Adding a 50 V battery to a circuit can potentially increase the intensity of the light bulbs, assuming the bulbs are designed to operate at a higher voltage.
The intensity of light bulbs is generally related to the power (P) consumed by the bulbs. In an electrical circuit, the power can be calculated using the formula:
P = V² / R
Where V is the voltage and R is the resistance of the bulb.
By increasing the voltage (V), the power consumed by the bulbs may increase if the resistance (R) remains constant. A higher power output generally results in increased brightness and intensity of the light bulbs.
However, it's important to note that if the light bulbs are not designed to handle the increased voltage, it could lead to overloading and potential damage.
Therefore, it is crucial to ensure that the light bulbs are compatible with the increased voltage before making any modifications to the circuit.
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Which of these is NOT considered an example of low EM energy?
A. infra-red
B. microwaves
C. ultra-violet
D. radio waves
ultra-violet is NOT considered an example of low Electromagnetic energy. Hence option C is correct.
Electromagnetic waves, which are synchronised oscillations of the electric and magnetic fields, are the traditional form of electromagnetic radiation. The electromagnetic spectrum is created at various wavelengths depending on the oscillation frequency. Electromagnetic waves move at the speed of light, typically abbreviated as c, in a vacuum. The oscillations of the two fields create a transverse wave in homogeneous, isotropic media when they are perpendicular to each other, perpendicular to the direction of energy and wave propagation, and perpendicular to each other. Either an electromagnetic wave's oscillation frequency or its wavelength can be used to describe its location within the electromagnetic spectrum. Because they come from different sources and have different effects on matter, electromagnetic waves of different frequencies are known by various names. These are listed in decreasing wavelength and increasing frequency order: sound waves, lower energy have lower frequency.
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A new planet called "Corus" was discovered by a team of astronomers that
is 60 x 106 km away from Earth. A satellite was launched by a rocket from
Earth to reach Corus. At a specific distance from Corus, the rocket releases
the satellite to the orbit of the planet. The satellite makes one complete
revolution around Corus in 15 Earth days. If Corus has a similar mass to
Mars, propose a suitable mass of the satellite and estimate:
i.
ii.
Distance between the satellite and the Corus's surface
Satellite's gravitational acceleration towards the core of Corus
Gravitational force between the satellite and the Corus
Minimum speed of the satellite to orbit Corus
iv.
Because the planet is so far away from Earth, we will assume that it has no effect on Corus. The satellite radius will be 121943.5927 km.
The mass of the Corus is precisely equivalent to the mass of the Mars, we take it M. We see that the rocket makes a total rotation about the planet in only 15 days, so we expect that the rocket was spinning all over the world about a radius r. In this way, the satellite will move with at his range in the wake of detaching from the rocket.
We know T = 2πr/v
mv²/ r = GMm/r ²
where m = mass of satellite
r = GMm/ mv² = GM /v²
r = GMT²/ 4 π²r² , putting the value of v
r³ = (GM / 4 π²r²) T²
r³ = ( GM / 4π² ) ¹/³ T²/³
G = 6.67 × 10 ⁻¹¹
M = 6.39 × 10 ²³ kg
T = 1296000
r = 10258.621 × 11886.94
r = 121943.5927 km
gravitational acceleration towards the core of corner = GM/ r²
a = 6.67 × 10 ⁻¹¹ ×6.39 ×10 ²³/ (121943592.7) ²
a = 2.89 × 10 ⁻³ m/s²
force between satellite and the Corus =mass of the satellite × acceleration of the satellite
iv) minimum speed = [GM/r(1+e)]¹/² e is the eccentricity of the satellite
How does gravitational acceleration work?Gravitational speed increase is portrayed as the article getting a speed increase because of the power of gravity following up on it. It is measured in m/s2, and its symbol is g. Gravitational acceleration is a vector quantity with a magnitude and a direction.
What does "gravitational" mean?The universe is governed by a force known as gravity, also referred to as gravitation. For any two items or particles having nonzero mass, the power of gravity will in general draw in them toward one another. Everything from subatomic particles to galaxies in a cluster is affected by gravity.
