Answer:
The specific heat capacity of water is 4.184 J/g°C, and we need to find the amount of heat required to raise the temperature of 10 kg (or 10,000 g) of water from 20°C to 100°C.
The formula for calculating the amount of heat required is:
Q = m * c * ΔT
where:
Q = heat energy (in Joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
We can start by calculating the change in temperature:
ΔT = 100°C - 20°C = 80°C
Then, we can plug in the values and solve for Q:
Q = 10,000 g * 4.184 J/g°C * 80°C
Q = 3,347,200 J
Therefore, Peter needs 3,347,200 Joules of heat to raise the temperature of his 10 kg vat of water from 20°C to 100°C.
Part F
Using your outline and the materials you've gathered, write a 250- to 500-word paper using word processing software. Be
sure to proofread and revise your writing to catch any errors in grammar, spelling, logic, or organization. Add a works
cited page at the end to give credit to your sources. Submit your completed paper and this activity to your teacher for
evaluation.
Please base it off of ‘compare and contrast analog and digital signals to determine which is more reliable for encoding and transmitting information’
Analog signals are continuous waves that keep changing over a time period, while digital signals are binary signals that consist of 0s and 1s1. Digital signals are more resistant to noise and interference, which makes them more reliable when compared to analog signals.
Digital signals are also easier to transmit and store, and they can be compressed without losing quality1. However, analog signals can carry an infinite amount of data, and they are less expensive to operate than digital signals.
In conclusion, digital signals are more reliable than analog signals for encoding and transmitting information because they are more resistant to noise and interference, and they are easier to transmit and store.
What is the difference between analog and digital signals?Analog signals are continuous waves that keep changing over a time period, while digital signals are binary signals that consist of 0s and 1s. Analog signals are described by the amplitude, period or frequency, and phase, while digital signals are described by bit rate and bit intervals. Analog signals have no fixed range, while digital signals have a finite number of values, i.e., 0 and 1. Analog signals are more prone to distortion, while digital signals are more resistant to noise and interference .
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How does horizontal forces on a car change when brake is applied?
Answer:
When a car brakes, it experiences a forward deceleration force, which is generated by the braking system. This force acts in the opposite direction to the car's motion and causes the car to slow down. As a result of this force, the horizontal forces acting on the car change.
When the brake is applied, the friction between the brake pads and the rotors generates a braking force that acts on the wheels. This force causes the wheels to slow down and, as a result, the car's forward motion is slowed down as well.
At the same time, the car's weight shifts forward, causing a decrease in the vertical force acting on the rear wheels and an increase in the vertical force acting on the front wheels. This increase in the front wheel's vertical force creates more friction between the tires and the road, which in turn generates a larger horizontal force that acts to slow down the car.
Additionally, as the car slows down, the aerodynamic drag acting on the car decreases, resulting in a smaller horizontal force acting on the car. However, the net effect of all these changes is a decrease in the car's forward motion and a corresponding decrease in the horizontal forces acting on the car.
The photos of an electromagnetic wave us 1. 5 × 10^-24 j of energy , what kind of electromagnetic waves are they ?
The electromagnetic waves with an energy of [tex]1.5 \times 10^{-24} J[/tex] are likely X-rays.
The energy of an electromagnetic wave is related to its frequency and wavelength by the equation:
E = hf = hc/λ
where:
E = energy of the electromagnetic wave
h = Planck's constant ([tex]6.626 \times 10^{-34} J s[/tex])
f = frequency of the electromagnetic wave
c = speed of light ([tex]3.00 \times 10^8 m/s[/tex])
λ = wavelength of the electromagnetic wave
Solving for λ:
λ = hc/E = [tex](6.626 \times 10^{-34} J s)(3.00 \times 10^8 m/s)/(1.5 \times 10^{-24} J)[/tex]
λ ≈ [tex]4.40 \times 10^-8 m[/tex]
This corresponds to a wavelength of about 44 nanometers, which falls in the range of X-rays. Therefore, the electromagnetic waves with an energy of [tex]1.5 \times 10^{-24} J[/tex] are likely X-rays.
