Answer:
The moon does not get pulled into the sun because of gravitational pull.
Explanation:
Gravitational pull is a force that pulls things down or into i guess you can say. Like are orbit, all of the planets (even the dwarf planet "pluto") are circling around are sun but we have things called moons that circle are planets. Are moon is orbiting us like we (are earth) are orbiting the sun. So to get into a little more detail, i will add that we circle the sun or the moon circles us because the action of earth pulling away from the suns gravitational pull is causing it to either rotate or revolve.So we are stuck in the gravitational force of the sun and the moon is stuck in ares. But as someone who LOVES astronamy will say that i watched a video about are earth, sun, and moon and it said that each year are moon is slowly pulling away from the earth. sooner or later we might not have a solar or lunar eclipse anymore.
Answer:
But the path of the Moon is always concave towards the Sun; the gravitational force exerted by the Sun on the Moon is always greater than the pull of the Earth on the Moon
Explanation:
The process of examining a change in one variable in a model while assuming that all the other variables remain constant is called _______.
Answer:
Ceteris Paribus
Explanation:
The process of examining a change in one variable in a model while assuming that all the other variables remain constant is called Ceteris Paribus.
Ceteris Paribus is a Latin phrase that means "all other things being equal" or "all other things held constant" in English. The phrase has found application in disciplines like Economics and Statistics. This phrase as being adopted as a process of examining a change in one variable in a model while assuming that all the other variables remain constant to ascertain the relationship between the variables or make deductions from an experimental study. An example of Ceteris Paribus application is the law of demand and supply in Economics. The law of demand states, Ceteris Paribus, the higher the price, the lower the quantity demanded and vice versa. Conversely, the law of supply states, Ceteris Paribus, the higher price, the higher the quantity supply and vice versa.
Johanna is studying what happens to the energy as a ball rolls down a ramp. What is she studying? Check all that apply.
1. gravitational potential energy
2. electrical energy
3. kinetic energy
4. mechanical energy
5. elastic potential energy
Answer:
Gravitational potential energy, kinetic energy and mechanical energy
Explanation:
The body is at a displacement to the gravity and therefore has energy stored in it as a result of its displacement against gravity. When this body is released the gravitational potential energy is converted to translational kinetic energy as well as rotational kinetic energy as it rolls down the the ramp. The sum of the kinetic energies and potential energy of the body amount to the mechanical energy of the ball down the ramp.
An energy pyramid, like the one below, represents which principle of ecology? Question 12 options:
a.Most producer organisms use photosynthesis to create organic compounds.
b.The ultimate source of energy in the biosphere is the sun.
c.Consumer organisms are more important than producer organisms.
d.Large amounts of energy are lost in the transfer between trophic levels.
Answer:
b.The ultimate source of energy in the biosphere is the sun.
Explanation:
The pyramid represents the energy that is initiated from the sun is then retained and stored in the new biomass and the each tropic level of the ecosystem there is 10% of the transfer of energy form one level to the next or higher level.Answer: D.Large amounts of energy are lost in the transfer between trophic levels.
Explanation:
A boat with a horizontal tow rope pulls a water skier. She skis off to the side so that the rope makes an angle of 10 degrees with the forward direction of motion. If the tension in the room is 200 N, how much work does the rope do on the skier during a forward displacement of 100 m?
Answer:
19696 J or 19700 J
Explanation:
COMPONENT OF FORCE IN DIRECTION OF DISPLACEMENT IS
F cosФ = 200 cos 10° = 196.96 N
work = component of force in direction of displacement × displacement
= 196.96 × 100
= 19696 J
= 19700 J
The work done by rope is equal to 19696 J on the skier during a forward displacement of 100 m
What is the work?Work can be described as the energy used when a force is applied in order to move an object through a specific displacement. The performed work defines both the applied force and the displacement of the object.
The force acting is to move an object in a line in the direction of the pushing force over a distance 'd'.
W= F ×d
Where 'F' is the force and 'd' is the displacement and W is work done by an object.
If the angle θ with the forward direction of motion then work = Fd cos θ
Given, the force acting, F = 200 N
The angle with the forward direction of motion, θ = 10°
Given, the displacement in the forward direction, d = 100 m
Work done by the rope, W = Fd cos θ
W = 200 × 100 cos 10°
W = 19696 J
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Submarine canyons were most likely formed by:Select one:a. scouring by glaciers during the last ice age.b. deposition of terrestrial sediment.c. earthquake activity.d. erosion by turbidity currents.e. erosion by major rivers in the past.
Answer:
Submarine canyons were most likely formed by erosion by major rivers in the past (option e)
Explanation:
A submarine canyon is a deep, relatively narrow, V-shaped depression whose bottom generally has a continuous slope. It can extend from the continental shelf to the great ocean depths.
