a hand pump is used to inflate a ball, the pump piston does 24 J of work on the air to compress it. the air in the pump loses 7 J of heat to the surroundings. what is the change in thermal energy of the air??
A Thermal energy of the air is 17 J of heat to the surroundings.
Thus, Thermal energy is produced by materials whose molecules and atoms vibrate more quickly as a result of a rise in temperature.
The atoms and molecules that make up matter are always in motion. The increase in temperature caused by heating a substance causes these particles to accelerate and collide.
The energy that arises from a heated substance is referred to as thermal energy. The more the substance's thermal energy and the more its particles travel at higher temperatures.
Thus, A Thermal energy of the air is 17 J of heat to the surroundings.
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A car travels a distance of 120 km in 4 hours. What is its average speed in kilometers per hour?
Answer:
60 kilometers per hr
Explanation:
5. Ametal block increases in temprature from 15°c to 60°c when supplied with 13500 joule of heat energy. a. Calculate the heat capacity of the metal. b. Calculate the specific heat capacity of the metal if this sample has a mass of 0.75kg.
The specific heat capacity of the substance is defined as the amount of heat energy supplied to the substance to increase the temperature of the substance by 1°C. The SI unit of specific heat is J/Kg.°C.
The heat energy, q = mC×ΔT, where m is the mass of the substance. C is the specific heat capacity of the material. ΔT is the change in temperature.
From the given,
a) heat supplied, q = 13500J
Initial temperature,T₁ = 15°C
Final temperature, T₂ = 60°C
Specific heat capacity, C=?
q = mCΔT
13500 = C(T₂ - T₁)
13500/(60-15) = C
13500/45 = C
C = 300 J/Kg.°C
Thus, the specific heat capacity is 300J/Kg.°C.
b) mass of the substance = 0.74kg
q = mCΔT
13500 = 0.75×C×(60-15)
13500/(0.75×45) = C
C = 400 J/Kg.°C
Thus, the specific heat capacity with heat energy of 13500 J is 400J/kg.°C.
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Part 3: Energy Conversions 7. Record your data in the chart and include at least 5 potential-kinetic energy conversions shown in your device's construction. Example Item Description of potential-kinetic energy conversion Example Book The book had gravitational potential energy when it was on the table. Then as the book fell off the table, it was in motion and had kinetic energy. 1 2 3 4 5
Here are five potential-kinetic energy conversions that could be shown in the construction of a device: Pendulum, Roller Coaster, Wind-up Toy, Elastic Slingshot, Windmill.
Pendulum: A pendulum consists of a weight attached to a string or rod, suspended from a fixed point. When the weight is lifted to a certain height, it possesses gravitational potential energy.
As the weight is released, it swings back and forth, converting the potential energy into kinetic energy. At the highest point of each swing, the weight briefly comes to a stop and has maximum potential energy, which is then converted back to kinetic energy as it swings downward.
Roller Coaster: In a roller coaster, potential-kinetic energy conversions occur throughout the ride. When the coaster is pulled up to the top of the first hill, it gains gravitational potential energy.
As the coaster descends, the potential energy is converted into kinetic energy, resulting in a thrilling and high-speed ride. Subsequent hills and loops continue to convert potential energy into kinetic energy and vice versa as the coaster moves along the track.
Wind-up Toy: Wind-up toys typically have a spring mechanism inside. When the toy is wound up, potential energy is stored in the wound-up spring. As the spring unwinds, it transfers its potential energy into kinetic energy, causing the toy to move or perform actions. The kinetic energy gradually decreases as the spring fully unwinds.
Elastic Slingshot: With an elastic slingshot, potential-kinetic energy conversions are evident when the slingshot is stretched. As the user pulls back on the elastic band, potential energy is stored.
Windmill: Windmills harness the kinetic energy of the wind and convert it into other forms of energy. As the wind blows, it imparts kinetic energy to the blades of the windmill. The rotating blades then transfer this kinetic energy into mechanical energy, which can be used for various purposes such as grinding grains or generating electricity.