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Christopher conducts an experiment in which he tests how much sugar dissolves at different temperatures of water. One step requires him to use water with a temperature of 32. 5°C. What is this temperature in kelvins? Record your answer to the nearest tenth
"The required kelvin temperature when celsius temperature is given is calculated to be 305.65 K."
There are three main scales to measure temperature. They are Celsius, Fahrenheit and Kelvin scales of measurement. Kelvin is said to be the official metric unit of temperature.
The temperature of water is given as 32.5 °C.
The same temperature in kelvins should be found out.
Converting celsius scale temperature to kelvin scale, we should add up 273.15 to the temperature in celsius scale.
Kelvin temperature = 273.15 + Celsius temperature
Kelvin temperature = 273.15 + 32.5 °C = 305.65 K
Thus, the kelvin temperature is calculated to be 305.65 K.
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A 6-kg ball traveling westward at 30 m/s hits a 20-kg ball at rest. The 6-kg ball bounces east at 9. 0 m/s. What is the speed and direction of the 20-kg ball? Use Given, Find, Equation. Show all your works. Write your answer in complete sentence
The speed of the 20-kg ball after collision is 11.7 m/s, and it is traveling in the westward direction. We can also say that the direction of the velocity is opposite to the direction of the initial velocity of the first ball.
Mass of the first ball ([tex]m_{1}[/tex]) = 6 kg
Velocity of the first ball before collision ([tex]v_{1}[/tex]i) = 30 m/s, traveling westward
Mass of the second ball ([tex]m_{2}[/tex]) = 20 kg
Velocity of the second ball before collision ([tex]v_{2}[/tex]i) = 0 m/s, at rest
Velocity of the first ball after collision ([tex]v_{1}[/tex]f ) = 9.0 m/s, traveling eastward
Velocity of the second ball after collision ( [tex]v_{2}[/tex]f)
Equation:Conservation of momentum:
[tex]m_{1}[/tex][tex]v_{1}[/tex]i+ [tex]m_{2}[/tex][tex]v_{2}[/tex]i=[tex]m_{1}[/tex] [tex]v_{1}[/tex]f + [tex]m_{2}[/tex] [tex]v_{2}[/tex]f
We can use the conservation of momentum equation to solve for the final velocity of the second ball (v2f). First, we need to find the initial velocity of the first ball in the eastward direction, which can be found using the velocity formula:
[tex]v_{1}[/tex]i,east = -[tex]v_{1}[/tex]i,west = -30 m/s
Now, we can substitute the given values into the conservation of momentum equation and solve for [tex]v_{2}[/tex]f:
[tex]m_{1}[/tex][tex]v_{1}[/tex]i+ [tex]m_{2}[/tex][tex]v_{2}[/tex]i= [tex]m_{1}[/tex][tex]v_{1}[/tex]f + [tex]m_{2}[/tex] [tex]v_{2}[/tex]f
(6 kg)(-30 m/s) + (20 kg)(0 m/s) = (6 kg)(9.0 m/s) + (20 kg)( [tex]v_{2}[/tex]f )
-180 kgm/s = 54 kgm/s + 20 kg [tex]v_{2}[/tex]f
-234 kgm/s = 20 kg [tex]v_{2}[/tex]f
[tex]v_{2}[/tex]f = (-234 kgm/s) / (20 kg)
[tex]v_{2}[/tex]f = -11.7 m/s
Therefore, the speed of the 20-kg ball after collision is 11.7 m/s, and it is traveling in the westward direction. We can also say that the direction of the velocity is opposite to the direction of the initial velocity of the first ball.
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Explain how velocity, displacement, angle, and time are related in projectile motion
Projectile motion is the motion of an object that is projected into the air and then follows a path determined by the forces acting on it.