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A fisherman with mass m stands at the center of a small boat which is stationary on the water. The boat also has
mass m and is a distance d from the shore. The fisherman walks on the boat toward the shore. Assume there is
no drag force between the boat and water, and that there is no net external force applied to the system.
What happens to the boat?
As the fisherman walks towards the shore on the boat, the boat moves away from the shore to maintain the center of mass of the fisherman-boat system.
When the fisherman (mass m) stands at the center of the small boat (also mass m) and walks towards the shore, the following occurs:
1. As the fisherman moves towards the shore, he exerts a force on the boat in the opposite direction, due to Newton's Third Law of Motion (action and reaction forces are equal and opposite).
2. The boat will move away from the shore in response to the force exerted by the fisherman's movement. This is because the fisherman-boat system is initially stationary, and there is no net external force acting on it.
3. The center of mass of the fisherman-boat system remains constant. This means that as the fisherman moves closer to the shore, the boat must move further away from the shore to maintain the same center of mass.
4. When the fisherman stops walking, the boat will also stop moving away from the shore, but at a greater distance than initially. The fisherman and the boat would have moved relative to each other, but their combined center of mass remains at the same distance (d) from the shore.
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Two students are given cubic boxes, measuring 10 cm on a side. robert puts a single glass marble with a diameter of 10 cm in the box. susan puts 1,000 1-cm glass marbles in her box. which box is heavier?
The total mass of the glass marbles is m = ρV = 2500 kg/m³ × 4.19×[tex]10^{-3}[/tex] m³ = 10.5 g. Susan's box is heavier than Robert's box because it contains more glass mass.
Assuming the density of the glass marbles is constant, the weight of each box will depend on the total mass of glass in the box.
The volume of the single glass marble is (4/3)πr³ = (4/3)π(0.05m)³ = 5.24×[tex]10^{-5}[/tex] m³. The volume of the box is 10 cm × 10 cm × 10 cm = [tex]10^{-3}[/tex] m³.
Therefore, only one glass marble can fit in the box, which has a total mass of m = ρV = 2500 kg/m³ × 5.24×[tex]10^{-5}[/tex] m³ = 0.13 g.
The volume of 1,000 glass marbles is 1000 × (4/3)π(0.01m)³ = 4.19×[tex]10^{-3}[/tex] m³. Therefore, the total mass of the glass marbles is m = ρV = 2500 kg/m³ × 4.19×[tex]10^{-3}[/tex] m³ = 10.5 g.
Thus, Susan's box is heavier than Robert's box because it contains more glass mass.
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You produce a wave by oscillating one end of the rope up and down 2.0 times a second .what is the frequency of this wave?
The frequency of the wave produced by oscillating one end of the rope up and down 2.0 times a second is also 2.0 Hz (Hertz).
Frequency is defined as the number of oscillations or cycles that a wave completes in one second. In this case, each oscillation of the rope creates one complete cycle of the wave.
Therefore, if the rope is oscillating 2.0 times per second, it is completing 2.0 cycles of the wave each second, which is equivalent to a frequency of 2.0 Hz.
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An assembly line has a staple gun that rolls to the left at 1. 5 m/s while parts to be stapled roll past it to the right at 2. 2 m/s. The staple gun fires 13 staples per second. How far apart are the staples in the finished part?
The distance between two consecutive staples in the finished part is approximately 0.28 meters or 28.46 centimeters.
Consider the relative velocity between the staple gun and the parts to be stapled.
The staple gun is rolling to the left at 1.5 m/s, while the parts are rolling to the right at 2.2 m/s. Therefore, the relative velocity between the staple gun and the parts is:
v_rel = v_parts - v_staple_gun = 2.2 m/s - (-1.5 m/s) = 3.7 m/s
The staple gun fires 13 staples per second, so the time between two consecutive staples is:
t = 1/13 s
During this time, the relative velocity between the staple gun and the parts causes the distance between the two consecutive staples in the finished part. Let's call this distance "d".
d = v_rel * t = 3.7 m/s * (1/13 s) = 0.2846 m
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A beam of light travels into a new denser medium causing the speed of light to change to 2.5 x 10 8 m/s. what is the index of refraction for the new medium?
