A satellite in space took a picture of a double eclipse when both Earth and the Moon moved between the satellite and the Sun at the same time. Some students claim that they could see a double eclipse from Earth if a lunar and a solar eclipse happened at the same time. They wonder if they could ever see that type of double eclipse from their town.
Answers
If a double eclipse were to occur with a lunar eclipse and a solar eclipse happening simultaneously, it would be possible to observe a double eclipse from a specific town on Earth. This would be an extraordinary and rare event, as it would require precise alignment and timing of both the Earth, Moon, and Sun. However, it is important to note that such a simultaneous occurrence of a lunar and solar eclipse is highly improbable in reality.
Related Questions
The solar wind is highly ionized plasma. Throughout its trajectory, it presents variations in temperature, density, pressure and speed, mainly due to the presence of eruptive phenomena such as Coronal Mass Ejections (CME). Assuming that on its journey between the Sun and the Earth (150X10^6 km), the temperature (1.3X10^5 K) and speed (450 km/s) remain constant. What kind of fluid is it? When an EMC is generated, the temperature (10^7 K) and the speed of the Solar Wind (10^3 km/s) increase and assuming that this increase remains constant along the same trajectory, what type of fluid will be the Solar Wind thanks to this disturbance?
laminar fluid <2100
transition 2,100-4,000
turbulent fluid >4000
Re=(Velocity)(Length)/ VISCOSITY
calculate the viscosity depending on the temperature
and explain in words the solution of the problem.
Answers
The solar wind, with a constant temperature of 1.3x10^5 K and a speed of 450 km/s, is a highly ionized plasma.
Due to the increase in temperature (10^7 K) and speed (10^3 km/s) during a Coronal Mass Ejection (CME), the solar wind becomes a turbulent fluid.The solar wind, which is a stream of charged particles (plasma) emitted by the Sun, exhibits characteristics of a highly ionized plasma. As it travels from the Sun to the Earth, the solar wind encounters variations in temperature, density, pressure, and speed. In this case, with a constant temperature of 1.3x10^5 K and a speed of 450 km/s, the solar wind can be classified as a highly ionized plasma.
During a Coronal Mass Ejection (CME), eruptive phenomena on the Sun's surface cause a sudden release of a large amount of plasma and magnetic fields into space. This disturbance leads to an increase in the temperature of the solar wind to around 10^7 K and an increase in its speed to approximately 10^3 km/s. This enhanced energy and velocity result in a significant disruption of the solar wind's flow, making it a turbulent fluid.
Turbulent fluids are characterized by chaotic and irregular motion, with strong fluctuations and mixing. The increased temperature and speed during a CME generate turbulent behavior within the solar wind, leading to the disruption and mixing of plasma particles along its trajectory.
To determine the type of fluid flow, the Reynolds number (Re) is often used. It relates the flow's characteristics, such as velocity and length, to the fluid's viscosity. In this case, since the solar wind is a plasma and has negligible viscosity, the Reynolds number is extremely high, well above the turbulent threshold of 4000. Therefore, the solar wind with increased temperature and speed due to a CME can be classified as a turbulent fluid.
In summary, the solar wind, with a constant temperature of 1.3x10^5 K and a speed of 450 km/s, is a highly ionized plasma. However, during a Coronal Mass Ejection (CME), the temperature and speed of the solar wind increase significantly, leading to turbulent fluid behavior.
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Describe the relationship between frequency and wavelength.
Answers
Answer:
frequency is equal to the wavelength multiply by the speed of light
Explanation:
wavelength multiply by the speed of light Will give you the frequency
with which of these statements would galileo have agreed? select one: one cannot be both a christian and a scientist. science is not fit for the minds of the common people. the geocentric model of the universe is superior to the heliocentric model. science has no practical applications. science is the enemy of religion.
Answers
The correct option is B, Science has no practical applications these statements would Galileo have agreed.