Velocity, displacement, angle, and time are all related in projectile motion. The velocity of a projectile determines its speed and direction. The initial velocity of the projectile is composed of two components, the horizontal velocity and the vertical velocity. The horizontal velocity remains constant throughout the projectile motion, while the vertical velocity changes due to the force of gravity. The displacement of the projectile is its change in position during the motion. The displacement is dependent on both the initial velocity and the time elapsed. The horizontal displacement is determined by the horizontal velocity and time, while the vertical displacement is determined by the vertical velocity and time. The angle of projection is the angle at which the projectile is launched into the air. The angle affects the initial velocity components and determines the shape of the projectile path. The time of flight is the total time that the projectile is in the air. The time of flight is dependent on the vertical velocity component, the angle of projection, and the height of the launch point. In summary, the velocity, displacement, angle, and time are all related in projectile motion, and changes in any one of these factors will affect the trajectory of the projectile.
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one treatment of cataracts is to surgically remove the variable lens of the eye. if we assume that the cornea's refractive power focuses objects at infinite distances onto the retina of a person who has had this surgery, what power correcting lenses would they need to be able to read text at a 22-cm near-point distance? again, give your answer in units of diopters, to the nearest tenth of a diopter and with the correct sign.
To read text at a near-point distance of 22 cm, a person who has had cataract surgery would need a corrective lens with a refractive power of +4.5 diopters.
If a person has had the surgical removal of the variable lens of the eye, they would have lost the ability to accommodate (change the shape of the lens to focus on objects at different distances). Therefore, they would require corrective lenses to see clearly at different distances.
To determine the power of the corrective lenses required to read text at a near-point distance of 22 cm, we can use the following formula:
[tex]1/f = 1/d_o + 1/d_i[/tex]
where:
f = focal length of the corrective lens
[tex]d_o[/tex] = object distance (distance from the eye to the object)
[tex]d_i[/tex]= image distance (distance from the eye to the image formed by the corrective lens)
We want to find the power of the corrective lens, which is given by:
[tex]P=1/f[/tex]
The near-point distance [tex](d_o)[/tex] is 22 cm = 0.22 m. Since the cornea's refractive power focuses objects at infinite distances onto the retina, we can assume that the object distance is effectively at infinity, i.e., [tex]d_o = \infty[/tex]
Therefore, the formula becomes:
[tex]1/f = 1/\infty + 1/d_i[/tex]
[tex]1/f = 0 + 1/d_i[/tex]
[tex]f = d_i[/tex]
We want to find the focal length [tex](d_i)[/tex] of the corrective lens required to form an image of the text at a distance of 22 cm from the eye.
Using the formula, we get:
[tex]f = d_i = \frac{1}{(1/d_o + 1/d_i)}[/tex]
[tex]d_i = \frac{1}{(1/d_o + 1/f)}[/tex]
[tex]d_i = \frac{1}{(1/0.22 + 1/\infty)}[/tex]
[tex]d_i = 0.22 m[/tex]
Now, we can calculate the power of the corrective lens required as follows:
[tex]P = 1/f[/tex]
[tex]P = 1/0.22[/tex]
[tex]P = +4.5[/tex] diopters (to the nearest tenth of a diopter)
Therefore, the power of the corrective lenses required to read text at a near-point distance of 22 cm is +4.5 diopters (convex lenses).
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Determining the distance to stars can be challenging. The parallax method is one way of finding the distance to many stars around us. Your research team measures the parallax of two stars that have a distance of 5 degrees from each other in the night sky: The first star has a parallax of 0.11 arcsec, and the second has a parallax of 0.13 arcsec. How far apart are the two stars from each other? Express your answer in light-years
The distance between the two stars is approximately 35.65 light-years.
What is the distance between the stars?
