ting from rest, a disk rotates about its central axis with constant angular acceleration. in 5.0 s, it rotates 25 rad. during that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) what is the instantaneous angular velocity of the disk at the end of the 5.0 s? (d) with the an- gular acceleration unchanged,
A change in speed leads to an acceleration. When an object is rotating and its velocity is changing, we say it is accelerating angularly. Angular acceleration is also called rotational acceleration. It quantitatively expresses the change in angular velocity per unit time. Angular acceleration is a pseudo scalar. The sign of angular acceleration is considered positive if the angular velocity increases counterclockwise and negative if the angular velocity increases clockwise. Angular acceleration is used to study the motion of rotating objects such as wheels, fans, and the earth.
For the given problem:
α = 2θ/t²
Where, α = angular acceleration
θ = angle of rotation (25 rad)
t = time (5 sec)
α = 2×25/5²
α = 50/25
α = 2 rad/s²
ω = Δθ/Δt
Where, ω = average angular velocity
θ = angle of rotation (25 rad)
t = time (5 sec)
ω = 25/5
ω = 5 rad/s
ωₓ = α × t
Where, ωₓ = instantaneous angular velocity
α = angular acceleration (2 rad/s²)
t = time (5 sec)
ωₓ = 2 × 5
ωₓ = 10 rad/s
θ = ω₀t + ¹/₂αt²
Where, θ = angle of rotation
ω₀ = initial angular velocity (0 rad/s)
t = time (10 sec)
α = angular acceleration (2 rad/s²)
Since, ω₀ = 0
θ = ¹/₂αt²
θ = ¹/₂ × 2 × 10²
θ = 100 rad
Δθ = 100 - 25
Δθ = 75 rad
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The complete question is as follows:
Starting from rest , a disk rotates about its central axis with constant angular acceleration. In 5.0 s, it rotates 25 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity (c) What is the instantaneous angular velocity of the disk at the end of the 5.0 s (d) With the angular velocity of the disk at the end of the 5.0 s (d) With the angular acceleration unchanged through what additional angle will the disk turn during the next 5.0 s
Which of these is not method of preventing the corrosion of metals?
A. galvanising
B. painting
C. plating
D. rusting
Answer:
D. rusting is answer....
sphygmo- (sphygmo/gram; sphygmo/meter) means:
Sphygmo- (sphygmo/gram; sphygmo/meter) means pulse.A sphygmomanometer is a device used to measure blood pressure. Sphygmography is a technique for recording the pulse wave produced by the beating of the heart.
Sphygmo- is a combining form that means pulse, beating. It derives from the Greek word σφυγμός (sphygmos), which means "pulse, beating."For example, a sphygmomanometer is a medical instrument that measures blood pressure and consists of an inflatable cuff that is wrapped around the upper arm and a pressure gauge. Sphygmomanometer is a composite word made up of three roots: sphygmo- (pulse), mano- (pressure), and -meter (measuring instrument).
Sphygmo- means pulse or beating. A sphygmomanometer is a device used to measure blood pressure. Sphygmography is a technique for recording the pulse wave produced by the beating of the heart.
Sphygmo- is a combining form that means pulse, beating. It derives from the Greek word σφυγμός (sphygmos), which means "pulse, beating." Sphygmo- is used to form a number of medical terms such as sphygmography, sphygmomanometer, and sphygmotonometer.
Sphygmography is a technique for recording the pulse wave produced by the beating of the heart. It is done by placing a special pen or stylus on a piece of paper that is moving at a fixed speed and allowing it to trace out a curve that represents the pulse wave. The resulting tracing is called a sphygmogram.A sphygmomanometer is a medical instrument that measures blood pressure. It consists of an inflatable cuff that is wrapped around the upper arm and a pressure gauge. The cuff is inflated to a pressure above the systolic pressure of the patient's blood pressure. Then, the pressure is gradually released until the blood flow is restored and the pulse is felt again. The pressure at which the pulse is felt again is recorded as the systolic pressure.The sphygmotonometer is a newer version of the sphygmomanometer. It is a fully automated device that measures blood pressure using an electronic sensor. The device inflates the cuff and records the blood pressure automatically, without requiring any manual intervention.Sphygmo- means pulse or beating. A sphygmomanometer is a device used to measure blood pressure. Sphygmography is a technique for recording the pulse wave produced by the beating of the heart.