The index of refraction for the new medium is approximately 1.19917.
To find the index of refraction for the new medium, we can use the formula:
n = c / v
Where:
n = index of refraction
c = speed of light in a vacuum (approximately 3 x 10⁸ m/s)
v = speed of light in the new medium (2.5 x 10⁸ m/s)
In this case, we know that the speed of light in the medium (v) is 2.5 x 10⁸ m/s. The speed of light in a vacuum (c) is 299,792,458 m/s.
So, we can calculate the index of refraction (n) as:
n = c/v = 299,792,458 m/s / 2.5 x 10⁸ m/s = 1.19917
Therefore, the index of refraction for the new medium is approximately 1.19917.
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A roller coaster passes over the top of the hill going 2. 7 m/s and reacts the bottom going 14m/s
A)How high is the hill?
B)What is spread half way down?
(Energy problem)
The height of the hill will be h' = (1/2)(v_i^2)/g
The spread halfway down is equal to half of the initial potential energy of the roller coaster at the top of the hill.
The spread halfway down is equal to half of the initial potential energy of the roller coaster at the top of the hill.
To solve this energy problem, we can utilize the principles of conservation of energy. The total mechanical energy of the roller coaster, consisting of its potential energy (PE) and kinetic energy (KE), remains constant throughout the ride.
A) To determine the height of the hill, we can equate the initial and final mechanical energies of the roller coaster at the top and bottom of the hill, respectively.
At the top of the hill:
Initial mechanical energy (E_i) = PE_i + KE_i = mgh + (1/2)mv_i^2
At the bottom of the hill:
Final mechanical energy (E_f) = PE_f + KE_f = mgh' + (1/2)mv_f^2
Since the roller coaster is at the top of the hill, its final kinetic energy (KE_f) is zero because it has come to a stop momentarily. Therefore, we have:
E_i = PE_i + KE_i = PE_f + KE_f = mgh + (1/2)mv_i^2 = mgh'
We are given that the roller coaster's initial velocity at the top of the hill (v_i) is 2.7 m/s, and its final velocity at the bottom (v_f) is 14 m/s.
Substituting these values into the equation, we get:
(1/2)mv_i^2 = mgh'
Simplifying and solving for h', the height of the hill, we have:
h' = (1/2)(v_i^2)/g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
B) To find the spread halfway down, we need to calculate the difference in potential energy between the top and halfway down the hill.
The potential energy at the top of the hill (PE_i) is given by mgh, and the potential energy halfway down (PE_half) is given by (1/2)mgh.
The spread halfway down is the difference between these two potential energies:
Spread halfway down = PE_i - PE_half = mgh - (1/2)mgh = (1/2)mgh
Therefore, the spread halfway down is equal to half of the initial potential energy of the roller coaster at the top of the hill.
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If I weigh 742. 32 N on Earth at a place where g = 9. 80 m/s2 and 5900. 91 N on the surface of another planet, what is the acceleration due to gravity on that planet?
The acceleration due to gravity on the surface of the other planet is approximately 77.98 m/s².
To find the acceleration due to gravity on another planet, we can use the formula:
Weight = Mass × Acceleration due to gravity
On Earth, your weight is given as 742.32 N, and the acceleration due to gravity is 9.80 m/s².
We can rearrange the formula to solve for mass:
Mass = Weight / Acceleration due to gravity
So, on Earth, your mass would be:
Mass on Earth = 742.32 N / 9.80 m/s²
Mass on Earth = 75.63 kg
Now, let's consider the surface of another planet where your weight is given as 5900.91 N.
We'll use the same formula and solve for the acceleration due to gravity on that planet:
5900.91 N = Mass × Acceleration due to gravity on the other planet
Substituting the value of mass we calculated earlier:
5900.91 N = 75.63 kg × Acceleration due to gravity on the other planet
Now, we can solve for the acceleration due to gravity on the other planet:
Acceleration due to gravity on the other planet = 5900.91 N / 75.63 kg
Acceleration due to gravity on the other planet ≈ 77.98 m/s²
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The diffraction grating has 50 slots per millimeter. At what angle is
the maximum of the first row seen when a wavelength of 400 nm
falls perpendicular to the grid?