Galileo Galilei (1564-1642) was an Italian astronomer, physicist, and mathematician who made significant contributions to the development of modern science. He is known for his support of the Copernican heliocentric theory, which posits that the sun is at the center of the solar system and the planets orbit around it. This challenged the traditional geocentric model that placed Earth at the center of the universe.
Galileo also made important discoveries using the telescope, such as the moons of Jupiter and the phases of Venus, which provided further evidence for the heliocentric theory. He also made significant contributions to the fields of physics and mathematics, including the laws of motion and the principle of inertia. Galileo's work was controversial and he was eventually tried by the Inquisition for his support of the heliocentric theory.
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Complete Question:-
With which of these statements would Galileo have agreed?
a. Science is not fit for the minds of the common people.
b. Science has no practical applications.
c. Science is the enemy of religion.
d. One cannot be both a Christian and a scientist.
e. The geocentric model of the universe is superior to the heliocentric model.
The electric force between two charged objects is 64 N. What will be the magnitude of the electric force if the distance between these two
charges is cut in half?
Answers
On the screenshot linked below, the answer to part one is .237141. How would I solve part two?
Answers
(a) The fraction of neutron's kinetic energy transferred to the atomic nucleus is 0.068.
(b) The final kinetic energy of the neutron is 2.48 x 10⁻¹³ J.
What is the final velocity of the atom?The final velocity of the atom is calculated by applying the principle of conservation of linear momentum as follows;
initial momentum of the neutron = final momentum of the atom
m₁u₁ = m₂u₂
where;
m₁ is the mass of the neutronu₁ is the initial velocity of the neutronm₂ is the mass of the atomic nucleusu₂ is the final velocity of the atomic nucleusThe mass of the atom = 14.8m₁
u₂ = m₁u₁ / m₂
u₂ = m₁u₁ / (14.8m₁)
u₂ = 0.0676u₁
The initial kinetic energy of the neutron is calculated as;
K.Ei = ¹/₂m₁u₁²
The final kinetic energy of the atom is calculated as;
K.Ef = ¹/₂m₂u₂²
K.Ef = ¹/₂(14.8m₁)(0.0676u₁)²
K.Ef = 0.068 (¹/₂m₁u₁²)
K.Ef = 0.068 (K.Ei)
The fraction of neutron's kinetic energy transferred to the atomic nucleus is calculated as;
= 0.068 (K.Ei) / K.Ei
= 0.068
= 6.8 %
The final kinetic energy of the neutron is calculated as follows;
K.E.f (neutron) = (1 - 0.068) x (2.66 x 10⁻¹³ J)
K.E.f (neutron) = 2.48 x 10⁻¹³ J
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What is the speed of a wave that has a wavelength of 25 m and a frequency of 10 hertz? Show your work
Answers
Answer:
see below
Explanation:
use the formula
speed = wavelength x frequency
speed = 25 x 10
speed = 250 m/s
hope this helps
How many sig figs are in 0.32
Answers
When the wheel of a cart rolls from a smooth sidewalk onto a plot of grass, the interaction of the wheel with the blades of grass slows the wheel. What slows light when it passes from air into glass or water
Answers
The light oscillates electrons which then absorbs some of the energy and reemit it thereby causing a delay.
What are Electrons?This is defined as a subatomic particle which has a negative charge. In the glass of water the light oscillates the electrons which absorbs some energy.
The re-emission is what slows light when it passes from air into glass or water.
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Describe the language of the following PDA (z is the stack end symbol) (the figure can be located under a, z/bbz X, z/z b,6/1 ۸, 2/2 90 91 92 a, b/bbb 1 b,b/1
Answers
The language of the given pushdown automaton (PDA) can be described as follows:
The PDA has a stack alphabet consisting of symbols 'a', 'b', 'z', '6', '1', '۸', '2', '9', '0', 'x', 'y'. 'z' represents the stack end symbol.