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Only refer from the community conservation article please. Will mark brainliest!!!!
1. Summarize the scientific information that leads to conservation in each of the articles.
2. What social issues affected the problem or its solution in each of the stories?
3. How did economics delay scientists' first attempts for conservation in each story?
4. Describe the political actions that led to successful conservation in both stories.
Trees are being felled at an alarming rate by people. Forests are home to numerous animals and people, and by removing trees from them, we are destroying their habitat and depriving them of a safe haven.
What is community conservation?Community conservation refers to a strategy for protecting the environment that acknowledges local residents' rights to manage wildlife or gain both direct and indirect benefits from it.
The gorillas' habitat, the forest, is being destroyed by humans. Additionally, people are harming these creatures, and if this continues, gorillas may become extinct.
There won't be as many resources from trees, like paper, if people cut down less of them, therefore prices could rise.
Prices for tourists will increase if fewer gorillas are still alive. That won't benefit visitors since they will have to pay more, and since some individuals don't have a lot of money, there will be fewer visitors.
Thus, this way, community conservation is mandatory.
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Two forces of magnitude 10n and 20n acting at an angle of 30 degree. what will be the x component of their resultant force?
The resultant of the forces is 29 N. Option D
What is the resultant force?The resultant force is the force that acts in a given direction. Now we have two forces as enumerated in the question.
Thus;
Resultant = √(10)^2 + (20)^2 - [2 * 10 * 20 * cos (60 - 30))
Resultant = 29 N
Thus, the resultant of the forces is 29 N.
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Missing parts;
Two forces of magnitude 10N and 20N act on a body in directions making angles 30° and 60° respectively with x-axis what is the resultant force
A.17N
B. 19N
C. 23N
D. 29N
E. 37N
A runner whose mass is 50 kg accelerates from a stop to a speed of 10 m / s in 3 s. (A good sprinter can run 100 m in about 10 s, with an average speed of 10 m / s.) (a) What is the average horizontal component of the force that the ground exerts on the runner’s shoes? (b) How much displacement is there of the force that acts on the sole of the runner’s shoes, assuming that there is no slipping? Therefore, how much work is done on the extended system (the runner) by the force you calculated in the previous exercise? How much work is done on the point particle system by this force? (c) The kinetic energy of the runner increases—what kind of energy decreases? By how much?
a) 167N
b) There is no work done on the extended system due to zero displacement and , the work done on the particle system is 2500 J.
c) Decrease in the internal energy by 2500 J.
The average horizontal component of the force that the ground exerts on the runner’s shoes is 167N. and The work done on the particle system is 2500 J. So, the internal energy decreases by 2500 J.
Mass of the runner = m = 50 kg
Initial velocity = u = 0 m/s
Final velocity = v = 10 m/s
Time taken = t = 3 s
Acceleration, a = (v - u) / t= (10 - 0) / 3= 3.33 m/s²
Horizontal force = ma= 50 kg × 3.33 m/s²= 167 N
The average horizontal component of the force that the ground exerts on the runner’s shoes is 167N.b) There is no work done on the extended system due to zero displacement and the work done on the particle system is 2500 J.
Work done = Force x Displacement
But in this case, no displacement takes place because the runner doesn't move in the horizontal direction. So, work done on the extended system is zero.
Work done by the force, W = change in kinetic energy
W = (1/2)mv² - (1/2)mu²
Here, initial velocity, u = 0 m/s
Final velocity, v = 10 m/s
Mass, m = 50 kg.
