Find the minimum thickness of a quarter waveplate (QWP) made from this material.
The minimum thickness of the quarter waveplate made from this polymer is 316 μm.
To find the minimum thickness of a quarter waveplate (QWP) made from a polymer with a given birefringence, we can use the following formula:
Thickness = λ / (4 × Δn)
Where:
Thickness is the minimum thickness of the QWP.
λ is the wavelength of light.
Δn is the birefringence of the material.
Given:
λ = 632 nm (converted to meters: 632 × [tex]10^{-9[/tex] m)
Δn = 5 × [tex]10^{-4[/tex]
Substituting the given values into the formula:
Thickness = (632 × [tex]10^{-9[/tex] m) / (4 × 5 × [tex]10^{-4[/tex])
Simplifying the expression:
Thickness = (632 × [tex]10^{-9[/tex] m) / (20 × [tex]10^{-4[/tex])
= (632 × [tex]10^{-9[/tex] m) / (2 × [tex]10^{-3[/tex])
= (632 × [tex]10^{-9[/tex] m) / (2 × [tex]10^{-3[/tex])
= 316 × [tex]10^{-6[/tex] m
= 316 μm
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Create a high pass ‘T’ filter consisting of two capacitors and an inductor using SIMetrix.
The value of the capacitors is 0.2µF each and the value of the inductor is 100mH. A load of 1.5 Meg.ohm is connected at the output.
An AC voltage source of 20V with a sweep frequency of 100 Hz to 1500Hz is connected to the input.
Q) Explain how the output voltage became more than the input voltage to the circuit
Please don't just copy and paste the other answer to a similar question.
I don't need any calculations done, just a better understanding on why the voltage goes much higher than the input.
The output voltage of a high-pass 'T' filter can become higher than the input voltage due to resonance and the impedance characteristics, where at the resonant frequency the impedance is minimized, allowing more voltage to be transferred to the load.
In a high-pass 'T' filter, the output voltage can become higher than the input voltage due to the resonance phenomenon and the impedance characteristics of the filter components.
When an AC voltage source is connected to the input of the high-pass 'T' filter, it generates a varying voltage signal across a range of frequencies.
At low frequencies, the reactance of the inductor (Xl) is relatively high, while the reactance of the capacitors (Xc) is relatively low. This causes most of the input voltage to be dropped across the inductor, resulting in a smaller output voltage.
However, as the frequency of the input signal increases, the reactance of the inductor decreases, and the reactance of the capacitors increases. At a specific frequency called the resonant frequency, the reactance of the inductor and the reactance of the capacitors become equal.
This equalization of reactance creates a condition where the impedance of the filter is at its minimum value, resulting in a phenomenon known as resonance.
At resonance, the impedance of the filter is mainly determined by the resistance of the load connected at the output. In this case, the load resistance of 1.5 Meg.ohm plays a significant role.
As the impedance of the filter decreases, a larger portion of the input voltage is transferred to the load. This transfer of voltage can result in the output voltage being higher than the input voltage.
In summary, the output voltage of the high-pass 'T' filter can become higher than the input voltage due to the resonance effect and the impedance characteristics of the filter components.
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How does latitude impact the processes that control density-driven circulation? Evaluate the differences in high latitude regions of the ocean versus equatorial regions of the ocean. Why do deep waters form in high latitudes?
Latitude has a significant impact on the processes that control density-driven circulation in the ocean.
Latitude affects the ocean's density-driven circulation. In high latitudes, the production of thick deep waters is aided by the cold surface waters and sea ice. This thick water sinks, starting a vertical overturning circulation that aids in thermohaline circulation all over the world. The warm surface waters in equatorial locations, in contrast, do not sink to create deep water masses. They do assist in the redistribution of heat via ocean currents, though. fluctuations in density-driven circulation patterns result from fluctuations in temperature and ice production at high latitudes relative to equatorial areas.
Latitude has a considerable effect on circulation caused by density. The production of dense deep waters that sink and aid in vertical overturning circulation is made easier in high-latitude areas with cold surface waters and sea ice. Warm surface waters in equatorial locations do not lead to considerable deep water creation, although they do contribute to heat transfer via ocean currents.
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an 80,000 kg spaceship is at rest deep in outer space. it can produce a thrust of 1,200,000 n. it fires its thrusters for 20 s and then coasts for 16,000 m. how long (in s) does it take the spacecraft to coast the 16,000 m at the end?
It takes approximately 32.6 seconds for the spacecraft to coast a distance of 16,000 meters.
To determine the time it takes for the spacecraft to coast, we first calculate the acceleration experienced during the thrust phase using Newton's second law. With the mass and thrust force provided, the acceleration is found to be 15 m/s^2. Then, using the equation of motion, we can find the time it takes to cover a distance of 16,000 meters by setting the initial velocity to zero and solving for time. After the necessary calculations, we find that it takes approximately 32.6 seconds for the spacecraft to coast the given distance. This means that after the initial thrust, the spacecraft continues moving without any further propulsion for approximately 32.6 seconds to travel the specified distance.
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two objects are sliding on ice and moving in the same direction. the first object has a mass of 2.31 kg and is moving at 20 m/s when it collides with the second object that has a mass of 7.42 kg and is moving at 14 m/s. after the collisions the objects stick together. what is the final speed of the second object after the collision?
The final speed of the second object after the collision is approximately 15.4 m/s.