Please I really need your help
The maximum of the first order is seen at an angle of approximately 0.001143 degrees.
To find the angle of the maximum of the first order for a diffraction grating, you can use the grating equation:
nλ = d * sin(θ)
where n is the order of the maximum (in this case, n=1 for the first order), λ is the wavelength, d is the distance between the slots (grating spacing), and θ is the angle we need to find.
First, we need to find the grating spacing (d). Since there are 50 slots per millimeter, the spacing would be:
d = 1 mm / 50 slots = 0.02 mm
We should convert this to meters for consistency with the wavelength unit (nm):
d = 0.02 mm * (1 m / 1000 mm) = 0.00002 m
Now, plug in the values into the grating equation:
(1)(400 * 10^(-9) m) = (0.00002 m) * sin(θ)
Divide both sides by 0.00002 m:
(400 * 10^(-9) m) / (0.00002 m) = sin(θ)
20 * 10^(-6) = sin(θ)
Now, find the angle θ by taking the inverse sine:
θ = arcsin(20 * 10^(-6))
θ ≈ 0.001143 degrees
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Two large speakers broadcast the sound of a band tuning up before an
outdoor concert. While the band plays an A whose wavelength is 0. 773 m,
Brenda walks to the refreshment stand along a line parallel to the speakers. If
the speakers are separated by 12. 0 m and Brenda is 24. 0 m away, how far
must she walk between the "loudspots"?
Brenda needs to walk a distance of 0.3865 meters to reach the next loud spot.
Brenda is walking along a line parallel to the speakers, the sound waves from each speaker will reach her in phase and interfere constructively, producing a loud spot. The distance between consecutive loud spots is equal to half the wavelength, so we can calculate this distance using the wavelength of the sound wave:
Distance between loud spots = 0.5 × wavelength
For an A note with a wavelength of 0.773 m, the distance between consecutive loud spots is:
Distance between loud spots = 0.5 × 0.773 m = 0.3865 m
Since Brenda is 24.0 m away from the speakers and the speakers are 12.0 m apart, she is equidistant from the two speakers and will hear the sound at its maximum intensity.
Therefore, she is currently at a loud spot. To find the next loud spot, she needs to walk a distance equal to the distance between consecutive loud spots:
Distance between loud spots = 0.3865 m
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what needs to happen to the index of refraction to produce a rainbow
To produce a rainbow, the index of refraction needs to vary with wavelength, which causes the different colors of light to refract at slightly different angles.
This occurs when light enters a water droplet and is bent, or refracted, as it slows down due to the higher index of refraction of water compared to air. The different colors of light then reflect off the inner surface of the droplet and are refracted again as they exit the droplet, creating a spectrum of colors. This process is called dispersion and is what creates the beautiful colors of a rainbow.
To produce a rainbow, the index of refraction needs to vary with the wavelength of light. This phenomenon, called dispersion, causes different colors (wavelengths) of light to bend at slightly different angles when passing through a medium like water droplets in the atmosphere. The variation in the index of refraction leads to the separation of colors and the formation of a rainbow.
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If the speed of light in some unknown material is 2. 00 × 108 m/s, what is the index of refraction of the medium?
The index of refraction of the unknown material is 1.5.
The index of refraction (n) of a medium is defined as the ratio of the speed of light in a vacuum (c) to the speed of light in the medium (v):
n = c / v
In this case, the speed of light in the unknown material is given as 2.00 × [tex]10^8[/tex] m/s. The speed of light in a vacuum is approximately 3.00 × [tex]10^8[/tex] m/s. Substituting these values into the formula:
n = (3.00 × [tex]10^8[/tex] m/s) / (2.00 × [tex]10^8[/tex] m/s)
Simplifying the expression:
n = 1.5
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The vast majority of stars in a newly formed star cluster are __________. red giants very high-mass, type o and b stars less massive than the sun about the same mass as our sun
The vast majority of stars in a newly formed star cluster are less massive than the sun, and about the same mass as our sun.