The transitions of the pushdown automaton (PDA) are as follows:
(a, z, z) -> (X, z): This transition allows the PDA to replace an 'a' at the input with an 'X' on the stack while maintaining the stack end symbol 'z'.(z, b, z) -> (z, z): This transition allows the PDA to pop a 'b' from the input without modifying the stack.(z, z, b) -> (6, 1): This transition allows the PDA to push '6' and '1' onto the stack when encountering a 'b' on the input.(6, 1, b) -> (۸, 2): This transition allows the PDA to replace the '6' and '1' on the top of the stack with '۸' and '2' respectively when another 'b' is read from the input.(x, y, b) -> (b, b, b): This transition allows the PDA to replace 'x' and 'y' on the top of the stack with 'b', 'b', and 'b' when a 'b' is read from the input.(b, b, b, b) -> (1, b): This transition allows the PDA to replace the 'b', 'b', and 'b' on the top of the stack with '1' and 'b' when another 'b' is encountered.(1, b, b) -> (1, 1): This transition allows the PDA to replace the '1' and 'b' on the top of the stack with '1' and '1' when another 'b' is read from the input.(1, 1, z) -> (z, z): This transition allows the PDA to pop '1' from the stack without modifying the input.Thus, the language accepted by this PDA is characterized by a sequence of 'a's followed by a sequence of 'b's, where the number of 'b's is three times the number of 'a's, and each 'b' is followed by a corresponding sequence of '90', '91', '92', 'a', 'b', 'b', 'b', '1', 'b', 'b', '1', and ending with '1'.
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How much work is done in moving a \( 25.0 \mathrm{~kg} \) book to a shelf \( 1.50 \mathrm{~m} \) high? What is the potential energy of the book as a result?
Answers
the work done to raise a 25kg book to a 1.5 m shelf is 367.5J and the potential energy of the book will be 367.5
Work done by a force is equal to the scalar product of the force applied and displacement. If d displacement happens on applying a force F, then work done, W by the force is
W = F. d
given:
mass, m = 25 kg
height, d = 1.50 m
work has to be done against the weight of the book, F = mg
work done W = F. d
W = mgd
W = (25) ×(9.8)×(1.50)
W = 367.5 J
This work is stored as the potential energy of the book.
Therefore, the work done to raise a 25kg book to a 1.5 m shelf is 367.5J and the potential energy of the book will be 367.5J
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What power is emitted by a circuit with a 9.00 volt battery and a 50.0 Ohm resistor? A)450 W B)0.00222 W C)1.62 W D)5.55 W E)0.180 W
Answers
Given,
Voltage, V=9 V
Resistor, R=50.0 ohm
To find
The power emitted by the battery.
Explanation
The power is:
\(\begin{gathered} P=\frac{V^2}{R} \\ \Rightarrow P=\frac{9^2}{50} \\ \Rightarrow P=1.62\text{ W} \end{gathered}\)Conclusion
The power is: C.1.62W
in general, what happens to the density and temperature of a gas as it expands?
Answers
The temperature of a gas will increase and the density will decrease as it
expands.
The expansion of gas happens as a result of an increase in the elastic
collision of the atoms. This is as a result of the increase in temperature which causes the atoms to be more spatially arranged.
The density of gas also reduces as a result of the same number of gas
molecules occupying more volume.
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when a 0.106 kg mass is suspended at rest from a certain spring, the spring stretches 3.80 cm. find the instantaneous acceleration of the mass when it is raised 6.30 cm, compressing the spring 2.50 cm.
Answers
The instantaneous acceleration of the mass when it is raised 6.30 cm and compresses the spring 2.50 cm is 16.26 m/s^2.