W = (1/2) × 50 × (10)²= 2500 J
The work done on the point particle system by this force is 2500 J. c)
The increase in the kinetic energy of the runner is due to the work done by the ground. So, the kinetic energy increases at the expense of some other form of energy. By doing work, some internal energy gets dissipated due to friction. So, there is a decrease in the internal energy.
Therefore, The work done on the particle system is 2500 J. So, the internal energy decreases by 2500 J. The work done on the particle system is 2500 J. So, the internal energy decreases by 2500 J. and The average horizontal component of the force that the ground exerts on the runner’s shoes is 167N.
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What height will the object reach? 12 points. Will give brainliest.
Answer:
12.7 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 56.7 Km/hr
Maximum height (h) =..?
First, we shall convert 56.7 Km/hr to m/s. This can be obtained as follow:
Initial velocity (m/s) = 56.7 x 1000/3600
Initial velocity (m/s) = 15.75 m/s
Next, we shall determine the time taken to get to the maximum height. This can be obtained as follow:
Initial velocity (u) = 15.75 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
v = u – gt (since the ball is going against gravity)
0 = 15.75 – 9.8 × t
Rearrange
9.8 × t = 15.75
Divide both side by 9.8
t = 15.75/9.8
t = 1.61 secs.
Finally, we shall determine the maximum height as follow
h = ½gt²
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 1.61 secs.
Height (h) =..?
h = ½gt²
h = ½ × 9.8 × 1.61²
h = 4.9 x 1.61²
h = 12.7 m
Therefore, the maximum height reached by the ball is 12.7 m
you are standing on a frictionless surface. you throw a heavy rock forward. will you accelerate, and if so, in which direction?
Yes, you will accelerate when you throw a heavy rock forward while standing on a frictionless surface. You will accelerate in the opposite direction to the rock.
The laws of motion developed by Sir Isaac Newton illustrate that an object in motion stays in motion unless an external force acts on it. So, when you toss a heavy rock forward on a frictionless surface, the rock will continue moving in the forward direction until something stops it.
The force that pushes the rock forward is equivalent in magnitude to the force that pushes you backward. Thus, the rock's mass is higher than yours; it will travel further than you, and you will accelerate in the opposite direction to the rock.
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As we know, the moon is a satellite of our earth, what is the
theoretical period of the moon? The average radius of the
moon's orbit is 3.84 108 m and the mass of the earth is 5.97 x
1024 kg (in hours, G = 6.67 x 10-9 N (m/kg) 3).
Answer:c
Explanation:c
This question involves the concepts of the time period, orbital radius, and gravitational constant.
The theoretical period of the moon is "658 hr".
The theoretical time period of the moon around the earth can be found using the following formula:
\(\frac{T^2}{R^3}=\frac{4\pi^2}{GM}\)
where,
T = Time Period of Moon = ?
R = Orbital Radius = 3.84 x 10⁸ m
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
Therefore,
\(\frac{T^2}{(3.84\ x\ 10^8\ m)^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\T^2=(9.91\ x\ 10^{-14}\ s^2/m^3)(56.62\ x\ 10^{24}\ m^3)\\\\T=\sqrt{561.34\ x\ 10^{10}\ s^2}\)
T = 2.37 x 10⁶ s\((\frac{1\ h}{3600\ s})\)
T = 658 hr
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The attached picture shows the derivation of the formula for orbital speed.
What are the three forms of water discussed in the passage?
Answer:
The three forms of water are gas, liquid, and ice, I'm not sure if these were mentioned in the passage since you did not attach a picture.
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A plane flying horizontally at a speed of 50m/s and at an elevation of 160m drops a package, and 2. 0 s later it drops a second package. How far apart will the two packages land on the ground if air resistance is negligible?
Answer:
100m
Explanation:
becuase velocity is constant
trajectory of the packages will be same
a person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.25 ms2.25 ms from an initial speed of 3.25 m/s3.25 m/s . what is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.55 kg1.55 kg ?