Submarine canyons are habitats that appear interrupting the apparent homogeneity of the continental shelf. They play an important role in channeling organic matter. They transport sediments from the continental shelf to the deep ocean. Submarine canyons are usually found at the mouths of the rivers and are formed by the gigantic erosive power of the water that is slowly sculpting it from the continent, to the abysmal depths.
Submarine canyons were most likely formed by erosion by major rivers in the past (option e)
The line of sight from a small boat to the light at the top of a 45-foot lighthouse built on a cliff 25 feet above the water makes a 20° angle with the water. To the nearest foot, how far is the boat from the cliff?
Answer:
The boat is 192 feet from the cliff.
Explanation:
Hi there!
Please see the attached figure for a graphical description of the problem.
Notice that the line of sight, the distance to the cliff and the height to the top of the lighthouse form a right triangle. Hence, we can apply trigonometric rules to find the distance from the boat to the cliff:
cos 20° = adjacent side / hypotenuse
sin 20° = opposite side / hypotenuse
The length of the opposite side is the height of the cliff plus the height of the lighthouse:
opposite side = 45 feet + 25 feet = 70 feet.
Using the equation of sin 20°, we can obtain the hypotneuse:
sin 20° = opposite side / hypotenuse
hypotenuse · sin 20° = opposite side
hypotenuse = opposite side / sin 20°
hypotenuse = 70 feet / sin 20°
hypotenuse = 205 feet
Now, using the equation of cos 20°, we can calculate the distance to the cliff (the length of the adjacent side):
cos 20° = adjacent side / hypotenuse
hypotenuse · cos 20° = adjacent side
205 feet · cos 20° = adjacent side
adjacent side = 192 feet (without rounding intermediate results)
The boat is 192 feet from the cliff.
Answer:
192 ft.
Explanation:
Tan (20) = Opp/Adj
Tan (20) = (45+25)/x
Tan (20) = 70/x
then x=70/Tan (20)
x= 192.3234 ft, and then rounded to the nearest foot...
x= 192 ft
A rocket is launched vertically from a point on the ground. An observer who is 1000 m away from the base of the launching pad, notices that the angle of elevation of the rocket is increasing at a rate of π/40 radians per second when the angle of elevation is π/3.
Find the speed of the rocket at that instant.
Answer:
dh/dt=18.75rad/sec
Explanation:
A rocket is launched vertically from a point on the ground. An observer who is 1000 m away from the base of the launching pad, notices that the angle of elevation of the rocket is increasing at a rate of π/40 radians per second when the angle of elevation is π/3.
Find the speed of the rocket at that instant.
you should have a right-angled triangle, where the angle of elevation is Ø, the adjacent is 1000 and the opposite is h, the height of the rocket.
tangent of an angle is opposite over adjacent side
tanØ = h/100 0
h = 1000tanØ
differentiate both sides with respect to dt,change in time
tanØ=sec^2 Ø
dh/dt = 1000sec^2 Ø (dØ/dt)
where Ø = 60° or π/3 rad
dØ/dt = π/40 radians per second
substituting the value into the equation
dh/dt=1000sin^260° *π/40 rad/sec
dh/dt=750*π/40 rad/sec
dh/dt=18.75rad/sec
dh/dt is the same as speed, change in height per time
Hole filling fasteners (for example, MS20470 rivets) should not be used in composite structures primarily because of the?
Answer:
We can cause delamination.
Explanation:
The reason why is because the probability of causing delamination increase considerably when we use Hole-filling fasteners. If we use a typical rivet, these tends to expands in order to fill the hole.
If we analyze the force applied by the expanded rod will cause that the matrial will be deteriorated and will cause that the material to delaminate around the edges of the hole and we can cause possible control and no protection to the material.
At a place where an object is thrown vertically downward with a speed of while a different object is thrown vertically upward with a speed of Which object undergoes a greater change in speed in a time of 2 s?
Answer:
Both objects will undergo the same change in velocity
Explanation:
m = Mass of the Earth = 5.972 × 10²⁴ kg
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
r = Radius of Earth = 6371000 m
m = Mass of object
Any object which is falling has only the acceleration due to gravity.
[tex]ma=\dfrac{GMm}{r^2}\\\Rightarrow a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^6)^2}\\\Rightarrow a=9.81364\ m/s^2[/tex]
The acceleration due to gravity on Earth is 9.81364 m/s²
So, the speeds of the objects will change at an equal rate of 9.81364 m/s² but the change will be negative when an object is thrown up.
Hence, both objects will undergo the same change in velocity.
Sequence of analytic functions that map unit disk to unit disk and converge pointwise to 1 at some point. prove sequence converges normally to 1.