To determine the distance between the two stars in light-years, we need to use the parallax formula:
distance = 1 / (parallax angle in arcseconds)
For the first star, the distance is:
distance = 1 / 0.11 = 9.09 parsecs
For the second star, the distance is:
distance = 1 / 0.13 = 7.69 parsecs
To determine the distance between the two stars, we can use the law of cosines:
c² = a² + b² - 2abcos(C)
where;
c is the distance between the two stars, a and b are the distances to each star, and C is the angle between the two stars.We know that the angle between the two stars is 5 degrees, or 300 arcminutes, or 18,000 arcseconds.
Converting the distances to light-years:
a = 9.09 x 3.26 = 29.59 light-years
b = 7.69 x 3.26 = 25.06 light-years
Substituting into the law of cosines:
c² = (29.59)² + (25.06)² - 2 x 29.59 x 25.06 * cos(18,000 arcseconds)
c² = 1,271.57
c = 35.65 light-years
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what is the purpose of the small propeller at the back of a helicopter that rotates in the plane perpendicular to the large propeller?
The small propeller located at the tail of a helicopter is called the tail rotor. Its purpose is to counteract the torque produced by the main rotor of the helicopter.
When the main rotor spins, it produces a torque that tries to rotate the body of the helicopter in the opposite direction. This is due to Newton's Third Law, which states that for every action, there is an equal and opposite reaction. The tail rotor, which is mounted perpendicular to the main rotor, produces a sideways thrust that counteracts the torque of the main rotor and keeps the helicopter stable.
The pilot can control the amount of thrust produced by the tail rotor through the helicopter's pedals. By pushing on the left pedal, the pilot increases the thrust of the tail rotor and causes the helicopter to rotate to the left. Similarly, pushing on the right pedal increases the thrust on the right side of the tail rotor and causes the helicopter to rotate to the right. This allows the pilot to control the direction and orientation of the helicopter in flight.
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A uniform electric field with a magnitude of 5.0 x 10^2 N/C is directed parallel to the positive x-axis toward the origin. What is the change in electrical energy of a proton (q= 1.60 x 10 ^-19 C ) as it moves from x = 5 m to x = 2 m?
The change in the electrical energy of a proton would be 2.40 x 10^-16 J.
Change in electrical energyThe change in electrical potential energy of a charged particle in a uniform electric field is given by the equation:
ΔPE = qEd
where q is the charge of the particle, E is the magnitude of the electric field, and d is the distance moved by the particle in the direction of the electric field.
In this case, the proton has a charge of q = 1.60 x 10^-19 C and moves a distance of d = 5 m - 2 m = 3 m in the direction of the electric field. The magnitude of the electric field is E = 5.0 x 10^2 N/C.
Substituting these values into the equation, we get:
ΔPE = (1.60 x 10^-19 C)(5.0 x 10^2 N/C)(3 m)
ΔPE = 2.40 x 10^-16 J
Therefore, the change in electrical energy of the proton as it moves from x = 5 m to x = 2 m is 2.40 x 10^-16 J.
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a block of mass m is attached to the end of a horizontal massless spring (spring constant k), resting on a frictionless table. the mass is then given an initial displacement x0 from equilibrium, and an initial speed v0. ignoring friction, use energy methods to finda. its maximum speed, andb. its maximum stretch from equilibrium, in terms of the given quantities.
The maximum speed of the mass is given by vmax = sqrt((k/m)xmax²) and the maximum stretch from equilibrium is given by xmax = ((2/m)(1/2)mv0²/k).
The mass is given an initial displacement x0 from equilibrium and an initial speed v0. The maximum speed and maximum stretch from equilibrium can be found using energy methods.Ignoring friction, the total energy of the system is given as E = (1/2)kx0² + (1/2)mv0², where (1/2)kx0² represents the potential energy of the spring, and (1/2)mv0² represents the kinetic energy of the mass.
At maximum speed, all potential energy is converted to kinetic energy. Therefore, we can say that (1/2)kxmax² = (1/2)mvmax², where xmax is the maximum stretch of the spring from equilibrium and vmax is the maximum speed of the mass. Therefore, we can say that (1/2) mvmax² = (1/2) kx0², where x0 is the initial displacement of the mass from equilibrium.