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A spring stretches 0.294-m when a 0.360-kg mass is gently suspended from it as in Fig. 11–3b. The spring is then set up horizontally with the 0.431-kg mass resting on a frictionless table as in Fig. 11–5. The mass is pulled so that the spring is stretched 0.250-m from the equilibrium point, and released from rest.
Determine:
(a) the spring stiffness constant k.
The spring constant k based on the information is 12.0 N/m.
How to calculate the valueFrom the information, a spring stretches 0.294-m when a 0.360-kg mass is gently suspended from it as in Fig. 11–3b. The spring is then set up horizontally with the 0.431-kg mass resting on a frictionless table.
The spring constant k is the force required to stretch or compress the spring by a unit distance. In this case, the spring is stretched by 0.294 m when a 0.360 kg mass is suspended from it.
This means that the force exerted by the spring is equal to the weight of the mass, which is 0.360 kg x 9.8 m/s^2 = 3.53 N.
Therefore, the spring constant k is:
= 3.53 N/0.294 m
= 12.0 N/m.
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A wagon, Initially traveling at a constant 3.6 m/s, starts going down a hill that creates an acceleration of
1.6 m/s2. What is the wagon's velocity 4.8 s after it starts accelerating down the hill?
QUESTION ❗️❗️❗️❗️❗️❗️
How could you measure the flow rate of various liquid
a.Place each one on a scale and measure its weight versus its volumes
b.Place them in a beaker and se which one floats to the top
c.Pour them down an incline and time how long it takes each one to reach the bottom
d.Burn each sample to create a deposit than can be analyzed
We can measure the flow rate of various liquid by pouring the liquid down an incline and time how long it takes each one to reach the bottom.
option C.
What is the flow rate of a liquid?
The flow rate of a liquid is how much fluid passes through an area in a particular time.
Flow rate can be articulated in either in terms of velocity and cross-sectional area, or time and volume. As liquids are incompressible, the rate of flow into an area must be equivalent to the rate of flow out of an area.
Generally, the best equipment to measure the flow rate of a liquid is flow meters. In the absence of flow meters, we can other methods such as the one given in the options.
We can pour the various liquid down an incline and time how long it takes each one to reach the bottom.
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Gas and plasma are phases of matter, yet has runs a car and plasma is part of your blood. Compare and contrast these terms and offer an explanation for the use of similar names.
Gas and plasma are indeed phases of matter, but they have distinct characteristics and applications.
Gas vs plasmaGas is a state of matter where particles have high energy and are free to move around, filling the space they occupy. Gaseous substances, like air, are typically composed of neutral atoms or molecules.
Plasma, on the other hand, is an ionized gas consisting of positively and negatively charged particles. It is formed when gas is heated to extremely high temperatures or exposed to a strong electric field. Plasma is found in stars, lightning, and fluorescent lights, and it also plays a crucial role in technologies like plasma TVs and fusion reactors.
The similarity in names can be attributed to the ionized nature of plasma. In plasma, particles become charged, similar to the positive and negative ions found in the human body's blood plasma. Both terms derive from the Greek word "plasma," meaning "something molded or formed."
This connection may have influenced the choice of naming the ionized state of matter and the component of blood plasma using similar terminology.
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Organisms belonging to the same species can have______traits
Answer:
similar or identical traits.
Fill in the blanks: the standard international (SI) unit for mass is the , The standard international (SI) unit for force is the .
The standard international (SI) unit for mass is the kilogram (kg). It is a fundamental unit of measurement used to quantify the amount of matter in an object. The standard international (SI) unit for force is the Newton (N)
The mass of the platinum-iridium cylinder known as the International Prototype of the Kilogramme, which is held at the International Bureau of Weights and Measures in France, is what is used to define the kilogramme.