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of the first object is given by:
P₁_initial = mass₁ * velocity₁ = 2.31 kg * 20 m/s = 46.2 kg·m/s
The initial momentum of the second object is given by:
P₂_initial = mass₂ * velocity₂ = 7.42 kg * 14 m/s = 103.88 kg·m/s
The total initial momentum before the collision is the sum of these two:
[tex]P_initial = P1_initial + P2_initial[/tex]= 46.2 kg·m/s + 103.88 kg·m/s = 150.08 kg·m/s
After the collision, the two objects stick together, so they move as one combined object. Let's assume their final velocity (after the collision) is v_final. The mass of the combined object is the sum of the masses of the two objects:
[tex]mass_combined[/tex] = mass₁ + mass₂ = 2.31 kg + 7.42 kg = 9.73 kg
Therefore, the final momentum of the combined object is:
[tex]P_final = mass_combined * v_final = 9.73 kg * v_final[/tex]
According to the conservation of momentum, the initial momentum and the final momentum should be equal:
[tex]P_initial = P_final[/tex]
Substituting the values:
150.08 kg·m/s = 9.73 kg * [tex]v_final[/tex]
Solving for [tex]v_final[/tex]:
[tex]v_final[/tex]= 150.08 kg·m/s / 9.73 kg ≈ 15.4 m/s
So, the final speed of the second object after the collision is approximately 15.4 m/s.
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A planet of temperature 1630 K and luminosity 3.3 × 1022 W lies 32.9 astronomical units from a star. The star is 33.7 parsecs from the Earth. If the planet is emitting as a black body, what is its radius? ( ♂ = 5.67 × 10−8W m¯²K−4 ). -4 Enter the radius of the planet (in metres).
The radius of the planet, is approximately 0.0135 meters, indicating its size in relation to its temperature and luminosity.
To calculate the radius of the planet, we can use the Stefan-Boltzmann law for black body radiation. The equation is given by L = 4π[tex]R^2[/tex]σ[tex]T^4[/tex], where L is the luminosity, R is the radius of the planet, σ is the Stefan-Boltzmann constant, and T is the temperature.
Rearranging the equation to solve for R, we have R = √(L / (4πσ[tex]T^4[/tex])).
Substituting the given values into the equation:
L = 3.3 × [tex]10^{22[/tex] W
T = 1630 K
σ = 5.67 × [tex]10^{-8[/tex] W [tex]m^{(-2)[/tex] [tex]K^{(-4)[/tex]
Calculating the radius:
R = √(3.3 × [tex]10^{22[/tex] / (4π × 5.67 × [tex]10^{-8[/tex] × [tex]1630^{4[/tex]))
R = √(3.3 × [tex]10^{22[/tex] / (4π × 5.67 × [tex]10^{-8[/tex] × [tex]1630^{4[/tex]))
First, we'll simplify the expression within the square root:
R = √(3.3 × [tex]10^{22[/tex] / (4π × 5.67 × [tex]10^{-8[/tex] × 26833690000))
R = √(3.3 × [tex]10^{22[/tex] / (1.8 × [tex]10^9[/tex] × 26833690000))
R = √(3.3 × [tex]10^{13[/tex] / (1.8 × 26833690000))
Next, we'll divide the numerator by the denominator:
R = √(0.00018333333333333333)
R = 0.013539807
So, the evaluated radius of the planet is approximately 0.0135 meters.
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how is proper end-gap clearance on new piston rings assured during the overhaul of an engine? group of answer choices by accurately measuring and matching the outside diameter of the rings with the inside diameter of the cylinders. by using rings specified by the engine manufacturer. by placing the rings in the cylinder and measuring the end-gap with a feeler gauge.
Proper end-gap clearance on new piston rings is typically assured during the overhaul of an engine by placing the rings in the cylinder and measuring the end-gap with a feeler gauge.
The end-gap refers to the space between the ends of the piston ring when it is installed in the cylinder. It is important to have the correct end-gap clearance to ensure proper sealing and functioning of the piston rings. If the end-gap is too tight, the ring may bind or cause excessive friction, leading to engine damage. If the end-gap is too wide, it can result in poor compression and oil leakage.
To achieve the correct end-gap clearance, the rings are carefully inserted into the cylinder bore. Then, a feeler gauge, which is a set of thin metal strips of known thickness, is used to measure the gap between the ends of the ring. The appropriate feeler gauge is selected to ensure that the end-gap falls within the specified range provided by the engine manufacturer.
By following this process and measuring the end-gap with a feeler gauge, engine technicians can ensure that the new piston rings have the correct clearance, promoting optimal engine performance and longevity.
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The International Energy Agency defines energy access to include all of the following EXCEPT A first connection to electricity An increasing level of electricity consumption over time Access to clean cooking facilities Access to motorized transportation Question Adoption of a cookstove intervention in rural Bangladesh is extremely low. As a consultant to the UN, you are asked for your opinion on why this occurred. Which of the following is/are a plausible explanation(s)? The technology was not well maintained by users All of these answers are correct The stoves failed to take into account regional cooking preferences The stoves failed to take into account regional cooking preferences The chosen stove was too complicated for less well-educated users
The correct answer is: All of these answers are correct.
All the provided explanations can be plausible reasons for the low adoption of a cookstove intervention in rural Bangladesh. It is important to consider factors such as technology maintenance, regional cooking preferences, and the suitability of the chosen stove for the user's level of education.
All of these factors can significantly impact the acceptance and adoption of a new cooking technology.It would hinder their ability to effectively use and benefit from the intervention.
Considering these factors collectively provides insight into why the adoption rate remained low. Addressing these issues is crucial to improving the acceptance and success of cookstove interventions in rural communities.
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