In a newly formed star cluster, most stars are categorized as low-mass or medium-mass stars, similar in size to our sun. This is because the process of star formation results in a mass distribution that follows a pattern called the initial mass function (IMF).
The IMF indicates that lower mass stars are much more abundant than high-mass stars.
High-mass, Type O and B stars, as well as red giants, are not as common in newly formed star clusters. Type O and B stars are very massive, hot, and luminous, but their rarity is due to the fact that they consume their nuclear fuel at a rapid rate, leading to shorter lifespans.
Red giants are also relatively rare in new star clusters, as they represent a later evolutionary stage of lower-mass stars, such as those with masses similar to our sun.
In summary, the vast majority of stars in a newly formed star cluster are less massive than the sun and about the same mass as our sun. High-mass, Type O and B stars, and red giants are less common in these clusters.
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Assuming birdman flies at a speed of 37m/s,how high should birdman fly to hit the bucket if the bucket is placed 118m from the start of the field
Birdman needs to fly at a height of 49.05m to hit the bucket placed 118m away from the start of the field, assuming he flies at a speed of 37m/s.
Birdman's required heightTo calculate the height at which Birdman needs to fly to hit the bucket, we need to use the equations of motion and consider the horizontal and vertical components separately.
Let's assume that Birdman is launching himself horizontally from the start of the field and needs to hit the bucket at a distance of 118m. We can use the horizontal distance, speed, and time to calculate the time it takes for Birdman to reach the bucket:
Horizontal distance = 118m
Horizontal speed = 37m/s
Time = Distance / Speed
Time = 118m / 37m/s
Time = 3.189s
Now that we know the time it takes for Birdman to reach the bucket horizontally, we can use the vertical component of motion to calculate the height at which he needs to fly.
We know that the only force acting on Birdman is gravity, and we can use the equation of motion for a vertically launched projectile to calculate the height:
Vertical distance = (Initial vertical velocity x Time) + (0.5 x Acceleration x Time^2)
Assuming that Birdman launches himself vertically with zero initial velocity, the equation simplifies to:
Vertical distance = 0.5 x Acceleration x Time^2
Where Acceleration is the acceleration due to gravity, which is approximately 9.81m/s^2.
Vertical distance = 0.5 x 9.81m/s^2 x (3.189s)^2
Vertical distance = 49.05m
Therefore, Birdman needs to fly at a height of 49.05m to hit the bucket placed 118m away from the start of the field, assuming he flies at a speed of 37m/s.
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A diverging lens has a focal length of -12. 8cm. An object is placed 34. 5cm from the len’s surface: Determine the image distance
The image distance formed by the diverging lens is 9.335cm.
To determine the image distance formed by a diverging lens with a focal length of -12.8cm, we can use the thin lens formula:
1/f = 1/do + 1/di
where f is the focal length of the lens, do is the distance from the object to the lens, and di is the distance from the lens to the image.
Substituting the given values, we get:
1/-12.8cm = 1/34.5cm + 1/di
Simplifying and solving for di, we get:
1/di = 1/-12.8cm - 1/34.5cm
1/di = -0.078125 cm^-1 - 0.02898550724637681 cm^-1
1/di = -0.1071105072463768 cm^-1
di = 9.335 cm
It's worth noting that the negative sign for the focal length indicates that the lens is a diverging lens.
The positive sign for the object distance indicates that the object is located on the same side of the lens as the incident light,
while the negative sign for the image distance indicates that the image is formed on the opposite side of the lens as the incident light, which is typical for a diverging lens.
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The curved path taken by the project object
The object is known as a projectile, and its course is known as its trajectory.
What is a projectile?A projectile is an object that moves freely under the effects of gravity and air resistance after being pushed by an external force. Although projectiles are any items in motion across space, they are most typically found in warfare and sports.
The curving route that an object takes when thrown is known as projectile motion.