The newton (N), on the other hand, serves as the standard international (SI) unit for force. The force needed to accelerate a one kilogramme mass by one square metre per second is measured in newtons. It is a derived unit that is frequently used to measure a variety of forces, including electromagnetic, mechanical, and gravitational forces.
Sir Isaac Newton, a distinguished scientist who made substantial advances to our knowledge of forces and motion, is honoured by having his name attached to the newton.
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What is the value of the universal gas constant (R) in Sl units?
7. Light of a frequency 6.8x10 ¹4Hz falls on a pair of slits that are 2.00x104 cm apart. The central bright spot is 50 cm from the screen. How far is the 1st order bright spot from the central bright spot?
The 1st order bright spot is located approximately 1.1025x10^-9 m away from the central bright spot.
To determine the distance of the 1st order bright spot from the central bright spot in a double-slit interference setup, we can use the formula for the position of bright fringes:
y = (m * λ * L) / d
where:
y is the distance from the central bright spot to the m-th order bright spot,
m is the order of the bright spot (in this case, m = 1 for the 1st order),
λ is the wavelength of light,
L is the distance from the slits to the screen (in this case, L = 50 cm = 0.5 m), and
d is the distance between the slits (d = 2.00x10^4 cm = 200 m).
Given that the frequency of light is 6.8x10^14 Hz, we can use the relationship between frequency and wavelength to calculate the wavelength (λ) using the formula:
c = λ * f
where c is the speed of light (approximately 3x10^8 m/s).
Rearranging the formula, we have:
λ = c / f
λ = (3x10^8 m/s) / (6.8x10^14 Hz)
Calculating the value of λ, we get:
λ = 4.41x10^-7 m
Now we can substitute the values into the formula for the position of the bright spot:
y = (1 * 4.41x10^-7 m * 0.5 m) / 200 m
Simplifying the equation, we have:
y = 1.1025x10^-9 m
In summary, the distance of the 1st order bright spot from the central bright spot in this double-slit interference setup is approximately 1.1025x10^-9 m.
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A war-wolf or trebuchet is a device used during the Middle Ages to throw rocks at castles and now sometimes used to fling large vegetables and pianos as a sport. A simple trebuchet is shown in the figure below. Model it as a stiff rod of negligible mass, d = 2.60 m long, joining particles of mass m1 = 0.115 kg and m2 = 68.5 kg at its ends. It can turn on a frictionless, horizontal axle perpendicular to the rod and 13.0 cm from the large-mass particle. The operator releases the trebuchet from rest in a horizontal orientation.
Find the maximum speed that the small-mass object attains when it leaves the trebuchet horizontally.
The maximum speed that the small-mass object attains when it leaves the trebuchet horizontally is approximately 28.3 m/s.
To find the maximum speed that the small-mass object attains when it leaves the trebuchet horizontally, we can apply the principle of conservation of mechanical energy.
Initially, the trebuchet is at rest, so its total mechanical energy is zero. As the small-mass object leaves the trebuchet horizontally, it gains kinetic energy. At this point, all of the potential energy of the system is converted into kinetic energy.
The potential energy of the system can be calculated as the sum of the gravitational potential energies of the two masses:
PE = m1 * g * h1 + m2 * g * h2
Since the trebuchet is released from rest in a horizontal orientation, the initial height h1 is zero. The height h2 can be calculated as the perpendicular distance between the pivot point and the center of mass of the larger mass m2:
h2 = 13.0 cm = 0.13 m
Therefore, the potential energy simplifies to:
PE = m2 * g * h2
The kinetic energy of the small-mass object can be calculated as:
KE = (1/2) * m1 * v^2
where v is the maximum speed of the small-mass object.