Answer:
Cause: If you have a disk that does not appear in File Explorer and is shown in Disk Management as Uninitialized, it may be because the disk does not have a valid disk signature. Basically, this means that it has never been initialized or formatted, or that the drive format has been damaged in some way.
It is also possible that the disk has hardware problems or problems connecting, but we will get to that in some paragraphs.
Solution: If the unit is completely new and should only be initialized, erasing the data it contains, the solution is easy: see Initialize new disks (Initialize new disks). However, there are many possibilities that you have already tried this and that it has not worked. Or maybe you have a disk full of important files and you don't want to erase it by initializing it.
There are a number of reasons why a disk may fail or that it did not have to be initially started, with a common reason why the disk is failing. There are many things that can be done to correct a disk error, but here are some steps to see if we get them to work again. If the disc works after performing one of these steps, do not worry about the following; just relax, celebrate it and, maybe, update the backups.
Examine the disk in Disk Management. If Disconnected appears as shown here, try to right-click on it and select Connected.
The disk appears as disconnected
If the disk appears in Disk Management as Connected and has a main partition that appears as Correct, as shown here, it is a good sign.
The disc is shown as connected with a correct volume
If the partition has a file system, but no drive letter (for example, E :), see Change a drive letter (Change a drive letter) to add a drive letter manually.
If you don't have a file system (NTFS, ReFS, FAT32 or exFAT) and you know the disk is empty, right-click on the partition and select Format. When formatting a disk, all the data it contains is erased, so don't do it if you are trying to recover files from the disk; instead, go to the next step.
If you have an external disk, disconnect it and reconnect it, and then select Action> Re-examine the disks.
Turn off your PC, disable the external hard drive (if it is an external drive with a power cable), and then reactivate your PC and disk. To deactivate your Windows 10 PC, select the Start button, then the power button, and then select Shut down.
Plug the disk into another USB port that you can find directly on your PC (not in a hub). Sometimes, USB disks do not have enough power from some ports or have other problems with certain ports. This is especially common with USB hubs, but sometimes there are differences between the ports of a computer, so you can try different ports, in case we have.
Try another cable. It might seem crazy, but the cables fail a lot, so you can try using another cable to connect the disk. If you have an internal disk on a desktop computer, you will probably need to turn off your PC before changing cables; Consult your PC manual for more information.
Composed by Device Manager to see if there are problems. Press and hold the Start button (or right-click on it), and then select Device Manager from the context menu. Search for devices with an exclamation point next to it or other problems, double-click on the device, and then read its status.
A PERT chart
a. is a numeric counterpart to the Gannt chart.
b. shows the variance of the critical path only.
c. is based on alpha distribution.
d. shows the variance of the critical path based on the alpha distribution
e. allows simulation of every path in the network, taking variances of each activity in consideration
Answer:
A.
Explanation:
A PERT chart is a project management tool that provides a graphical representation of a project's timeline. The Program Evaluation Review Technique (PERT) breaks down the individual tasks of a project for analysis. But PERT charts are considered preferable to Gantt charts because they identify task dependencies, but they're often more difficult to interpret.
An oxygen sensor is being tested with a digital multimeter using the MIN/MAX function. The readings are: minimum = 78 mV; maximum = 932 mV; average = 442 mV. Technician A says that the engine is operating normally. Technician B says that the oxygen sensor is skewed too rich. Which technician is correct?
Answer:
Technician A
Explanation:
An oxygen sensor is being tested with a digital multimeter using the MIN/MAX function. The readings are: minimum = 78 mV; maximum = 932 mV; average = 442 mV. Technician A says that the engine is operating normally. Technician B says that the oxygen sensor is skewed too rich. Which technician is correct?
an oxygen sensor detects te amount of oyen in an engine at the exhaust pipe.
if the oxygen sensor detects oxygen at the above boundary condition, ten the engine is in good working condition
Two long thin parallel wires 13.0 cm apart carry 25-A currents in the same direction.