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Chayse is working on analyzing the electromagnetic spectrum, focusing on ultraviolet rays. He determines that a uv ray has a frequency of 1. 53 × 10^16 hz and a wavelength of 1. 96 × 10^-8 m. Are his results reasonable? explain your answer.
His results appear to be accurate and reasonable.
Yes, Chayse's results are reasonable. The frequency and wavelength of ultraviolet (UV) rays fall within the appropriate range of values for this type of electromagnetic radiation.
UV rays have higher frequencies and shorter wavelengths than visible light, and their frequencies typically range from about 7.5 × 10^14 Hz to 3 × 10^16 Hz, while their wavelengths range from about 10 nm to 400 nm. Chayse's measured frequency of 1.53 × 10^16 Hz and wavelength of 1.96 × 10^-8 m are consistent with these values for UV rays.
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Circle the letter of each sentence that is true about how a psychrometer works.
a. The dry-bulb thermometer is cooled by evaporation when the wind blows.
b. The higher the humidity, the faster water evaporates from the bulb.
c. The wet-bulb thermometer reading is always higher than the dry-bulb reading.
d. When relative humidity is high, there is no difference between the wet-bulb and dry-bulb thermometer readings. (PLEASE HELP!!!)
A statement that is true about how a psychrometer works is "The higher the humidity, the faster water evaporates from the bulb". Therefore, the correct answer is b.
(a) is false because the dry-bulb thermometer is not cooled by evaporation when the wind blows. The dry-bulb thermometer measures the temperature of the air, while the wet-bulb thermometer measures the temperature of the air cooled by the evaporation of water from its wick.
(b) is true because the rate of evaporation from the wet-bulb thermometer depends on the humidity of the air. In humid air, there is less difference between the wet-bulb and dry-bulb readings because less evaporation occurs, while in dry air, more evaporation occurs and the wet-bulb temperature is lower.
(c) is false because the wet-bulb thermometer reading is always lower than the dry-bulb reading. The wet-bulb thermometer is cooled by the evaporation of water from its wick, which causes its temperature to be lower than that of the dry-bulb thermometer.
(d) is false because the difference between the wet-bulb and dry-bulb thermometer readings is greatest when the relative humidity is low. When the relative humidity is high, there is less evaporation from the wet-bulb thermometer, and the difference between the two readings is smaller.
In summary, a psychrometer works by measuring the difference in temperature between a dry-bulb thermometer and a wet-bulb thermometer, which is cooled by evaporation from its wick.
The rate of evaporation from the wet-bulb thermometer depends on the humidity of the air, and the difference between the two thermometer readings is greatest when the air is dry.
The wet-bulb thermometer reading is always lower than the dry-bulb reading, and the difference between the two readings is smaller when the relative humidity is high. Therefore, the correct answer is b.
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2. A woman lifts up a laundry basket 1. 5m and carries it 20m across the room. This takes 15s.
Work is done on the laundry basket_*
(20 Points)
in walking across the room
during the entire 15s
work is not done
in lifting the basket
The woman did work on the laundry basket by lifting it and carrying it. The total work done was 547J when she lifted it 1.5m and carried it 20m in 15 seconds.
The work done by the woman on the laundry basket can be calculated by finding the force required to lift the basket and carry it across the room, and then multiplying that force by the distance covered. Work is defined as the product of force and displacement in the direction of the force.
To lift the laundry basket up 1.5m, the woman needs to exert a force equal to the weight of the basket, which can be calculated as mass times gravity. Assuming the basket has a mass of 10kg, the force required to lift it is 98N. The work done in lifting the basket is therefore W = Fd = 98N x 1.5m = 147J.
To carry the basket 20m across the room, the woman needs to exert a force equal to the weight of the basket plus any additional force required to overcome friction.
Assuming the coefficient of friction between the basket and the floor is 0.2, the force required is approximately 20N. The work done in carrying the basket is therefore W = Fd = 20N x 20m = 400J.
The total work done by the woman on the laundry basket is the sum of the work done in lifting it and the work done in carrying it, which is 147J + 400J = 547J.
Therefore, the total work done by the woman on the laundry basket as she lifts it up 1.5m and carries it 20m across the room in 15 seconds is 547J.
