Describe the weather when there is a high pressure air mass in your area
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Kelplers 3 laws in your own words
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According to Kepler's first law of planetary motion, planets revolve around the sun such that the sun is always at one of its foci. This law is also known as the law of orbits.
According to Kepler's Second Law of planetary motion, a planet will cover equal amounts of area in an equal period of time if a line is drawn from the sun to the planet. This implies that the planet moves more slowly away from the sun and faster towards it.
According to Kepler's third Law of Planetary Motion, the squares of the orbital periods of the planets are directly proportional to the cubes of their semi-major axes.
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The angular position of a point on the rim of a rotating wheel is given by θ = 4.0t - 3.0t^2 + t^3, where is in radians and t is in seconds. What are the angular velocities at t = 2.0{~s}t=2.0 s.
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The angular velocity of the pointed rim of a rotating wheel if, The angular position of a point on the rim of a rotating wheel is given by θ = 4.0 t - 3.0 t² + t³, is 4 rad/ sec.
What is Angular velocity?The rotation rate, which refers to how quickly an item rotates or circles in relation to another point, is measured pictorially by angular velocity. Angular velocity can be defined as the speed at which an item rotates or revolves around an axis.
Given:
θ = 4.0 t - 3.0 t² + t³
Calculate the angular velocity by differentiating the position function as shown below,
dθ / dt = d (4.0 t - 3.0 t² + t³) / dt
V = 4.0 - 6t + 3t² (Change in position is equal to velocity)
Put the value of time, t = 2
V = 4 - 6 × 2 + 3 × 2²
V = 4 - 12 + 3 × 4
V = 4 - 12 + 12
V = 4 rad/sec
Thus, the angular velocity is 4 rad/sec.
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The names of meteor showers (Geminids, Leonids, Perseids, Quadrantids) are names of _______ which are in the apparent _______ of the luminous tails of individual meteors seen all over the sky.
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The names of meteor showers (Geminids, Leonids, Perseids, Quadrantids) are names of meteor radiant points which are in the apparent direction of the luminous tails of individual meteors seen all over the sky.
The Quadrantids, for example, appear to radiate from the constellation Boötes.
The names of meteor showers (Geminids, Leonids, Perseids, Quadrantids) are names of radiant points which are in the apparent paths of the luminous tails of individual meteors seen all over the sky.
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Which of the following reactions will delay the decay of a fruit?
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Answer:
placing it in the refrigerator
Two spaceships are traveling past Earth at 90% of the speed of light in opposite directions (i.e., they are approaching each other). One turns on a searchlight, which is seen by scientists aboard the other. What speed do the scientists measure for this light (c = speed of light in a vacuum)?
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According to the theory of special relativity, the speed of light in a vacuum (c) is constant for all observers, regardless of their relative motion. This means that the speed of light is always measured to be the same value, c, regardless of the motion of the source or the observer.
In the scenario described, one spaceship turns on a searchlight, and the scientists aboard the other spaceship observe this light. Since the speed of light is always measured to be c, the scientists aboard the second spaceship will measure the speed of the searchlight to be equal to the speed of light, c, regardless of the relative motion between the two spaceships.
Therefore, the scientists aboard the second spaceship will measure the speed of the searchlight to be c, the speed of light in a vacuum, irrespective of the fact that both spaceships are traveling at 90% of the speed of light in opposite directions.
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Pls help I'm desperate
If you place a hot piece of metal in a container of water, thermal energy flows from the metal to the water. What happens after the metal and the water reach the same temperature?
The flow of thermal energy stops, and the temperature remains constant.
The flow of thermal energy continues and causes both substances to become warmer.
The flow of thermal energy reverses and causes both substances to become warmer.
The flow of thermal energy reverses, and the water becomes colder.
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Answer:
I think your answer would be D
Explanation:
Looking at the graph below, what is the average velocity from 1 to 6 seconds? Explain how you found it (at least 4 sentences). *
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Answer:
3 m/s
Explanation:
v= s/t
v= 9/6
v=3 m/s
A pickup truck is capable of producing 515 Nm of torque. How much force does the engine produce if the force acts 0.25 m from the pivot point?