The magnitude of the average contact force exerted on the leg by the person is 2238.8N.
A person slaps her leg with her hand which results her hand to come to rest in a time interval of 2.25 millisecond from an initial speed of 3.25m/s.
According to second law of motion of Newton,
The average force applied by a body can be given by the change in momentum per unit time of the body.
So, we can write,
F = M∆V/t
Putting values,
F = 1.55 x 3.25 x 10³/2.25
F = 2238.8 N.
So, the magnitude of the average contact force exerted on the leg is 2238.8N.
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In lightning , light is seen first and sound is heard later it is due to
1)optical illusion
2) variation in temperature
3) position of observer
4) difference of velocities of light and sound
Answer:
option 4
Explanation:
Light's velocity in air ( 3 × 10^8 m/s ) is much greater than sound's velocity in air ( 343 m/s )
Hence due to difference in velocities , during lightning light is seen first & sound is heard later
you are using a 50-mm-focal-length lens to photograph a tree. if you change to a 100-mm-focal-length lens and refocus, the image height on the detector changes by a factor of
The image height on the detector will change by a factor of 2 if you change from a 50-mm-focal-length lens to a 100-mm-focal-length lens and refocus.
The magnification of a lens is given by the ratio of the image height to the object height. Since the object height remains the same, the change in magnification is solely determined by the change in focal length.
The magnification of a lens is given by the formula:
Magnification = - (image distance / object distance).
Since we are only interested in the ratio of image heights, we can ignore the negative sign.
For the 50-mm lens, the magnification is:
Magnification1 = 50 mm / object distance.
For the 100-mm lens, the magnification is:
Magnification2 = 100 mm / object distance.
Taking the ratio of the two magnifications:
Magnification2 / Magnification1 = (100 mm / object distance) / (50 mm / object distance) = 100 mm / 50 mm = 2.
Therefore, the image height on the detector changes by a factor of 2 when switching from a 50-mm-focal-length lens to a 100-mm-focal-length lens and refocusing.
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It takes a 50 N horizontal force to pull a 20 kg object along the ground at a constant velocity. What is the coefficient of kinetic friction?
The coefficient of kinetic friction is equal to 0.255.
Given the following data:
Horizontal force = 50 NewtonMass = 20 kgTo calculate the coefficient of kinetic friction:
Mathematically, the coefficient of kinetic friction is given by the formula;
\(F_k = umg\)
Where;
Fk represents the force of kinetic friction.μ represents the coefficient of kinetic friction.m represents the mass.g is the acceleration due to gravity.Making u the subject of formula, we have:
\(u = \frac{F_k}{mg}\)
Substituting the given parameters into the formula, we have;
\(u = \frac{50}{20 \times 9.8}\\\\u = \frac{50}{196}\)
Coefficient of kinetic friction, u = 0.255.
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A 5 kg object oscillates with simple harmonic motion. Its position as a function of time varies according to the following equation x(t) = 2 sin(( π/2)t + ( π/6))
At what time (in seconds) after t = 0 seconds is the kinetic energy first at a maximum?
The kinetic energy is first at a maximum after approximately 3.67 seconds after t=0.
To determine the time when the kinetic energy is first at a maximum, we need to find the time at which the velocity of the object is at a maximum. The kinetic energy of an object undergoing simple harmonic motion is maximum when the velocity is maximum.
The equation for velocity as a function of time in simple harmonic motion is given by:
v(t) = Aω cos(ωt + φ)
where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase constant.
In this case, the equation for position is x(t) = 2 sin((π/2)t + (π/6)), which is in the form x(t) = A sin(ωt + φ).
Comparing the two equations, we can see that ω = π/2.
The velocity equation becomes:
v(t) = (π/2)A cos((π/2)t + φ)
To find the time at which the velocity is maximum, we need to find when the cosine term is at its maximum value of 1. This occurs when (π/2)t + φ = 0 or (π/2)t + φ = 2π.