Therefore, the maximum speed of the mass is given by vmax = ((k/m)xmax²) and the maximum stretch from equilibrium is given by xmax = ((2/m)(1/2)mv0²/k). These equations can be used to find the maximum speed and initial displacement and velocity.
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How far will the driver be behind the vehicle in front at 30mph ? Give your answer to 2 significant figures.
The driver should be at least 39 meters behind the vehicle in front at 30mph. Rounded to 2 significant figures, this is 39 meters
How did we arrive at this value?The recommended safe following distance while driving is at least 3 seconds behind the vehicle in front of you. At 30mph (or approximately 13 meters per second), a 2-second following distance would be:
3 seconds x 13 meters/second = 39 meters
Therefore, the driver should be at least 39 meters behind the vehicle in front at 30mph. Rounded to 2 significant figures, it could then be concluded that the driver should be at least 39 meters.
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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?
Waves that vibrate upward and downward can be thought of as having a frequency of 2 Hz, and waves that move a certain distance in a certain amount of time—1 second—have a wave speed of 20 m/s.
The wave is what?A wave is an energetic disturbance in such a medium that doesn't include any net particle motion. Elastic deformation, a change in pressure, an electric or magnetic strength, an electric potential, or a change in temperature are a few examples.
What is With an illustration, define a wave?Waves are aberrations that spread or migrate away from their source. Waves can transport energy between sites, but they do not always transport mass. Ocean waves, light waves, and sound waves are typical types of waves.
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The distance of east-west highway is 1.030 x 10^6m. How much is the value of mantissa in this number?
(a) 1.030
(b) 0.030
(d) 6
(c) 10
The mantissa distance, which is referred to as the decimal portion of a logarithm, has a value of 0.030 in this number.
Is the decimal portion of a number's logarithmic value referred to as the mantissa?A common logarithm's integral portion is known as the characteristic, and its non-negative decimal portion is known as the mantissa. If log 39.2 equals 1.5933, then 1 is the characteristic and 5933 is the logarithm's mantissa.
What is the number's logarithmic mantissa?The base-10 logarithm's mantissa, which represents the digits of the provided integer but not its magnitude, is a common logarithm's fractional component. For instance, both ㏒10201.3010 and ㏒102002.3010 have a mantissa of 0.3010.
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If a door within a wall is open sound produced on one side of the wall will be able to reach a person anywhere on the other side of the wall due to
It is generally knowledge that sound waves may bend around corners or pass through doorways, enabling us to hear conversations taking place in nearby rooms.
Why do we hear noise on the opposite side of the barrier?The air on the opposing side of the barrier also vibrates as a result of the audio hitting the wall. You will be able to hear some sound through a sturdy, solid wall since it won't spread the vibrations too much.
What is the term for when sound penetrates a wall?A reflecting wave is created as a sound crosses a room and collides with a wall, reintroducing some of that waveform back into the space. The originating sound will endeavor to go through the wall and into the next space. Sound Transmission is the name for the energy that remains after this transmission.
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The inner orbit is farthest from the nucleus.
O True
O False
HELP PLEASE
Answer: False
Explanation:
a car is traveling up one side of a hill and down the other side. the top of the hill is a circular arc with a radius of 45.0 m. determine the maximum speed the car can have without losing contact with the road.
When a car travels up one side of a hill and down the other, it can lose contact with the road if it goes too fast. The maximum speed the car can have without losing contact with the road is 21.8 m/s.
The weight of the car is equal to its mass times gravity, which is equal to m * g. The normal force, on the other hand, is the force that the ground exerts on the car, and it is equal in magnitude to the weight of the car but in the opposite direction.
When the normal force is zero, the car loses contact with the road, which is why we're looking for the maximum speed that will still produce a normal force greater than zero.