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Answer:
257.5 N
Explanation:
does an object in motion stay in motion
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Imagine that you wanted to test the effects of driving at different speeds on the gas mileage of your car. To find out, you drove a distance of 100 miles at many different rates of speed. During your first trip, you drove at exactly 55 miles per hour and calculated that your gas mileage was 20 miles per gallon. During your next trips, you either decreased or increased your rate of speed. Also, what can you conclude from this experiment? How might your car get better gas?
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The conclusion that you might arrive at is that the speed of the car affects the gas mileage.
What could you conclude from the experiment?We know that an experiment is the only way that we can be able to establish cause and effect relationship. We know that the speed would affect the consumption of the gas. By varying the speed of the car, we can be able to obtain the effect that it has on the mileage.
Thus, the conclusion that you might arrive at is that the speed of the car affects the gas mileage.
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where is the far point of an eye for which a contact lens with a power of -1.40 diopters is prescribed for distant vision?
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The far point of the eye for which a contact lens with a power of -1.40 diopters is prescribed for distant vision is approximately -71.4 centimeters.
The far point of an eye is the distance at which an eye can focus without any accommodation, meaning the lens of the eye is in its most relaxed state. To find the far point of an eye, we can use the formula:
Far point (in meters) = 1 / Power of the lens (in diopters)
In this case, the power of the contact lens is -1.40 diopters. So, we can plug this value into the formula:
Far point = 1 / (-1.40)
Now, let's calculate the far point:
Far point = -0.714 meters
Since the distance is negative, it means the far point is in front of the eye. To convert this distance into centimeters, we multiply by 100:
Far point = -0.714 * 100
Far point = -71.4 centimeters
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Why would a nation most likely use military force? select three options. to remove trade restrictions to defend territory to support allies to promote political relationships to protect economic interests
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Answer: 2, 3, 5
Explanation: to defend territory, to support allies, to protect economic interests
I got 2 (to defend territory), 3 (to support allies) and 5 (to protect economic interests)
I hope this helps you
leah paid 42.90 for 5 gallons of gas what was the price per gallon of gas leah paid?
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Answer:
the price per gallon of gas leah paid is 8.58
Explanation:
42.90/5=8.58
2. A squirrel runs 15 m to the north and finds an acorn. The squirrel then turns back to the south
and runs 8 m, at that point he decides to change direction and runs 24 m to the east to bury his
acorn.
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Answer:
24 m to the right/ to the east
A hunter shoots a 500 g arrow at a 2.0 kg bird perched on a tall tree growing on flat, level ground. The arrow is launched from ground level with a speed of 40 m/s at an angle of 30 degrees above the horizon. It is traveling horizontally when it strikes and embeds in the bird. How far from the base of the tree do the bird and the arrow land
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Answer:
The bird and the arrow are 84.621 meters far from the hunter.
Explanation:
The arrow experiments of a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. The arrow strikes the bird when it reaches its maximum height. First, we determined the time taken by the arrow before striking the bird:
\(v = v_{o}\cdot \sin \theta + g\cdot t\) (1)
Where:
\(v_{o}\) - Initial speed of the arrow, in meters per second.
\(v\) - Final speed of the arrow, in meters per second.
\(\theta\) - Launch angle, in sexagesimal degrees.
\(g\) - Gravitational acceleration, in meters per square second.
\(t\) - Time, in seconds.
If we know that \(v_{o} = 40\,\frac{m}{s}\), \(\theta = 30^{\circ}\), \(g = -9.807\,\frac{m}{s^{2}}\), \(v = 0\,\frac{m}{s}\), then the time taken by the arrow is:
\(t = \frac{v-v_{o}\cdot \sin \theta}{g}\)
\(t = \frac{0\,\frac{m}{s}-\left(40\,\frac{m}{s}\right)\cdot \sin 30^{\circ} }{-9.807\,\frac{m}{s^{2}} }\)
\(t = 2.039\,s\)
And the initial horizontal distance of the arrow (\(x_{i}\)) is determined by this kinematic formula:
\(x_{i} = (v_{o}\cdot \cos \theta)\cdot t\) (2)
If we know that \(v_{o} = 40\,\frac{m}{s}\), \(\theta = 30^{\circ}\) and \(t = 2.039\,s\), then the initial horizontal distance is:
\(x_{i} = \left[\left(40\,\frac{m}{s} \right)\cdot \cos 30^{\circ}\right]\cdot (2.039\,s)\)
\(x_{i} = 70.633\,m\)
There is an inelastic collision between the arrow and the bird, the initial velocity of the bird-arrow system is:
\(v = \frac{m_{A}\cdot v_{o}\cdot \cos \theta}{m_{A}+m_{B}}\) (3)
Where:
\(v\) - Initial velocity of the bird-arrow system, in meters per second.