Since the total mechanical energy is conserved, we have:
PE = KE
m2 * g * h2 = (1/2) * m1 * v^2
Plugging in the given values, such as g = 9.8 m/s^2, m1 = 0.115 kg, m2 = 68.5 kg, and h2 = 0.13 m, we can solve for v:
(68.5 kg * 9.8 m/s^2 * 0.13 m) = (1/2) * 0.115 kg * v^2
Solving for v, we find:
[tex]v^2 = (68.5 kg * 9.8 m/s^2 * 0.13 m) / (0.115 kg)[/tex]
[tex]v^2 = 800[/tex]
v ≈ 28.3 m/s
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if 1inch is 2.5cm then 1.0in^2 of surface area is
1.0 square inch of surface area is equal to 6.4516 square centimeters.
An inch is a unit of length commonly used in the United States and some other countries that have not adopted the metric system. It is denoted by the symbol "in" or double prime ("). One inch is equal to exactly 2.54 centimeters. It is subdivided into smaller units such as fractions (e.g., 1/2 inch, 1/4 inch) or decimals (e.g., 0.25 inches, 0.5 inches) for more precise measurements. The inch is primarily used for measuring shorter distances, such as the length of objects, fabric, or paper.
To convert square inches to square centimeters, we need to know the conversion factor for converting inches to centimeters.
Since 1 inch is equal to 2.54 centimeters (not 2.5 centimeters as mentioned in your statement), we can use this conversion factor to calculate the surface area in square centimeters.
To convert 1 square inch to square centimeters, we square the conversion factor:
1 inch^2 = (2.54 cm)^2 = 6.4516 square centimeters (approximately).
Therefore, 1.0 square inch of surface area is approximately equal to 6.4516 square centimeters.
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Alice throws a ball on the ground,and it bounces back to her hand, there is no net change in the kinetic energy. What is the type of collision
Answer: the type of collision is elastic collision because both momentum and kinetic energy are conserved.
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A spring stretches 0.145-m when a 0.233-kg mass is gently suspended from it as in Fig. 11–3b. The spring is then set up horizontally with the 0.456-kg mass resting on a frictionless table as in Fig. 11–5. The mass is pulled so that the spring is stretched 0.192-m from the equilibrium point, and released from rest.
Determine:
(c) the magnitude of the maximum velocity vmax.
The maximum velocity of oscillation of the spring is 1.57m/s.
Displacement of the spring, x = 0.145 m
Mass of the object suspended from the spring, m = 0.233 kg
The spring constant of the spring is given by,
k = mg/x
k = 0.233 × 9.8/0.145
k = 15.74 N/m²
The angular frequency of the oscillation of the spring is given by,
ω = √(k/m)
ω = √(15.74/0.233)
ω = 8.21 rad/s
Amplitude of the horizontal oscillation of the spring, A = 0.192 m
Therefore, the maximum velocity of oscillation of the spring is given by,
v(max) = Aω
v(max) = 0.192 x 8.21
v(max) = 1.57 m/s
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What can be said about the speed ofa particle if the net work done on it is zero?
If the net work done on a particle is zero, the particle will move with a constant speed.
The principle of work and kinetic energy, often known as the work-energy theorem, states that the change in kinetic energy of a particle is equal to the sum of the entire work done by all of the forces acting on it.
So,
W = ΔKE
Thus, we can say that the kinetic energy of the particle will not change if the net work done on it is equal to zero.
As a result, the state of motion of the particle will not change, and thus the speed of the particle will also remain constant.
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A test rocket is launched by accelerating it along a 200.0-m incline at 1.60 m/s2
starting from rest at point A (the figure (Figure 1).) The incline rises at 35.0 ∘
above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).
Question: Find the greatest horizontal range of the rocket beyond point A.
Figure 1 attached.
The greatest horizontal range of the rocket beyond point A is approximately 17.89 meters.
To find the greatest horizontal range of the rocket beyond point A, we need to analyze the projectile motion of the rocket after it leaves the incline.