To solve this problem, we can use the equation for the force exerted by a spring:
F = -kx
When the mass is suspended at rest from the spring, the force exerted by the spring balances the weight of the mass, so we can write:
kx = mg
Solving for the spring constant, we get:
k = mg / x
Substituting the given values, we have:
k = (0.106 kg)(9.81 m/s^2) / 0.0380 m = 27.36 N/m
When the mass is raised 6.30 cm, the displacement of the spring is x = -0.0250 m (since the spring is compressed by 2.50 cm). The force exerted by the spring is:
F = -kx = -(27.36 N/m)(-0.0250 m) = 0.684 N
By Newton's second law, the net force on the object is:
Fnet = ma
where a is the instantaneous acceleration of the object.
The net force is the sum of the force exerted by the spring and the weight of the object:
Fnet = F + mg = 0.6875 N + (0.106 kg)(9.81 m/s^2) = 1.7239 N
Solving for the acceleration, we get:
a = Fnet / m = 1.7239 N / 0.106 kg = 16.2632 m/s^2
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Which types of rocks (igneou, sedimentary, metamorphic) tend to have crystalline textures? How does this texture form?
Answers
Answer:
metamorpic rock
Explanation:
it has a crystallin textures
2. Which positions of the pendulum would have the most potential energy?
A1 and 2
B2 and 3
C 1 and 5
D 4 and 5
Answers
Explanation:
im listening to metallica
turn the page
A motorized float in a Thanksgiving Day parade is traveling
north at a speed of 16.0 km/h. Part of a "Pilgrim" woman's role
In the thematic vignette is to walk across the float to call the
men to dinner. She walks at a speed of 3.22 km/h east as
the float passes the judge's booth, which is on the west side of the street. What is the woman's velocity with respect to the
Answers
you are using a microscope with a 10×× eyepiece. part a part complete what approximate focal length of the objective lens will give a total magnification of 200 ×× ? assume a length ll = 160 mmmm .
Answers
An approximate focal length of 8.42 mm for the objective lens will give a total magnification of 200× when using a 10× eyepiece.
To calculate the approximate focal length of the objective lens that will give a total magnification of 200×, we need to consider the magnification equation for a compound microscope
Total Magnification (M) = Magnification of the Objective Lens (\(M_{o}\)) × Magnification of the Eyepiece (\(M_{e}\))
Magnification of the Eyepiece (\(M_{e}\)) = 10×
Total Magnification (M) = 200×
From the equation, we can derive the magnification of the objective lens (\(M_{o}\)) as
\(M_{o}\) = M / \(M_{e}\)
Substituting the given values, we have
\(M_{o}\) = 200 / 10
\(M_{o}\) = 20×
Now, let's use the formula for the magnification of a compound microscope
\(M_{o}\) = 1 + (ll / \(f_{o}\))
Where:
ll = Length of the tube (160 mm)
\(f_{o}\) = Focal length of the objective lens (to be determined)
Rearranging the equation, we can solve for the focal length of the objective lens
\(f_{o}\) = ll / (\(M_{o}\) - 1)
Substituting the values
\(f_{o}\) = 160 mm / (20 - 1)
\(f_{o}\) = 160 mm / 19
\(f_{o}\) ≈ 8.42 mm
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A car moves with an average speed of 45 mph. How long does a car take to travel 90 miles?
Answers
Answer:
It took 2 hours
Explanation:
\({ \bf{time = \frac{distance}{speed} }} \\ \)
distance is 90 miles
speed is 45 miles per hour
\(t = \frac{90}{45} \\ \\ t = 2 \: h\)
Answer:
2 hours
Explanation:
45 times 2 is 90
What materials do you need to make a eco-friendly
house?
How do you build an eco-house?
What are simple ways to create a sustainable
house?
Answers
An eco-house is a house that is built to be environmentally friendly and sustainable. The materials used in the construction of an eco-friendly house are designed to be renewable, sustainable, and biodegradable.