\(m_{A}\) - Mass of the arrow, in kilograms.
\(m_{B}\) - Mass of the bird, in kilograms.
If we know that \(m_{A} = 0.5\,kg\), \(m_{B} = 2\,kg\), \(v_{o} = 40\,\frac{m}{s}\) and \(\theta = 30^{\circ}\), then the initial velocity of the bird-arrow system is:
\(v = \frac{(0.5\,kg)\cdot \left(40\,\frac{m}{s} \right)\cdot \cos 30^{\circ}}{0.5\,kg + 2\,kg}\)
\(v = 6.928\,\frac{m}{s}\)
And the maximum height reached by the arrow (\(y\)), in meters, is:
\(y = y_{o} + (v_{o}\cdot \sin \theta)\cdot t +\frac{1}{2}\cdot g\cdot t^{2}\) (4)
Where \(y_{o}\) is the initial height of the arrow, in meters.
If we know that \(y_{o} = 0\,m\), \(v_{o} = 40\,\frac{m}{s}\), \(\theta = 30^{\circ}\), \(g = -9.807\,\frac{m}{s^{2}}\) and \(t = 2.039\,s\), then the maximum height reached by the arrow is:
\(y = 0\,m + \left(40\,\frac{m}{s} \right)\cdot (2.039\,s)\cdot \sin 30^{\circ} +\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}}\right)\cdot (2.039\,s)^{2}\)
\(y = 20.394\,m\)
Time needed by the bird-arrow system to land is determined by this expression based on (4):
\(y = y_{o} + (v_{o,y})\cdot t +\frac{1}{2}\cdot g\cdot t^{2}\)
If we know that \(y_{o} = 20.394\,m\), \(y = 0\,m\), \(v_{o,y} = 0\,\frac{m}{s}\) and \(g = -9.807\,\frac{m}{s^{2}}\), then the time needed to land is:
\(0\,m = 20\,m + \left(0\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}\)
\(t = 2.019\,s\)
And the horizontal distance travelled by the bird-arrow system (\(x_{ii}\)), in meters, is calculated from a formula based on (2):
\(x_{ii} = v_{o, x}\cdot t\)
If we know that \(v_{o,x} = 6.928\,\frac{m}{s}\) and \(t = 2.019\,s\), then the distance travelled by the bird-arrow system is:
\(x_{ii} = \left(6.928\,\frac{m}{s} \right)\cdot (2.019\,s)\)
\(x_{ii} = 13.988\,m\)
The final distance of the bird-arrow system from the hunter is:
\(x = x_{i} + x_{ii}\)
\(x = 70.633\,m + 13.988\,m\)
\(x = 84.621\,m\)
The bird and the arrow are 84.621 meters far from the hunter.
Naoki's bicycle has a mass of 8 kg. If Naoki sits on her bicycle and starts pedaling with a force of 140.7 N, causing an acceleration
of 2.1 m/s2, what is Naoki's mass?
Tools
Answers
Naoki's mass is 67 kg, due to which the force calculated to pedal is 140.7 N at acceleration of 2.1 m/s².
What is Newton's second law of motion?Newton's second law of motion is a fundamental principle in physics that describes the relationship between force, mass, and acceleration. It states that an object's acceleration is proportional to the force applied to it and inversely proportional to its mass. We can rewrite this equation to find the mass:
F = m * a
where F is the net force applied to an object, m is its mass, and a is the resulting acceleration.
To solve this problem, we can use Newton's second law of motion, which states that the force applied to an object is equal to the product of its mass and acceleration:
F = m * a
where F is the applied force, m is the object's mass, and an is the acceleration.
The second law is mathematically stated as follows:
m = F / a
Substituting the given values, we get:
m = 140.7 N / 2.1 m/s²
m = 67 kg
Naoki's mass is 67 kg.
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david popenoe's research suggests that the weakest families on earth may be found in
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David Popenoe's research suggests that the weakest families on earth may be found in Western, industrialized nations.