Solving for t, we get:
(π/2)t = -φ or (π/2)t = 2π - φ
t = (-2φ/π) or t = (4π - 2φ)/π
Since the phase constant φ in the given equation is π/6, we substitute this value into the equations:
t = (-2(π/6)/π) or t = (4π - 2(π/6))/π
t = -1/3 or t = 11/3
Therefore, the kinetic energy is first at a maximum at t = 11/3 seconds.
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Find the modulation index of an AM wave having 8V variation where maximum amplitude of the AM wave i 9V
Correct option is (A) 0.8
What is Modulation Index ?
-- Modulation index is defined as the ratio of the amplitude of the modulating wave to that of the carrier wave
- Modulation index: m = Am / Ac
: Given 2Am = 8 Am + Ac = 9
⇒ Ac = 5
∴ m = 4/5 = 0.8.
What is Electromagnetic Wave ?
-'Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field.
-- In other words, EM waves are composed of oscillating magnetic and electric fields.
What is Magnetic Field?
-- A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials.
--A moving charge in a magnetic field experiences a force perpendicular to its own velocity and to the magnetic field.
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Liquid ammonia flows through a pipe at a mass flow rate of 100
kg/s .
If the cross-sectional area of the pipe is 0.01
m^2, determine the flow rate of momentum through the
pipe.
The momentum flow rate through the pipe carrying liquid ammonia is 1 × \(10^6\) kg·m/s.
The flow rate of momentum (Ṁ) through the pipe can be calculated by multiplying the mass flow rate (ṁ) by the velocity (v). The speed can be determined using the equation v = ṁ / (ρA), where ρ is the density of the liquid ammonia and A is the pipe's cross-sectional area.
Given:
ṁ = 100 kg/s
A = 0.01 m²
Assuming the density (ρ) of liquid ammonia is 700 kg/m³, we can calculate the velocity (v):
v = ṁ / (ρA)
v = 100 kg/s / (700 kg/m³ × 0.01 m²)
v = 10000 m/s
Now, we can calculate the flow rate of momentum (Ṁ):
Ṁ = ṁv
Ṁ = 100 kg/s × 10000 m/s
Ṁ = 1 × \(10^6\) kg·m/s
Therefore, the momentum flow rate through the pipe is 1 × \(10^6\) kg·m/s.
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Please answer the question for gravitational field attached in the image.
We are given:
Radius of Venus= 6 * 10⁶ m
Height above the planet's ground = 9.4 * 10⁶ m
Mass of Venus = 4.867 * 10²⁴ kg
Finding the Gravitational Field:
We first need to calculate the distance between the center of masses
Distance between center of masses:
Since the point is above the surface of Venus, the radius of Venus and the height will add
Distance between the center of masses = 9.4*10⁶ + 6 * 10⁶
Distance (r) = 15.4 * 10⁶ m
Now back to Gravitational field!
We know the formula for Gravitational Field:
g = GM / r² [where g is the gravitational field, M is the mass of the planet, r is the distance between the center of masses and G is the gravitational constant]
plugging the known values in the formula:
g = (6.67 * 10⁻¹¹ * 4.867 * 10²⁴) / (15.4*10⁶)² [G = 6.67*10⁻¹¹ Nm²/kg²]
g = (32.46 * 10¹³) / (237 * 10¹²)
g = (32.46 * 10¹³) / (23.7 * 10¹³)
g = 32.46/23.7 [the 10¹³ got cancelled out]
g = 1.367 N/kg² (or m/s²)
Which method of measurement is most likely to be both accurate and
precise?
A. Measuring the temperature of a pond using a digital thermometer
OB. Measuring the temperature difference between two objects by
comparing how they feel
C. Measuring the speed of a runner by comparing it to that of other
runners
O
D. Measuring time using a clock that has slowed down
The method of measurement most likely to be both accurate and precise is measuring the temperature of a pond using a digital thermometer, as listed in Option A, as there is less chance of mistakes when measured by machines.