The following are some materials needed to construct an eco-friendly house:
Bamboo: This is a versatile and sustainable material that is widely used in eco-friendly home building. It is durable, lightweight, and has a low carbon footprint. It can be used for flooring, walls, and furniture.Reclaimed
wood: Using reclaimed wood is another way to ensure that your eco-home is sustainable. It reduces the amount of waste in landfills and requires less energy than new wood to produce. Reclaimed wood can be used for floors, walls, and other parts of the home that require wood.
Cob: Cob is an eco-friendly building material made of clay, sand, and straw. It is an ancient technique that is gaining popularity because it is low cost and sustainable. It can be used to construct walls, floors, and even roofs.There are several ways to build an eco-house, but they all involve using sustainable materials, reducing energy consumption, and conserving water.
Here are some steps to building an eco-house:Design the house to take advantage of natural light and air circulation.Incorporate energy-efficient appliances, such as solar panels, into the home's design.Install water-saving fixtures such as low-flow toilets and showerheads.Use sustainable building materials like bamboo, reclaimed wood, and cob.Use insulation made from recycled materials.Reduce waste by using materials efficiently and recycling.Simple ways to create a sustainable house include:Switching to energy-efficient light bulbs.
Reducing water usage by fixing leaks, using low-flow fixtures, and collecting rainwater.Installing a programmable thermostat to reduce energy usage when you are not at home.Using natural cleaners instead of harsh chemicals.Reducing waste by recycling and composting.Installing double-pane windows to reduce heat loss during winter.A sustainable home can be created by following these simple steps. These steps not only reduce energy usage, but they also help to protect the environment and create a healthier living space for you and your family.
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can u give me the answers to this?
Answers
Answer:
Hope this helps.
Explanation:
Let's start b writing down coordinates of all points:
A(0,0,0)
B(0,5,0)
C(3,5,0)
D(3,0,0)
E(3,0,4)
F(0,0,4)
G(0,5,4)
H(3,5,4)
a.) When we reflect over xz plane x and z coordinates stay same, y coordinate changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.
A(0,0,0)
Reflecting
A(0,0,0)
B(0,5,0)
Reflecting
B(0,-5,0)
C(3,5,0)
Reflecting
C(3,-5,0)
D(3,0,0)
Reflecting
D(3,0,0)
b.)
A(0,0,0)
Moving
A(-2,-3,1)
B(0,-5,0)
Moving
B(-2,-8,1)
C(3,-5,0)
Moving
C(1,-8,1)
D(3,0,0)
Moving
D(1,-3,1)
A force of 20 N acts upon a 5 kg block. Calculate the acceleration of the object.
Answers
Answer:
The answer is 4 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
\(a = \frac{f}{m} \\ \)
where
f is the force
m is the mass
From the question we have
\(a = \frac{20}{5} \\ \)
We have the final answer as
4 m/s²Hope this helps you
Calculate the net force of the object in the image shown 20N 15 N
Answers
One summer afternoon, Kim heard on the news that the humidity in Ocala, Florida was 80%. How would the air feel when Kim goes outside?
Group of answer choices
cold and dry
warm and wet
cold and wet
warm and dry
Answers
A hockey player skates across a pond with constant velocity v and hits a puck straight ahead toward the goal.
We can assume there is no friction.
After the puck is hit, what is the velocity of the center of mass of the hockey player-puck system?
Choose 1 answer:
It is equal to the velocity of the hockey puck.
It is equal to half the velocity of the hockey player.
C
It is equal to the original velocity of the hockey player.
It is equal to the new velocity of the hockey player,
Answers
The correct answer is A. After the puck is hit, the velocity of the center of mass of the hockey player-puck system is equal to the velocity of the hockey puck.
We can assume that there is no friction in the given problem.
Therefore, the total momentum of the system remains constant.
It implies that if the hockey player and the puck are at rest, their total momentum will be zero.
After the player hits the puck, they move together as one system.
As the player hits the puck, he exerts a force on the puck in a particular direction.
The puck moves in the same direction as that of the player with the same speed but opposite in direction.