1. Determine the magnitude of the magnetic field vector at a point 10.0 cm from one wire and 6.0 cm from the other.
2. Determine the direction of the magnetic field vector at that point.
Answer:
3.33 x 10⁻⁵ T
Explanation:
given,
current carried by two parallel wire = 25 A
distance between two wire = 13 cm
1) Let p be the distance between two wire
p = 13 cm = 0.13 m
a = 10 cm = 0.1 m
b = 6 cm = 0.06 m
using cosine law
[tex]p^2 = a^2 + b^2- 2ab cos\theta[/tex]
[tex]cos\theta =\dfrac{a^2+b^2}{2ab}[/tex]
[tex]\theta =cos^{-1}(\dfrac{0.1^2+0.06^2}{2\times 0.1 \times 0.06})[/tex]
θ = 105.96°
θ = 106°(approximately)
now, magnetic field on one wire
[tex]B_1 = \dfrac{\mu_oI}{2\pi r}[/tex]
[tex]B_1 = \dfrac{4\pi \times 10^{-7}\times 25}{2\pi (0.1)}[/tex]
B₁ = 5 x 10⁻⁵ T
now, magnetic field on another wire
[tex]B_2= \dfrac{\mu_oI}{2\pi r}[/tex]
[tex]B_2 = \dfrac{4\pi \times 10^{-7}\times 25}{2\pi (0.06)}[/tex]
B₂ = 8.33 x 10⁻⁵ T
resultant field,
= [tex]\sqrt{B_1^2+B_2^2 - 2 B_1B_2 cos \theta}[/tex]
= [tex]\sqrt{(5\times 10^{-5})^2+(8.33\times 10^{-5})^2 - 2\times 5 \times 10^{-5}\times 8.33 \times 10^{-5} cos 106^0}[/tex]
= 3.33 x 10⁻⁵ T
1) The magnitude of the magnetic field vector at the given point is; -3.195 * 10⁻⁵ T
2) The direction of the magnetic field vector at the given point is; 129.71° from the x-axis
What is the magnitude and direction of the magnetic Field?We will start with finding the direction of the field for each wire from the tangent to the circle around the wire.
For their magnitudes, we have;
B₁ = (µ₀/4π) * 2I₁/L₁
B₁ = (4π * 10⁻⁷/4π) * 2 * 25/0.1
B₁ = 5 * 10⁻⁵ T
B₂ = (µ₀/4π) * 2I₂/L₂
B₂ = (4π * 10⁻⁷/4π) * 2 * 25/0.06
B₂ = 8.33 * 10⁻⁵ T
If we draw a vector diagram, we will have;
B = B₁(-sin θ₁ i + cos θ₁ j) + B₂(-sin θ₂ i + cos θ₂ j)
B = (5 * 10⁻⁵ )(-sin 26.34 i + cos 26.34 j) + (8.33 * 10⁻⁵)(-sin 47.7 i + cos 47.7 j)
B = -(8.38 * 10⁻⁵) i + (10.09 * 10⁻⁵) j
Direction of the field is;
θ = tan⁻¹((10.09 * 10⁻⁵)/-(8.38 * 10⁻⁵)
θ = 129.71° from the x-axis
Thus, magnitude is;
B_mag = B₁ * cos 129.71°
B_mag = -3.195 * 10⁻⁵ T
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"Which of the following is most like the rotation" of stars in the disk of the Milky Way? cars moving at a constant speed on a circular race track the planets in our solar system a DVD in a player the hands of an analog clock the tire of a car moving
Answer:
cars moving at a constant speed on a circular race track.
Explanation:
The rotation of stars in the disk of milky way is more like that of cars moving at a constant speed on a circular race track.
That is all stars are rotating at fixed point in a fixed orbit , such that no two stars are crossing each other. Speed of all stars are different from each other but all stars are rotating at a constant speed.
A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal. Find the magnitude of the truck's horizon- tal component of displacement. Answer in units of m.
Answer:
1422.8 m
Explanation:
Given:
Displacement of the truck is, [tex]\vec S=1430\ m[/tex]
The direction of the truck's displacement is 5.76° with the horizontal.
A vector inclined at angle [tex]\theta[/tex] with the horizontal is resolved into 2 components which are mutually perpendicular to each other. One of the component is along the horizontal and the other is along the vertical.
If a vector 'A' is inclined at an angle [tex]\theta[/tex] with the horizontal, then its horizontal and vertical components are given as:
[tex]Horizontal:\\A_x=A\cos \theta\\\\Vertical:\\A_y=A\cos \theta[/tex]
Here, the vector is 'S' and its horizontal component is needed.
Therefore, the horizontal component is given as:
[tex]S_x=S\cos \theta\\S_x=1430\times \cos(5.76\°)\\S_x=1422.8\ m[/tex]
5. (II) A fisherman’s scale stretches 3.6 cm when a 2.4-kg fish hangs from it. (a) What is the spring stiffness constant and (b) what will be the amplitude and frequency of oscillation if the fish is pulled down 2.1 cm more and released so that it oscillates up and down?
The spring stiffness constant of the fisherman's scale, based on Hooke's Law, is 653.3 N/m. If a fish is pulled down further by 2.1 cm, it will oscillate with an amplitude of 5.7 cm and a frequency of approximately 0.816 Hz.
Explanation:To answer part (a) of your question about what the spring stiffness constant is, we use Hooke's Law. This law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. This relationship is formally shown as F = k * x, where F is the force (in this case, the weight of the fish influences this force), k is what we are looking for (the spring stiffness constant), and x is the displacement (how much the spring stretches due to the force).