Thus, the maximum speed can be found using the equation: v² / R = g; where v is the maximum speed of the car,
R is the radius of the circular arc, and
g is the acceleration due to gravity.
Substituting the given values gives: v² / 45.0 = 9.81
Solving for v, we get: v = √(45.0 * 9.81) = 21.8 m/s
Therefore, the maximum speed the car can have without losing contact with the road is 21.8 m/s.
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A truck accelerates at a 1.25 m/s^2. How far does it go as it accelerates uniformly from 63 km/h to 84 km/h
The distance covered by the truck would be 188.4 meters.
Distance-time problemInitial velocity, u = 63 km/h = (63 x 1000 m) / (3600 s) = 17.5 m/s
Final velocity, v = 84 km/h = (84 x 1000 m) / (3600 s) = 23.3 m/s
We can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, s is the distance traveled.
Substituting the values we get:
(23.3)^2 = (17.5)^2 + 2(1.25)s
Simplifying and solving for s, we get:
s = [(23.3)^2 - (17.5)^2] / (2 x 1.25) = 188.4 meters (rounded to one decimal place)
Therefore, the truck travels a distance of 188.4 meters as it accelerates uniformly from 63 km/h to 84 km/h with an acceleration of 1.25 m/s^2.
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Galaxies are some of the most beautiful objects in the universe and are observable in many differ- ent shapes, colours and sizes. Astronomers have classified galaxies into different groups: spiral (SA), intermediate spiral (SAB), barred spiral (SB), lenticular (SO), elliptical (E), and irregular (Irr). Which galaxy classes are illustrated by the shapes below (A1-A4)? Find the correct class (B1-B4) and name (C1-C4) of each galaxy shown in the images: NGC 2337, NGC 300, NGC 1365, Messier 110
We can see here that the galaxy classes illustrated by the shapes are:
A1: Irregular (Irr)
A2: Lenticular (SO)
A3: Spiral (SA)
A4: Intermediate spiral (SAB)
B1: Lenticular (SO)
B2: Irregular (Irr)
B3: Intermediate spiral (SAB)
B4: Elliptical (E)
What is a galaxy?A galaxy is a vast system of stars, gas, dust, and other matter that is held together by gravity. It is one of the fundamental building blocks of the universe.
Galaxies come in a variety of shapes and sizes, ranging from small dwarf galaxies to massive elliptical galaxies. Our own Milky Way galaxy contains billions of stars and is just one of countless galaxies in the observable universe.
Galaxies play an important role in the evolution of the universe, as they are the sites of star formation, supernovae explosions, and the merging of smaller galaxies into larger ones.
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he frequency of a wave does not change as it passes from one medium to another. What will most likely happen if a light wave moves from the air into a solid? The wavelength of the light wave will increase. The speed of the light wave will decrease. The wavelength of the light wave will remain the same. The speed of the light wave will remain the same.
Answer:
Generally. the index of refraction of the solid is greater than 1
V(air) / V(solid) = N (index of refraction)
Since V (solid) = V (air) / N and V(soiid) = λ(solid) * f
V(solid) / V(air) = [V (air) / N] / (V air) = 1 / N
The velocity is less in the solid and the wavelength is also less
(PLS HELP ME NOW!!!!!)
Answer:
D
Explanation:
suppose that an ion has an absorption line at a rest wavelength of 1000.0 nm. this line is shifted to 1000.1 nm in the spectrum of a star. how fast is the star moving?
The speed of the star is approximately 29979.2458 m/s.
The speed of star can be solved using the Doppler shift formula:
Δλ/λ = v/c
where Δλ is the shift in wavelength,
λ is the rest wavelength,
v is the velocity of the object (in this case, the star), and
c is the speed of light.