Therefore, the puck’s velocity is equal to that of the player but in the opposite direction.
Since the puck is light in weight and moves with a high velocity, it has a higher kinetic energy than the player does.
It means that the puck moves faster than the player.
Therefore, the velocity of the center of mass of the hockey player-puck system is equal to the velocity of the hockey puck.
The velocity of the player is equal to the velocity of the puck, but in the opposite direction.
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Answer: Equal to the original velocity of the hockey player
Explanation: Khan
A ball with mass 0.25 kg is thrown upward with initial velocity 24 m/s from the roof of a building 26 m high. Neglect air resistance. (a) Find the maximum height above the ground that the ball reaches. (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. (c) Plot the graphs of velocity and position versus time. 23. Assume that conditions are as in Problem 22 except that there is a force due to air resistance of ∣v∣/30, where the velocity v is measured in meters per second. (a) Find the maximum height above the ground that the ball reaches. (b) Find the time that the ball hits the ground. (c) Plot the graphs of velocity and position versus time. Compare these graphs with the corresponding ones in Problem 22. 24. Assume that conditions are as in Problem 22 except that there is a force due to air resistance of v2/1325, where the velocity v2 is measured in meters per second. (a) Find the maximum height above the ground that the ball reaches. (b) Find the time that the ball hits the ground. (c) Plot the graphs of velocity and position versus time. Compare these graphs with the corresponding ones in Problems 22 and 23.
Answers
(a) The maximum height above the ground that the ball reaches is approximately 28.18 meters.
(b) The time that the ball hits the ground is approximately 5.03 seconds.
(c) The graphs of velocity and position versus time will show specific patterns that illustrate the ball's motion.
(a) To find the maximum height, we can use the kinematic equation for vertical motion. Since the ball is thrown upward, its initial velocity is positive, and the acceleration due to gravity is negative.
Using the equation for displacement, we can find that the maximum height reached by the ball is when its velocity becomes zero. By substituting the known values into the equation, we can calculate the maximum height as approximately 28.18 meters.
(b) To determine the time the ball hits the ground, we need to find when its position becomes zero. Again, using the kinematic equation for vertical motion, we can set the displacement equal to zero and solve for time.
By substituting the given values, we can calculate the time of impact as approximately 5.03 seconds.
(c) The graphs of velocity and position versus time will exhibit distinct patterns. The velocity graph will show a parabolic shape, with the maximum velocity occurring at the initial point and decreasing to zero at the maximum height.
On the way down, the velocity will become negative and increase in magnitude until it reaches a maximum negative value just before hitting the ground.
The position graph will also follow a parabolic shape, with the maximum height corresponding to the maximum point on the graph. As time increases, the position decreases until it reaches zero when the ball hits the ground.
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ApplyStreets A and L run parallel to each other. BoulevardN forms a 75° angle with Street L south of (below) theirintersection. What angle does Boulevard N make withStreet L north of (above) their intersection? Justifyyour answer.
Answers
The angle that Boulevard N makes with Street L north of their intersection is also 75°.
What is intersection?Intersection is a point where two or more lines, curves, surfaces, or objects meet or cross each other. It is a common concept in mathematics, engineering, and everyday life. In mathematics, an intersection is a point where two or more sets of objects meet. In engineering, an intersection is a point where two or more roads, railways, or other transportation routes meet. In everyday life, an intersection is a point where two or more paths, streets, or other routes meet. Intersections are important in transportation, as they are the points where vehicles can change direction or turn around. Intersections are also important in mathematics, as they are the points where two or more lines, curves, or surfaces meet.
This is because the angle formed by two parallel lines is always equal. Since A and L run parallel to each other, the angle formed by Boulevard N and Street L south of their intersection is equal to the angle formed by Boulevard N and Street L north of their intersection. Therefore, the angle that Boulevard N makes with Street L north of their intersection is also 75°.