In this problem, the fish's weight becomes the force, which you can calculate as mass * gravity, or 2.4 kg * 9.8 m/s² = 23.52 N. The scale stretches 3.6 cm, or 0.036 m. Substituting these into Hooke's law gives us 23.52 N = k * 0.036 m, and solving for k provides a value of approximately 653.3 N/m. Thus, the spring stiffness constant is 653.3 N/m.
As for part (b) of the question, determining the amplitude and frequency of the oscillation if the fish is pulled down a further 2.1 cm and released, we know that the amplitude simply the displacement from the equilibrium position. Hence, total displacement from equilibrium is 3.6 cm (initial stretching) + 2.1 cm (additional pull) = 5.7 cm, or 0.057 m. This is the amplitude of the oscillation.
For the frequency of the oscillation, we need to understand the fish-spring system as a simple harmonic oscillator. The frequency (f) of a simple harmonic oscillator system is given by f = 1 / (2π) √(k/m), where k is the spring constant and m is the mass. Substituting our known values gives a frequency of approximately 0.816 Hz. Therefore, the oscillations will have an amplitude of 5.7 cm and a frequency of 0.816 Hz.
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A student is pushing a 50 kg cart, with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s^2. How much friction must be acting on the cart?
Answer:
450N
Explanation:
1) F=ma F2=50kg*3m/s^2=150N
2)Ff=F1-F2=600N-150N=450N
A student is pushing a 50 kg cart, with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s². Friction force acting on the cart is 450 N.
What is force?
A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Given in the question a student is pushing a 50 kg cart, with a given force of [tex]F_1[/tex] = 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s²,
F = ma
[tex]F_2[/tex] = 50 kg*3 m/s²
=150 N
Frictional force,
[tex]F_f[/tex] = [tex]F_1 - F_2[/tex]
= 600 N-150 N
= 450 N
A student is pushing a 50 kg cart, with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s². Friction force acting on the cart is 450 N.
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A block floats partially submerged in a container of liquid. When the entire container is accelerated upward, which of the following happens? Assume that both the liquid and the block areincompressible.
(A) The block descends down lower into the liquid.(B) The block ascends up higher in the liquid.(C) The block does not ascend nor descend in the liquid.
(D) The answer depends on the direction of motion of the container.(E) The answer depends on the rate of change of the acceleration
When the entire container is accelerated upward, the block does not ascend nor descend in the liquid.
Explanation:The answer to the question is (C) The block does not ascend nor descend in the liquid.
When the entire container is accelerated upward, the liquid and the block inside it will experience the same acceleration. Since the block is partially submerged, both the liquid and the block will move together without any relative change in their positions. The buoyant force on the block remains the same, and the block does not ascend nor descend in the liquid.
A car traveling at 100 km/hr strikes an unfortunate bug and splatters it. The force of impact is:______
A) greater on the bug.
B) greater on the car.
C) the same for both.
Answer:
The force of impact is same for both objects.
Explanation:
It is given that,
Speed of the car, v = 100 km/h
It is understood that the mass of car is more than that of the bug. When one object strikes another object, they both applies a force on each other. One object applies the same force on another object when they strikes as per Newton's third law of motion.
In this case, when a car strikes a bug, the force of impact is the same for both. Hence, the correct option is (c).
According to Newton's third law, for every action, there is an equal and opposite reaction. Therefore, the force of impact when a car strikes a bug is the same for both. However, due to its smaller mass, the impact has a much larger effect on the bug.
Explanation:The subject of this question is Physics and it relates to the principle of Newton's Third Law which states that 'for every action, there is an equal and opposite reaction'. When the car and the bug collide, they apply forces to each other of equal magnitude and opposite direction. Thus, the force of impact is actually the same for both, as suggested by option C.
However, the consequences of this force are more notable for the bug due to its much smaller mass compared to the car. The high acceleration experienced by the bug (according to Newton's second law 'Force = Mass x Acceleration') causes it to splatter, while the car continues largely unaffected.
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A person in bare feet is standing under a tree during a thunderstorm, seeking shelter from the rain. A lightning strike hits the tree. A burst of current lasting 43 µs passes through the ground; during this time the potential difference between his feet is 21 kV. If the resistance between one foot and the other is 550 Ω,
what is the current through his body and how much energy is dissipated in his body by the lightning?
Answer:
[tex]I=38.181\ A[/tex] is the current through the body of the man.
[tex]E=34.5\ J[/tex] energy dissipated.
Explanation:
Given:
time for which the current lasted, [tex]t=43\times 10^{-6}\ s[/tex]potential difference between the feet, [tex]V=21000\ V[/tex]resistance between the feet, [tex]R=550\ \Omega[/tex]Now, from the Ohm's law we have:
[tex]I=\frac{V}{R}[/tex]
[tex]I=\frac{21000}{550}[/tex]
[tex]I=38.181\ A[/tex] is the current through the body of the man.