We can break down the rocket's motion into horizontal and vertical components. The horizontal component remains constant, while the vertical component is influenced by gravity. Since the rocket is subject only to gravity after leaving the incline, the horizontal velocity remains constant throughout the motion.
First, let's calculate the initial velocity of the rocket in the horizontal direction. We can use the acceleration and the distance traveled along the incline to find the time taken to reach the end of the incline.
Using the equation of motion: distance = initial velocity × time + (1/2) × acceleration × time^2, we can substitute the given values:
200.0 m = 0 × t + (1/2) × 1.60 m/s^2 × t^2.
Simplifying the equation, we get:
[tex]1.60 t^2 = 200.0,\\t^2 = 200.0 / 1.60,\\t^2 = 125,[/tex]
t = √125,
t ≈ 11.18 s.
Now that we have the time taken to reach the end of the incline, we can calculate the horizontal distance traveled by the rocket using the formula: distance = velocity × time.
Since the horizontal velocity remains constant at 1.60 m/s, the horizontal distance is:
distance = 1.60 m/s × 11.18 s,
distance ≈ 17.89 m.
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A virtual satellite orbits the earth at an altitude h = 1600km with an altitude v = 7.1km / s. The amperage of the centrifugal force is F ’= 3151N. Calculate the satellite mass. It is known that the radius of the earth R = 6400 / km.
An object is placed at several different distances to the left of the lenses and mirrors (focal length = f). For each case, draw the 3 principal rays to locate the image and then (S.A.L.T.) [Size, Attitude, Location, Type] the image.
To determine the image characteristics using the 3 principal rays and SALTS (Size, Attitude, Location, Type), we'll consider both lenses and mirrors separately. Here's how you can analyze each case:
Lenses:
Place an object at different distances to the left of a lens with a focal length (f).
a) Object placed beyond 2f:
In this case, the object is placed far beyond twice the focal length of the lens.
Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.
Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.
Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.
The image will be formed on the opposite side of the lens, between the focal point and twice the focal length.
SALTS:
Size: The image will be smaller than the object.
Attitude: The image will be inverted.
Location: The image will be located between the focal point and twice the focal length.
Type: The image will be real.
b) Object placed at 2f:
In this case, the object is placed at twice the focal length of the lens.
Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.
Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.
Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.
The image will be formed on the opposite side of the lens at twice the focal length.
SALTS:
Size: The image will be the same size as the object.
Attitude: The image will be inverted.
Location: The image will be located at twice the focal length.
Type: The image will be real.
c) Object placed between f and 2f:
In this case, the object is placed between the focal point and twice the focal length of the lens.
In this case, the object is placed far beyond twice the focal length of the mirror.
Principal ray 1: A ray parallel to the principal axis will reflect through the focal point on the same side.
Principal ray 2: A ray passing through the focal point on the object side will reflect parallel to the principal axis.
Principal ray 3: A ray passing through the center of curvature will reflect back along the same path.
The image will be formed on the opposite side of the mirror, between the focal point and twice the focal length.
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In order to measure the potential difference across one of the bulbs in the circuit, the voltmeter must be connected in parallel with it. So, option D.
When two points in a circuit have different electric potentials, a voltmeter is a tool or instrument that measures their potential difference.
We are aware that a voltmeter is a tool that measures the same potential drop in all configurations that are in parallel.
The potential difference between two points in a circuit is thus always measured by connecting a voltmeter in parallel across the conductor's ends.
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Why does sound travel more quickly through a solid than through a liquid or a gas?
Sound travels more quickly through a solid than through a liquid or a gas because the particles in a solid are closer together than the particles in a liquid or a gas
What more should you know about the speed of sound?The speed of sound in a material is said to be determined by the density of the material and the elasticity of the material.
The density of a material is a measure of how much mass is contained in a given volume.
The elasticity of a material is a measure of how much the material can be stretched or compressed without breaking.
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what is the voltage supplied to a wire that has a resistance of 1200 Q and a current of 0.10 amps
The voltage supplied to the wire is 120 volts.