What is a precise measurement?Accuracy refers to how close the measured value is to the true value, while precision refers to how reproducible the measurements are, and using digital thermometers is typically accurate and precise because there is a lower chance of human-like mistakes.
Hence, the method of measurement most likely to be both accurate and precise is measuring the temperature of a pond using a digital thermometer, as listed in Option A.
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which of the following changes when light is refracted? A. Frequency B. Phase C. Speed D. Wavelength
When light is refracted, the speed and wavelength of light change.
When light travels from one medium to another, such as from air to water or from air to glass, it undergoes refraction. Refraction is the bending of light as it passes through different substances with different optical densities.
During refraction, the speed and wavelength of light can change, while the frequency remains constant.
Speed of light: The speed of light is different in different media. It is generally slower in denser media like water or glass compared to its speed in vacuum or air. The speed of light in a medium is given by the equation:
v = c/n,
where v is the speed of light in the medium, c is the speed of light in vacuum (approximately 3 × 10^8 meters per second), and n is the refractive index of the medium.
Wavelength of light: The wavelength of light also changes when it undergoes refraction. The wavelength is the distance between two consecutive peaks or troughs of a light wave.
The relationship between the wavelength of light in a medium and its wavelength in vacuum is given by:
λ = λ₀/n,
where λ is the wavelength of light in the medium, λ₀ is the wavelength of light in vacuum, and n is the refractive index of the medium.
When light is refracted, the speed and wavelength of light change, while the frequency remains constant.
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Not having to worry about your health, job, or being attacked are related to this level
of needs.
Answer:
This level of needs is known as safety needs. Safety needs are the need for security, stability, and protection from physical and emotional harm. These needs include physical safety, financial security, health and well-being, and protection from discrimination and harassment.
Explanation:
A soccer player kicks a ball with initial velocity of 10m/s at an angle of 30 degrees above the horizontal.....what is the magnitude of the vector
Answer: 10m/s^2
Explanation:
Given the following :
Initial Velocity(U) = 10m/s
Angle of inclination = 30°
In other to calculate the magnitude of the vector,
We need to resolve into verti AL and horizontal component.
Vertical component of a vector(Uy) = U × Sinθ
Horizontal component(Ux) = U × Cosθ
Uy = 10 × sin30 = 10 × 0.5 = 5m/s
Ux = 10 × Cos30 = 10 × 0.866 = 8.66m/s
Therefore the magnitude equals :
U = √(Uy)^2 + (Ux)^2
U = √5^2 + 8.66^2
U = √25 + 74.9956
U = √99.9956
U = 9.9997799
U = 10m/s^2
Now see if you can tell the difference between inertia and momentum. How could you change the momentum of a
moving car?
Explanation:
Inertia is the measure of the mass of an object. It is the tendency of an object to change the direction of an object.
Momentum is the product of mass and velocity. When an object having mass m moves with a particular velocity it will have momentum.
If the speed of an object is changed from u to v, its momentum can be changed and it can be calculated as :
Δp = m(v-u)
Hence, this is the required solution.
In the Vector Addition Lab, the following data was collected for determining the displacement from the door of the Physics classroom to another location in the building: 2 m, West; 14.0 m, South; 25.0 m, East; 11.0 m, North; and 2 m, West. The magnitude of the resultant displacement from the physics classroom to the assigned location is closest to ___ meters. (Round to the nearest integer.)
Answer:
21 m
Explanation:
Since the displacement from the door is 2 m West, we have our vector as -2i. The vector representing 14.0 m South is -14.0j. The vector representing 25.0 m East is 25.0i. The vector representing 11.0 m North is 11.0j. And, the vector representing 2.0 m West is -2.0i.
So, to get our position vector at the other location, we add all the vectors together.