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Complete Question:
What is the total work done by the woman on the laundry basket as she lifts it up 1.5m and carries it 20m across the room in 15 seconds?
An engine is received that hunts and surges at top no-load speed only. to determine whether the carburetor or the governor system is causing the symptom, a specific troubleshooting process can be followed. technician a says that to separate the governor system from the carburetor, simply hold the throttle plate still and see if the engine continues to hunt & surge. technician b says that hunting and surging is caused exclusively by a lean mixture in the carburetor. which technician is correct? technician a technician b both technicians a and b neither technicians a nor b
Technician A is correct in their suggestion to separate the governor system from the carburetor by holding the throttle plate still, while Technician B is incorrect in stating that hunting and surging is caused exclusively by a lean mixture in the carburetor. Therefore, the correct answer is Technician A.
Technician A is correct. To determine whether the carburetor or the governor system is causing the hunting and surging symptom at top no-load speed, holding the throttle plate still is a useful troubleshooting process. By holding the throttle plate still, the engine can be tested to see if it continues to hunt and surge, which will help determine if the governor system or the carburetor is causing the issue. This method allows for the separation of the governor system from the carburetor, making it easier to identify the cause of the problem.
On the other hand, Technician B is not entirely correct. While a lean mixture in the carburetor can cause hunting and surging, it is not the only possible cause. Other factors such as a malfunctioning governor system can also result in these symptoms. Therefore, it is essential to follow the troubleshooting process outlined by Technician A to accurately identify the cause of the problem and address it effectively.
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1. Harry is playing on a swing set at a park. It takes 17. 3 seconds for him to swing back and forth 5 times. What is the swing's period?
2. What is the frequency of a wave that occurs 278 times every 20 seconds?
3. The lowest frequency that the average human ear can hear is 20 Hz. This sound wave travels at a speed of 331 m/s through the air. What is the wavelength of this sound wave?
The lowest frequency that the average human ear can hear is 20 Hz. This sound wave travels at a speed of 331 m/s through the air, the wavelength of this sound wave is 16.55 meters
1. To determine the swing's period, we need to divide the total time it takes for Harry to swing back and forth by the number of oscillations. In this case, it takes 17.3 seconds for him to swing 5 times. The period (T) can be calculated as follows: T = 17.3 seconds / 5 oscillations. The swing's period is 3.46 seconds.
2. To find the frequency of a wave, we need to divide the number of occurrences by the time interval. In this case, the wave occurs 278 times every 20 seconds. The frequency (f) can be calculated as follows: f = 278 occurrences / 20 seconds. The frequency of the wave is 13.9 Hz.
3. The average human ear can hear a frequency as low as 20 Hz. Given that the speed of sound in air is 331 m/s, we can find the wavelength (λ) of this sound wave using the formula: speed = frequency × wavelength, or λ = speed / frequency. Plugging in the values, λ = 331 m/s / 20 Hz. The wavelength of this sound wave is 16.55 meters.
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Ice (the solid phase of water) has a lower density than water because:
importance of pressure in our daily life
Answer:
used in the ideal gas law to describe the energy of a gas, and many more situations.
We can observe total internal reflection when light travels (n_glass = 1 50.flower = 1.33) a. from glass to water b. from water to glass c. from air to glass
We can observe total internal reflection when light travels from air to glass, but not from glass to water or from water to glass. This is because in those cases, the light is traveling from a higher refractive index medium to a lower one, and thus there is no opportunity for internal reflection.
Total internal reflection occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index, and the angle of incidence is greater than the critical angle. In this case, n_glass = 1.50 and n_water = 1.33.
a. From glass to water: Total internal reflection can occur as the light is moving from a higher refractive index (glass) to a lower refractive index (water).
b. From water to glass: Total internal reflection cannot occur as the light is moving from a lower refractive index (water) to a higher refractive index (glass).
c. From air to glass: Total internal reflection cannot occur as the light is moving from a lower refractive index (air) to a higher refractive index (glass).
Therefore, total internal reflection can be observed when light travels from glass to water (option a).