To calculate the voltage supplied to a wire, we can use Ohm's Law, which states that voltage (V) is equal to the product of current (I) and resistance (R). Mathematically, this relationship is expressed as V = I * R.
In this case, the wire has a resistance of 1200 Ω (ohms) and a current of 0.10 amps. We can substitute these values into the formula to find the voltage:
V = I * R
V = 0.10 A * 1200 Ω
V = 120 A * Ω
Therefore, the voltage supplied to the wire is 120 volts.
It's important to note that Ohm's Law holds true for resistors and other components in a circuit that obey Ohm's Law. In real-world scenarios, there may be other factors to consider, such as the presence of non-ohmic devices or components with varying resistance.
Additionally, in an AC (alternating current) circuit, the relationship between voltage, current, and resistance may involve complex quantities and phase differences. However, for a simple DC (direct current) circuit with a linear resistor, Ohm's Law provides an accurate relationship between voltage, current, and resistance.
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3. Fulcrum left
Explanation:
t is the relationship between the ping pong ball's release height and its bounce height, in this
timent? In your answer make sure to:
Restate the question and then Answer it by identifying a relationship shown in the data.
Cite three pieces of numerical evidence to fully show the relationship in the data.
Explain how each piece of evidence supports your claim. Be as specific as possible
Describe how the relationship in the data connects to the following concept:
"Potential energy can be converted into kinetic energy. Kinetic energy can also be
converted back into potential energy."
Force varies directly with pressure. A force of 22,000 newtons acts on an object, causing 110 N/m² of pressure. The
force decreases to 18,000 newtons. What is the new amount of pressure caused by the force?
○ 45 N/m²
90 N/m²
O 75 N/m²
60 N/m²
The final amount of pressure caused by the force is 90 N/m².
Initial amount of force, F₁ = 22 x 10³ N
Initial amount of pressure produced, P₁ = 110 N/m²
Final amount of force exerted, F₂ = 18 x 10³ N
Pressure is defined as the amount of force acting on an object per unit area of the object.
So, we can say that the force and pressure are directly proportional.
F ∝ P
So, F₁/P₁ = F₂/P₂
Therefore, the final amount of pressure caused by the force is,
P₂ = F₂P₁/F₁
P₂ = 18 x 10³x 110/22 x 10³
P₂ = 18/0.2
P₂ = 90 N/m²
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Distinguish between mass and weight. Include the definitions, units of measurement, how they are measured, and what they depend on
Mass is the measure of the amount of matter in an object and remains constant regardless of location, measured in units like kilograms or grams, while weight represents the gravitational force exerted on an object, varying with the strength of the gravitational field and measured in units like newtons or pounds.
Mass and weight are distinct concepts in physics, differing in their definitions, units of measurement, how they are measured, and what they depend on. Here's a breakdown of their differences:
Mass:
Definition: Mass refers to the amount of matter in an object. It is an intrinsic property and remains constant regardless of the object's location or gravitational environment.
Units of measurement: The standard unit of mass in the International System of Units (SI) is the kilogram (kg). Other common units include grams (g) and metric tonnes (t).
Measurement: Mass can be measured using various techniques, including balances and scales. These instruments compare the unknown mass to known masses and determine the equilibrium or balance point.
Dependence: Mass is independent of gravity and remains the same regardless of the gravitational force acting on the object.
Weight:
Definition: Weight is the force exerted on an object due to the gravitational pull of a celestial body (usually Earth). It represents the measure of the object's gravitational attraction towards that body.
Units of measurement: The standard unit of weight in the SI system is the newton (N). However, weight is commonly expressed in units of force, such as pounds (lb) or kiloponds (kp).
Measurement: Weight is typically measured using a spring scale or a device known as a weighing scale. These instruments rely on the deformation or stretching of a spring to determine the gravitational force acting on an object.