Energy dissipated in the body:
[tex]E=I^2.R.t[/tex]
[tex]E=38.181^2\times 550\times 43\times 10^{-6}[/tex]
[tex]E=34.5\ J[/tex]
In Times Square in New York City, people celebrate on New Year's Eve. Some just stand around, but many move about randomly. Consider a system comprised of all of these people. Approximately, what is the total linear momentum of this system at any given instant?
Answer:
The total linear momentum at any time is zero.
Explanation:
Hi there!
The momentum of an object is a vector that is calculated by multiplying the mass of the object times its velocity.
The momentum of the system is calculated by adding the momenta of each object that comprise the system (in this case, all the people in Times Square).
Since the people move randomly in all directions, it is most probable that the total momentum of the system is zero because the momentum vectors of people moving in one direction will surely be canceled by the momentum vectors of people moving in the opposite direction. In total, the sum of momentum vectors will be zero (the standing people have null momentum because their velocities are zero).
Final answer:
The total linear momentum of a crowd in Times Square on New Year's Eve can be approximated as zero due to the random and varied movements of the people in the crowd, balancing each other out.
Explanation:
When considering the total linear momentum of a system like the crowd in Times Square on New Year's Eve, we must consider that the system is made up of individuals moving in various directions with different velocities. According to the physical definition of momentum, which is the product of mass and velocity (P = mv), every person has their own momentum. If we assume an idealized situation where movement is random and there is as much motion in one direction as in any other, the total momentum of the system can approximatively be zero at any given instant due to the cancellation of people's momentums in opposite directions. Moreover, because people are not likely to all be moving in the same direction with the same speed at the same time, the net velocity (and thus momentum) of the system may be negligible if averaged over time.
According to Newton's second law in terms of momentum, in the absence of an external force, there would be no change in the total momentum of the system. Therefore, assuming Times Square is a closed system with no significant external forces acting on it, and given the randomness of movement, it can be approximated that the total linear momentum of the crowd is approximately zero.1
2. According to research, what is the most reliable form of identifying potentially effective reinforcers?
The most reliable form of identifying potentially effective reinforcers is known as Preference Assessment. It is an adaptive procedure vastly used in the field of reinforcers.
Explanation:
Preference Assessment identifies items that are likely to be effective as reinforcers by identifying a particular learner's preference for them.
Reinforcement is an important part of being an effective behavior technician and the preference assessments are crucial.Plenty of research has shown that when you give the learner an opportunity to show you what they want that is a much better indication of what will work as a reinforcer rather than asking family.Preference assessments identify attention, objects or activities that have a strong potential for serving as reinforcers for target clients. There are ethical reasons to use them.Beyond the scope of the guide is the exhaustive review of the Preference assessment methodology and literature. However, interested readers go through Virus-Ortega et al(2014) for decision- making process and selecting an appropriate method of preference assessment for individuals with disabilities."As an object falls into a black hole, it":
a. falls into a stable orbit around the black hole.
b. is broken up and heated before it reaches the event horizon.
c. passes into a location somewhere else.
d. can emit high energy (gamma-ray) electromagnetic radiation that can leave the black hole.
e. passes the event horizon intact.
Answer:D
Explanation:
Hawking showed that quantum effects allow black holes to emit exact black-body radiation. The
electromagnetic radiation is produced as if emitted by a black body with a temperature inversely proportional to the mass of the black hole. One of the pair falls into the
black hole while the other escapes
If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to the new value of the electric field at that same point?
a) 1:2
b) 1:3
c) 1:6
d) 1:9
Answer:
b)1 :3
Explanation:
Lets that
The value of a positive charge = q
As we know that electric filed on a point charge given as
[tex]E=\dfrac{Kq}{r^2}[/tex]
Where ,K=Constant
q=Charge ,r=Distance
If the value of the charge gets tripled ,q'= 3 q
Then electric filed E'
[tex]E'=\dfrac{Kq'}{r^2}[/tex]
[tex]E'=\dfrac{3Kq}{r^2}[/tex]
E' = 3 E
Therefore we can say that
[tex]\dfrac{E}{E'}==\dfrac{1}{3}[/tex]
therefore the answer will be --
b)1 :3
When the magnitude of a positive charge is tripled, the ratio of the original value of the electric field to the new value at the same point becomes 1:3. This is because the electric field is directly proportional to the charge. Therefore, if the charge is tripled, the electric field also triples.
If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to the new value of the electric field at that same point?