Substituting the given values, we get:
(1000.1 nm - 1000.0 nm) / 1000.0 nm = v/c
Simplifying, we get:
0.1 nm / 1000.0 nm = v/c
v = (0.1 nm / 1000.0 nm) * c
v = 299792458 m/s * (0.1 nm / 1000.0 nm)
v = 29979.2458 m/s
Therefore, the speed of the star is approximately 29979.2458 m/s.
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define moment with complete in depth explanation
Answer: In its most basic form, a moment is the product of the distance to a point, raised to a power, and a physical quantity (such as force or electrical charge) at that point: is the physical quantity such as a force applied at a point, or a point charge, or a point mass, etc.
The Principle of Moment says that when a system is in equilibrium the sum of its clockwise moments will be equal to the sum of its anticlockwise moments. Some examples where moments i.e. turning effects are applicable will involve levers, like seesaws, opening and closing doors, nutcrackers, can openers, and crowbars.
Explanation:
At what height above the earth is an object which has U•g of 1200J and a mass of 20kg?
The potential energy of an object at a height h above the earth’s surface is given by the formula U = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height above the earth’s surface.
Calculation-Given that the object has a potential energy of 1200 J and a mass of 20 kg, we can rearrange the formula to find the height h:
h = U / (mg)
Substituting the given values, we get:
h = 1200 J / (20 kg x 9.81 m/s^2) = 6.12 m
Therefore, the object is at a height of 6.12 meters above the earth’s surface.
If something had a mass of 30 kg on the moon, what would its mass be on Earth?Given that mass is independent of everything, the first part of the question's response is straightforward: As mass does not change as you move from one location in the universe to another, the boy on the moon will still have the same mass as the boy on Earth.
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which constellation appears to have the shape of a big ladle or a question mark?
Answer:
Ursa Major
Explanation:
The constellation that appears to have the shape of a big ladle or a question mark is Ursa Major (also known as the Great Bear).
In a resistive circuit if the current is increased to two times, the percentage change in the amount of heat dissipated in the circuit would be
Answer:
P = I V = I*2 R power used in resistive circuit
Doubling the current will cause a 4X increase in power consumed
How do I write this in scientific Notation? Help quick
Answer:
2,500,000,000,000 = 2,5 x 10^12
0.00023 = 2.3 x 10^-4
101 = 1.01 x 10^2
0.0000000032 = 3.2 x 10^9
Explanation:
Hope it helps :) !!!!!!!!!!!!
Need help this is a grade
Answer: C
Explanation: When you have a tomato for example, it absorbs every color but red and therefore, appears red.
how does the president fufill the role of a economic leader
The president plays the part of an economic leader by planning the government budget.
What does an economic leader do?examining data on monetary movements and trends. examining federal policies and initiatives to make sure they continue to support strong economic principles. advocating for economic measures that will help American workers.
What function is the president performing?The President of the United States, who also serves as head of state and commander-in-chief of the armed forces, is given control over the Executive Branch. The president's responsibility as chief legislator is to have an impact on the creation of laws. Through his State of the Union Address, the president contributes to setting the country's policy agenda.
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A baseball is batted. It's a long fly ball. 5. 0 seconds later the ball reaches the outfield 100 meters away and returns to the height from which it left the bat. A. What maximum height did the baseball reach?
The maximum height reached by the baseball is 24.5 meters.
We can use the kinematic equation for the vertical motion:
h = vi*t + (1/2)at^2
where h is the maximum height reached by the baseball,
vi is the initial vertical velocity,
a is the acceleration due to gravity (-9.8 m/s^2),
and t is the time taken for the ball to reach its maximum height.
At the maximum height, the vertical velocity of the ball is zero. Therefore, we can write:
h = (1/2)at^2
We know that the total time for the baseball to travel to the outfield and return is 5 seconds, so the time taken to reach the maximum height is half of that or 2.5 seconds.
After substituting the values we know, we get:
h = (1/2)(-9.8 m/s^2)(2.5 s)^2
h = 24.5 meters
Therefore, the maximum height the baseball reached is approximately 24.5 meters.
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