So, r = -2i + (-14.0j) + 25.0i + 11.0j - 2i
= -4i + 25.0i - 14.0j + 11.0j
= 21.0i -3j m
Now, if we assume the position vector for the door is at the origin, we have r₀ = 0i + 0j m
So, our displacement from the door is r - r₀ = 21.0i - 3.0j - (0i + 0j) = 21.0i - 3.0j
So, the magnitude of the resultant displacement |r - r₀| = √(21.0² + 3.0²)
= 3.0√(7.0² + 1)
= 3.0√(49 + 1)
= 3.0√50
= 3.0 × 7.0711
= 21.2
≅ 21 m to the nearest integer
One forecasting model was used to forecast demand for a product. The forecasts and the demand are shown in the table below. B Actual Forecast 11 40 41 35 38 3 38 35 33 30 IX Calculate Moan Absolute Deviation (MAD) and Mean Squared Error (MSE). Show all details and use 1 decimal in your answer For the toolbar, press ALT+F10 (PC) or ALT+FN-F10 (Mac). BI V S Paragraph Arial 14px V QUESTION 1 The department manager is using a combination of methods to forecast sales of tonsters at a local department store. The demanders shown in the be Week Actu Demand 11 24 bo 2 bas x III A Using trend projection, calculate foresting values for week and week & Show details of your answer For the toolbar, pro ALT.F10 PC) O ALT.FN.F10 Mac BIS Paragraph Arial 14 Focus Chile we state
The estimated demand for Week 4 is 36.3
MAD(Mean Absolute Deviation) is used to calculate the average difference between forecast values and actual values. It calculates the deviation by taking the absolute value of the difference between actual and forecasted demand. The formula to calculate Mean Absolute Deviation is:
MAD= Sum of| Actual demand - Forecast demand | / number of periods
In the given table, the Actual demand is shown as B and the forecast demand is shown as F.
B Actual Forecast 11 40 41 35 38 3 38 35 33 30
Calculation of MAD:
Actual (B) Forecast (F) |B-F|11 40 29.041 35 5.043 38 0.053 3 35.054 38 3.055 35 0.056 33 3.057 30 3.058 0.0 30.0Total 103.0
The number of periods is 9 as shown in the table.
MAD= 103/9MAD= 11.44
Mean Squared Error (MSE) measures the average squared difference between the actual and forecasted values. The formula for MSE is:
MSE= Sum of (Actual demand - Forecast demand)^2 / number of periods.
Calculation of MSE:
Actual (B) Forecast (F) (B-F)^2 11 40 841 35 25 625 38 0 0 3 35 484 38 0 0 35 33 4 30 0 900Total 2854
The number of periods is 9 as shown in the table.
MSE= 2854/9MSE= 317.1
Therefore, the calculated MAD is 11.44 and MSE is 317.1.
Trend Projection formula is given by:
Y = a + bx
where Y is the estimated demand for a particular period.
a is the Y-intercept
b is the slope of the regression line x is the period number
In the given table, the Week number is shown as X and the Actual demand is shown as Y.
Week number Actual Demand 11 24 22 29
Using trend projection for Week 3, we can calculate the demand as follows:
Slope (b) = (nΣ(xy) - Σx Σy) / (nΣ(x^2) - (Σx)^2) =(2*22 - 1*24)/(2*3 - 1*1) = 20/5 = 4
Intercept (a) = Σy/n - b(Σx/n) =(24+22)/2 - 4(2/2) = 23Y = a + bx = 23 + 4(3) = 35
Therefore, the estimated demand for Week 3 is 35.
Using trend projection for Week 4, we can calculate the demand as follows:
Slope (b) = (nΣ(xy) - Σx Σy) / (nΣ(x^2) - (Σx)^2) =(2*29 - 1*24)/(2*5 - 1*1) = 34/9 = 3.78
Intercept (a) = Σy/n - b(Σx/n) =(24+22+29)/3 - 3.78(2.0) = 21.5Y = a + bx = 21.5 + 3.78(4) = 36.3
Therefore, the estimated demand for Week 4 is 36.3.