The electric field (E) created by a point charge (q) at a distance (r) is given by the equation:
[tex]E = k * q / r^2[/tex]
where k is the Coulomb constant. If the original charge is q and this creates an electric field E at distance r, then:
[tex]\text{Original Electric Field} (E_1) = k * q / r^2[/tex]
Now, if the charge is tripled, q becomes 3q. The new electric field (E2) is:
[tex]\text{New Electric Field} (E_2) = k * 3q / r^2[/tex]
Thus, the ratio of the original electric field (E1) to the new electric field (E2) is:
[tex]E_1 / E_2 = (k * q / r^2) / (k * 3q / r^2) = 1/3[/tex]
The electrical force between an inner electron and the nucleusof an atom is:_________a. larger for atoms of low atomic numberb. largerfor atoms of high atomic numberc. the same for all atoms
Answer:
b. Larger for atoms of high atomic number.
Explanation:
The atomic number of a chemical element is the number of protons found in the nucleus of every atom of that element.
The electrical force between an inner electron and the nucleus of an atom is described by the Coulomb law
[tex]Fe = \frac{kq_{1}q_{2} }{r^{2}}[/tex]
where k: Coulomb constant
q1 and q2 are the electric charges that interact (in our case electron and protons)
r: Distance between electron and each proton
Then, larger number of protons in the nucleus, more the electrical force will be.
The answer is:
b. Larger for atoms of high atomic number.
A helicopter (m = 3250 kg) is cruising at a speed of 56.9 m/s atan altitude of 185 m. What is the total mechanical energy of thehelicopter?A) 6.91 x 107 JB) 5.26 x 107 JC) 1.12x 108 JD) 2.27 x 108 JE) 3.91 x 107 J
Answer:
The mechanical energy of the helicopter is [tex]1.12\times 10^7\ J[/tex].
Explanation:
It is given that,
Mass of the helicopter, m = 3250 kg
Speed of the helicopter, v = 56.9 m/s
Position of the helicopter, h = 185 m
The energy possessed by an object due to its motion is called its kinetic energy. It is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
[tex]E=\dfrac{1}{2}\times 3250\times (56.9)^2[/tex]
[tex]E=5.26\times 10^6\ J[/tex]
The energy possessed by an object due to its position is called its potential energy. It is given by :
[tex]E=mgh[/tex]
[tex]E=3250\times 9.8\times 185[/tex]
[tex]E=5.89\times 10^6\ J[/tex]
The sum of kinetic and potential energy is called mechanical energy of the system. It is given by :
[tex]M=5.26\times 10^6+5.89\times 10^6[/tex]
[tex]M=11.15\times 10^6\ J[/tex]
or
[tex]M=1.12\times 10^7\ J[/tex]
So, the mechanical energy of the helicopter is [tex]1.12\times 10^7\ J[/tex]. Hence, this is the required solution.
An athlete notices that, after lifting a 100 kilogram weight, he has to increase the weight to 105 kilograms before he feels that the weight is actually heavier. If the athlete later lifts a 200 kilogram weight, which of the following coincides with the just noticeable difference for this weight?
a. A 190 kg weight
b. A 205 kg weight
c. A 215 kg weight
d. A 220 kg weight
Answer:
option A
Explanation:
given,
lifting weight = 100 Kg
increase in the weight = 105 Kg
later athlete lift = 200 Kg
Weber's law states that Just noticeable difference of weight is equal to ratio of change to the original.
Just noticeable difference for weight will be
=[tex]\dfrac{105-100}{100}[/tex]
= 5 %
since second weight is 200 Kg
so, Just noticeable difference of 200 Kg will be equal to ± 5 % of 200 Kg which is equal to ± 10 Kg.
now,
Just noticeable difference for 200 Kg
= 200 + 10 or 200 -10
= 210 Kg or 190 Kg
Hence, the correct answer is option is option A
Which heat transfer process is important in the transfer of energy from the Sun to Earth?
Answer: Radiation
Explanation:
The heat transfer process that is important in the transfer of energy from the Sun to Earth is Radiation.
Radiation is one of the mode of heat transfer which transfer energy/heat from one medium to another without any intervening medium. For example, heat from the sun reaches the earth through radiation. The sun hits the earth directly without any medium between the sun itself and the earth.
Radiation is the key heat transfer process by which energy is transferred from the Sun to Earth. It allows the transmission of energy through the vacuum of space and is integral to Earth's climate regulation, especially with respect to the natural greenhouse effect.
Explanation:The process of heat transfer that is important in the transfer of energy from the Sun to Earth is radiation. Unlike conduction and convection, radiation does not require a medium to transfer heat. This means that energy in the form of electromagnetic waves can travel through the vacuum of space from the Sun to the Earth. The Sun emits a tremendous amount of energy through nuclear reactions at its core, and this energy is transferred to Earth as electromagnetic radiation, primarily in the form of visible light and infrared radiation. When this radiation reaches Earth, it is absorbed and warms the planet's surface, playing a crucial role in Earth's climate and enabling life to exist.