Dependence: Weight depends on the strength of the gravitational field where the object is located. The weight of an object will vary depending on the celestial body it is interacting with, as gravitational forces differ.
Therefore, mass refers to the amount of matter in an object and is measured in units like kilograms or grams. It remains constant regardless of location and is determined using balances or scales. Weight, on the other hand, represents the gravitational force exerted on an object and is measured in units like newtons or pounds. It varies based on the strength of the gravitational field and is measured using spring scales or weighing instruments.
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Polaris has a fixed position in the sky as it lies directly over the celestial North Pole. Therefore, it is useful to be able to identify Polaris, to help you get your bearings. First identify the two stars in the Great Dipper that point to Polaris. Then identify the position of Polaris in this image of the night sky.
Answer:
Explanation:
Unfortunately, as an AI text-based model, I cannot directly view or interpret images. However, I can still provide you with the information you need to identify Polaris and its position in the night sky.
To locate Polaris using the stars of the Big Dipper (also known as the Great Dipper or Ursa Major), you can follow these steps:
1. Locate the Big Dipper: The Big Dipper is a prominent asterism, or a recognizable pattern of stars, within the constellation Ursa Major (the Great Bear). It is visible in the northern hemisphere during most of the year.
2. Identify the pointer stars: The two stars on the outer edge of the Big Dipper's bowl, farthest from the handle, are called the pointer stars. These stars are named Dubhe and Merak.
3. Extend the line between the pointer stars: Mentally extend an imaginary line that passes through Dubhe and Merak, extending it for approximately five times the distance between the pointer stars.
4. Locate Polaris: The extended line will lead you to Polaris, also known as the North Star. Polaris is relatively bright and appears as the last star in the handle of the Little Dipper (Ursa Minor constellation). It lies almost directly above the North Pole of the Earth and remains nearly fixed in the sky while other stars appear to rotate around it as the Earth rotates.
By following these steps, you should be able to identify Polaris and its position in the night sky, even without an image.
Polaris is positioned directly above the celestial North Pole in the sky, making it a useful navigation tool. The easiest way to locate it is by identifying the Great Dipper constellation and using its two pointer stars, Dubhe and Merak, to lead to the North Star.
Explanation:The star Polaris, also known as the North Star, is beneficial for navigation due to its fixed position in the sky above the celestial North Pole. The best way to locate it is by first finding the Great Dipper constellation. Two stars in the bowl of this Dipper, named Dubhe and Merak, form a line that leads directly to Polaris.
In the given image, without the benefit of visual reference, it is difficult to identify the specific position of Polaris. However, remember that in actual practice, you would find the two pointer stars of the Great Dipper and follow a line from these stars to locate Polaris.
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If the strong pull illustration below , a gradual pull of the lower string results in the op le strong breaking. Does this occur because of the balls weight or it’s mass?
Answer:
In the string pull illustration you described, the gradual pull of the lower string causes the top string to break. This occurs because of the tension that is created in the top string as a result of the pull. The weight or mass of the ball is not the primary cause of the breakage in this case.
Initial State: A sled and child are in motion halfway down a hill.
Final State: The sled and child are at rest at the bottom of the hill.
Notes: The system includes the sled, the child, and the Earth. The sled glides
freely until finally stopped by a rough patch of snow.
We have a sled and a child in motion halfway down a hill, and the final state is that both the sled and the child are at rest at the bottom of the hill. The system includes the sled, the child, and the Earth. The sled glides freely until it is stopped by a rough patch of snow.
The sled and child are in motion halfway down the hill. At this point, both the sled and the child possess kinetic energy due to their motion. The sled's motion is initiated by the force applied by the child or by the gravitational force acting on it.
As the sled and child continue down the hill, they experience a gravitational force pulling them towards the Earth. The sled glides freely, meaning there are no external forces acting on it apart from gravity and any frictional forces present on the hill. The child's weight is also acting on the sled, contributing to the force pushing it downhill.
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