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2. A woman lifts up a laundry basket 1. 5m and carries it 20m across the room. This takes 15s.
Work is done on the laundry basket_*
(20 Points)
in walking across the room
during the entire 15s
work is not done
in lifting the basket
A woman lifts up a laundry basket 1.5m and carries it 20m across the room in 15s.
Work is done on the laundry basket both in lifting the basket and in walking across the room during the entire 15s.
Lifting the basket:
The work done in lifting the basket is equal to the force applied multiplied by the distance lifted. The force applied is the weight of the basket, which is equal to the mass of the basket multiplied by the acceleration due to gravity (9.8 m/s^2).
Let's assume the mass of the basket is 'm'. The work done in lifting the basket vertically is given by:
Work_lift = force_lift × distance_lift = (m × 9.8) × 1.5
Carrying the basket horizontally:
When the woman carries the basket across the room, work is done against the force of friction between the basket and the floor. The work done is equal to the force of friction multiplied by the distance traveled horizontally.
The force of friction can be calculated using the coefficient of friction (μ) and the normal force (N). The normal force is equal to the weight of the basket since it is on a horizontal surface.
Let's assume the coefficient of friction between the basket and the floor is 'μ'. The work done in moving the basket horizontally is given by:
Work_horizontal = force_friction × distance_horizontal = (μ × m × 9.8) × 20
The total work done on the basket during the entire 15s is the sum of the work done in lifting and the work done horizontally:
Total work = Work_lift + Work_horizontal
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The coolant water used for nuclear fission reactions is usually
The coolant water used for nuclear fission reactions is: crucial in the process of generating electricity.
This water serves multiple functions, such as absorbing heat generated during the fission process, moderating the neutrons, and maintaining the temperature within a safe range. By circulating around the reactor core, the coolant water collects the heat produced and transfers it to a heat exchanger, which converts it into steam. The steam then drives a turbine connected to a generator, ultimately producing electricity.
Overall, the coolant water plays an essential role in the safe and efficient operation of nuclear power plants, ensuring the continuous generation of electricity through nuclear fission reactions.
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As soil particle size decreases from silt to clay, the field capacity __________ and the available water __________.
As soil particle size decreases from silt to clay, the field capacity typically increases and the available water decreases.
This is because as particle size decreases, the pore spaces between particles also decrease, which in turn decreases the amount of water that can be held in the soil.
However, the smaller pore spaces also increase the surface area available for water to adhere to soil particles, resulting in a higher field capacity.
Field capacity is the amount of water held in the soil after excess water has drained away, and it is affected by factors such as soil texture, structure, and organic matter content.
Available water is the amount of water that plants can extract from the soil, and it is influenced by factors such as the depth of the plant roots and the water-holding capacity of the soil.
Overall, understanding the relationship between soil particle size and water retention is important for effective irrigation and soil management practices.
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a car goes from 16 m/s to 2m/s in 3.5s. what is the cars acceleration
Ans. 4 m/s2
we know that,
acceleration = change in velocity/ total time
putting values we get,
16-2/3.5
= 14/3.5
=4
thus, the car's acceleration = 4 m/s2
What angular acceleration would be required in order to stop the earth from rotating over a period of 30.0 minutes
The angular acceleration required to stop the rotation of the Earth over a period of 30 minutes would be equal to the final angular velocity divided by the time interval.
The Earth's rotation is an example of rotational motion, which is described by angular velocity and angular acceleration. Angular velocity is the rate of change of angular displacement with respect to time, and angular acceleration is the rate of change of angular velocity with respect to time.
The final angular velocity would be zero, since the Earth would have stopped rotating, and the initial angular velocity can be calculated by dividing the circumference of the Earth (40,075 km) by the time period of 24 hours or 1,440 minutes, which gives a value of approximately 0.28 degrees per minute.
Therefore, the initial angular velocity would be (0.28 degrees/minute)(2pi radians/360 degrees) = 0.00489 radians/minute. Dividing this value by 30 minutes gives an angular acceleration of approximately 0.000163 radians/(minute²).
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