The greenhouse effect also plays an important role in Earth's temperature regulation. Gases in Earth's atmosphere, such as carbon dioxide and water vapor, trap some of the heat that is radiated from the Earth's surface, keeping the planet warmer than it would be otherwise. Without this natural greenhouse effect, Earth's average temperature would be significantly lower, making it less hospitable for life as we know it.
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April works at the grocery store. She needs to move a heavy crate of bananas (m = 65 kg) to the produce section. She drags the crate along the floor by applying a force of 280 N at an angle of 35° above the horizontal.
a. If there is no friction, what is the acceleration of the crate? (5 points)
280/65=4.3
b. If the coefficient of kinetic friction between the crate and the floor is 0.4, what is the acceleration of the crate? (5 points)
a) The acceleration in absence of friction is [tex]3.5 m/s^2[/tex]
b) The acceleration in presence of friction is [tex]0.6 m/s^2[/tex]
Explanation:
a)
First of all, we notice that the crate accelerates only along the horizontal direction, since the forces in the vertical direction are balanced (in fact, the weight is balanced by the normal reaction + the vertical component of the applied force).
This means that we can consider only the horizontal component of the applied force, which is
[tex]F_x = F cos \theta = (280)(cos 35^{\circ})=229.4 N[/tex]
where
F = 280 N is the applied force
[tex]\theta=35^{\circ}[/tex] is the angle
Now we can find the acceleration of the crate, by using Newton's second law:
[tex]F_x = ma[/tex]
where
[tex]F_x = 229.4 N[/tex] is the net force along the horizontal direction
m = 65 kg is the mass of the crate
a is the acceleration
Solving for a, we find
[tex]a=\frac{F_x}{m}=\frac{229.4}[65}=3.5 m/s^2[/tex]
b)
In this case, we also have the force of friction, so we have to compute the force of friction.
We start by writing the equation of the forces along the vertical direction:
[tex]R+F sin \theta - mg = 0[/tex]
where
R is the normal reaction
[tex]F sin \theta[/tex] is the vertical component of the applied force
[tex]mg[/tex] is the weight
We can re-write this as
[tex]R=mg-Fsin \theta[/tex] (1)
Now we can write the equation of the forces along the horizontal direction
[tex]F cos \theta - \mu_k R = ma[/tex] (2)
where
[tex]F cos \theta[/tex] is the horizontal component of the applied force
[tex]\mu_K R[/tex] is the force of friction, with [tex]\mu_k = 0.4[/tex] is the coefficient of friction
By substituting (1) into (2) and solving for a, we find the new acceleration of the crate:
[tex]F cos \theta - \mu_k (mg-F sin \theta) = ma\\a=\frac{F cos \theta - \mu_k mg +\mu_k F sin \theta}{m}=\frac{(280)(cos 35)-(0.4)(65)(9.8)+(0.4)(280)(sin 35)}{65}=0.6 m/s^2[/tex]
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Answer:
3.53 m/s^2
Explanation:
Use the net x value.
Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point of an eye for which a contact lens with a power of __.
Answer:
(a). The eye's near point is 68.98 cm from the eye.
(b). The eye's far point is 33.33 cm from the eye.
Explanation:
Given that,
Power = 2.55 D
Object distance = 25 cm for near point
Object distance = ∞ for far point
Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?
(a) We need to calculate the focal length
Using formula of power
[tex]f =\dfrac{1}{P}[/tex]
Put the value into the formula
[tex]f=\dfrac{100}{2.55}[/tex]
[tex]f=39.21\ cm[/tex]
We need to calculate the image distance
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}[/tex]
[tex]\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{1421}{98025}[/tex]
[tex]v=-68.98\ cm[/tex]
The eye's near point is 68.98 cm from the eye.
(b). We need to calculate the focal length
Using formula of power
[tex]f =\dfrac{1}{P}[/tex]
Put the value into the formula
[tex]f=-\dfrac{100}{3.00}[/tex]
[tex]f=-33.33\ cm[/tex]
We need to calculate the image distance
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}[/tex]
[tex]-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{1}{33.33}[/tex]
[tex]v=-33.33\ cm[/tex]
The eye's far point is 33.33 cm from the eye.
Hence, (a). The eye's near point is 68.98 cm from the eye.
(b). The eye's far point is 33.33 cm from the eye.
The near point for which a lens of +2.55 diopters is prescribed is 71 cm.
We have to obtain the focal length of the lens as follows;
f = 100/P
Where P is the power of the lens in diopters.
f = 100/2.55 = 39.2 cm
Now;
Given that;
1/f = -1/v + 1/u
f = focal length
v = image distance
u = object distance
1/v = 1/u - 1/f
u = Near point of the normal eye
1/v = 1/25 - 1/39.2
v = 71 cm
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