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Which other subatomic particle has the same mass as a neutron
Answer:
.................... protons :)
Answer:
protons
The two subatomic particles that have the same mass are protons and neutrons.
Describe the ways the reading addresses how some people use ""uncertainty"" as a way to undermine science:
The reading addresses how some people use "uncertainty" as a way to undermine science: by pointing out that these individuals often exploit the inherent limitations and ambiguities present in scientific research.
They tend to focus on the fact that scientific findings are often provisional, and conclusions can change with the introduction of new evidence or advancements in methodologies. By emphasizing the uncertainty aspect, they attempt to discredit or cast doubt on the overall credibility of scientific results.
Furthermore, these individuals may cherry-pick data or studies that support their preconceived beliefs, while ignoring the broader consensus within the scientific community. This selective presentation of evidence contributes to misinformation and public confusion about scientific issues, such as climate change or vaccinations.
In some cases, those who use uncertainty to undermine science may have ulterior motives, such as protecting vested interests or promoting a specific political or ideological agenda. By sowing doubt and mistrust in scientific findings, they can manipulate public opinion and obstruct the development or implementation of evidence-based policies.
To counteract these attempts to undermine science, it is essential for the scientific community and the public to recognize and appreciate the inherent uncertainties within the scientific process.
By fostering critical thinking, encouraging open and transparent communication, and supporting continued research and education, we can help build resilience against misinformation and promote a better understanding of the role of uncertainty in scientific progress.
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If it take light 8⅓ minute to travel from the Sun to Earth and if the Earth-Sun ditance i 150,000,000 km, calculate the peed of light in meter per econd. [Hint: Firt do ide calculation to figure out how many econd i equal to 8⅓ minute and how many meter i equal to 150,000,000 km. Then divide the ditance given in meter (m) by the time you found in econd () to get the anwer in unit of m/. ]
The speed of light is 1428571.428 m/s.
Speed is defined as the rate of change of position of an object in any direction. Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.
The time taken is \(8\dfrac{1}{3}\) minutes. The time taken in seconds is,
\(8\dfrac{1}{3} \times 60\) = 10500 seconds.
The distance between Earth-Sun is, 150,000,000 km , which is equal to \(15 \times 10^{7}\ km\).
1 kilometers = 100 meters.
\(15 \times 10^{9}\ m\)
The speed is given by distance divided by the time.
Speed = \(\dfrac{15 \times 10^{9}}{10500}\)
Speed = 1428571.428 m/s
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--The complete question is, If it take light 8⅓ minute to travel from the Sun to Earth and if the Earth-Sun distance is 150,000,000 km, calculate the speed of light in meter per second. [Hint: First do calculation to figure out how many second is equal to 8⅓ minute and how many meter is equal to 150,000,000 km. Then divide the ditance given in meter (m) by the time you found in second to get the anwer in unit of m/s. ]--
At position B where the ball just exactly before it hit the ground, how fast is the ball at point B?
980 m/s
31 m/s
980 m/s2
31 m/s2
The speed of the ball at point B, just before it hits the ground, is 31 m/s.
At the point just before the ball hits the ground, its vertical velocity will be the final velocity just before impact. The acceleration due to gravity, denoted as "g," is approximately 9.8 m/s². As the ball falls freely under gravity, its velocity increases by 9.8 m/s every second.
Using the equation of motion:
v² = u² + 2as
where:
v is the final velocity (which we want to find)
u is the initial velocity (which we assume to be zero, as the ball is dropped)
a is the acceleration (which is -9.8 m/s² due to gravity)
s is the displacement (the distance fallen, which we don't know but is not relevant to finding the speed)
Plugging in the values, we have:
v² = 0 + 2 * (-9.8) * s
Since we are only interested in the speed (magnitude of velocity), we can take the square root of both sides:
v = √(2 * 9.8 * s)
Given that s is the distance fallen at point B, the speed of the ball at point B just before it hits the ground is 31 m/s.
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