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The driver of a car traveling at 26.1 m/s applies the brakes and undergoes a constant
deceleration of 1.97 m/s^2.
How many revolutions does each tire make
before the car comes to a stop, assuming that
the car does not skid and that the tires have
radii of 0.21 m?
Answers
Each tire makes 131.9 revolutions before the car comes to a stop.
How to calculate?We will calculate the distance covered for by the car after the brakes are applied. We will apply the following equation of motion:
V^2=U² +2as where V is the final velocity, U is the initial velocity, a is the acceleration and s is the distance covered during that acceleration.
0=(26.1)² -2*1.97 *S (This negative because it is decelerating).
0 = 681.21 - 3.94 * S
S = 172.895
Circumference of the tire = 2πr where r is the radius.
Circumference = 1.31
Distance / circumference = Number of revolutions.
Number of revolutions. = 172.895/ 1.31
Number of revolutions = 131.9
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Details The force on a particle is described by 10x³ - 5 at a point x along the x-axis. Find the work done in moving the particle from the origin to x = 2.
Answers
Answer:
To find the work done in moving the particle from the origin to x = 2, we need to integrate the force over the given interval.
The work done (W) is calculated by integrating the force function with respect to displacement (dx) from the initial position (0) to the final position (2):
W = ∫(0 to 2) (10x³ - 5) dx
Integrating the force function, we get:
W = ∫(0 to 2) (10x³ - 5) dx = [2.5x⁴ - 5x] evaluated from 0 to 2
Now, substituting the upper limit (2) and lower limit (0) into the equation:
W = [2.5(2)⁴ - 5(2)] - [2.5(0)⁴ - 5(0)]
= [2.5(16) - 10] - [0 - 0]
= 40 - 10
= 30
Therefore, the work done in moving the particle from the origin to x = 2 is 30 units of work.
Explanation:
A 745 kg race car experiences an applied force from the engine of 38900 N and a frictional force of 34000 N. How long does it take to go from rest to 21 m/s ?
Answers
The time taken for the car to go from rest to 21 m/s is 3.2 seconds.
What is the acceleration of the car?The acceleration of the car is determined by applying Newton's second law of motion as shown below.
Mathematically, the formula for Newton's second law of motion is given as;
F (net) = ma
where;
m is the mass of the cara is the acceleration of the carThe net force acting on the car is calculated as follows;
F (net) = F - Ff
where;
F is the applied forceFf is the force of frictionF (net) = 38,900 N - 34,000 N
F (net) = 4,900 N
The acceleration of the car is calculated as follows;
a = F (net) / m
a = ( 4,900 N ) / ( 745 kg )
a = 6.58 m/s²
The time of motion of the car is calculated as follows;
v = u + at
v = 0 + at
v = at
t = v / a
t = ( 21 m/s ) / ( 6.58 m/s² )
t = 3.2 s
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What is the frequency of light when the energy of a single photon is 2. 23 × 10-21 j?.
Answers
When a single photon has an energy of 2.23*10-21, light has a frequency of *E=hv 2.23 10-21 J = (6.626... 3.03 10-19 J (6.022*1023 photons).
A photon is a tiny energy packet of electromagnetic radiation, also referred to as a light quantum.
Given energy, what is a photon's frequency?A single photon's energy is determined by the equation E = h v, where E is its energy (measured in SI units of J), h is the Planck constant (6.626 x 10-34 J s), and is the frequency of the radiation (measured in SI units of s-1 or Hertz, Hz) (see figure below).E = h is the energy equation.In the formula, E stands for energy, h for Planck's constant (6.626 x 10 -34 J s), and v for frequency.Since frequency rises as energy falls, frequency and energy are related directly in the energy equation. Because h is a constant, this is conceivable. between frequency and energy since frequency and energy are inversely correlated. Because h is a constant, this is conceivable.To learn more about Photon frequency refer to:
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