Industrialized countries such as America, France, the United Kingdom, and other Western nations have weaker family values and structures than developing nations, according to David Popenoe's research. The family's crucial role in society's economic, social, and emotional stability has long been recognized by social scientists.
Popenoe's research supports the notion that Western, industrialized societies may be experiencing a family crisis.Popenoe says that family disintegration and weakening is a serious social problem in the Western world. Popenoe's studies found that Western societies, which place a high value on independence, individualism, and personal satisfaction, are at odds with the family's objectives of commitment, obligation, and selflessness.
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ntally Check for Understanding
POSSIBLE POINTS: 5
In a football game, the kicker kicks the football a horizontal distance of 39 m. If the ball lands 3.9 s later, what is the football's horizontal velocity just before
it lands?
0 10 m/s
41 m/s
0 34 m/s
O 14 m/s
Answers
Answer: 34
Explanation:
34
consider the titration of 100.0 ml of 0.200 m acetic acid (ka by 0.100 m koh. calculate the ph of the result- ing solution after the following volumes of koh have been added.
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Answer: The pH of the solution after 50.5 mL of 0.100 M KOH has been added is 16.85.
The reaction between acetic acid and potassium hydroxide (KOH) is given by the equation:
CH3COOH + KOH → CH3COOK + H2O.
It can be observed that for every mole of KOH added, one mole of H+ from CH3COOH reacts with one mole of OH- from KOH to produce one mole of H2O. Therefore, after the addition of n moles of KOH, the number of moles of H+ remaining is (0.200 - n) moles (since 0.200 M is the initial concentration of acetic acid). On the other hand, the number of moles of CH3COOK produced is also n moles; thus the final concentration of CH3COOK is n/V (where V is the total volume of the solution).
The reaction produces a salt (CH3COOK), which is neutral. Therefore, the final pH is given by: pH = pKa + log([A-]/[HA]) where A- is CH3COOK and HA is CH3COOH.
To find the pH of the resulting solution after the following volumes of KOH have been added, we need to calculate the number of moles of KOH added, the number of moles of CH3COOK produced, the number of moles of CH3COOH remaining, and the total volume of the solution. Then we can use the Henderson-Hasselbalch equation above to find the pH.
Here are the calculations: Initial number of moles of CH3COOH = M × V = 0.200 × 0.100 = 0.020 moles. For each mole of KOH added, one mole of CH3COOH reacts with one mole of KOH to produce one mole of CH3COOK.Number of moles of KOH added Volume of KOH added (mL)Moles of KOH added0.02 0.00 (initial) 0.000.02 10.10 0.010.02 20.20 0.020.02 30.30 0.030.02 40.40 0.040.02 50.50 0.05Number of moles of CH3COOK produced = number of moles of KOH added = n moles. Concentration of CH3COOK produced = n/V = n/(Vinitial + V added) = n/(0.100 L + V added)
Number of moles of CH3COOH remaining = initial number of moles - number of moles of KOH added = 0.020 - n moles. Total volume of the solution after the addition of KOH = V initial + V added = 0.100 L + V added. We can then use the Henderson-Hasselbalch equation to find the pH: pH = pKa + log([A-]/[HA])where A- is CH3COOK and HA is CH3COOH.pKa of acetic acid = 4.76
Let's tabulate the values of n, [A-], [HA], and pH for each volume of KOH added: Number of moles of KOH added n (moles)Concentration of CH3COOK produced [A-] (M)
Number of moles of CH3COOH remaining [HA] (M)pH0.000.000.020.76 (initial)0.010.010.019.84 0.020.020.018.98 0.030.030.018.20 0.040.040.017.49 0.050.050.016.85.
The pH of the solution after 50.5 mL of 0.100 M KOH has been added is 16.85.
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Q1. How much gravitational potential energy do you gain when youwalk up the CN Tower stairs (346 m)?What type of energy/energies transformed into GPE in this scenario?
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Given data:
* The distance traveled is 346 m.
Solution:
Let m be the mass of the body.
Then, the gravitational potential energy of the body on reaching the top of the tower is,
\(U=\text{mgh}\)where g is the acceleration due to gravity, and h is the height of the tower,
As mg is the weight of the body.
Substituting the known value,
\(U=w\times346\)where w is the weight of the body,
When a person walks to the certain height, the internal energy of the person's body is converted into the gravitational potential energy.
The mass of the earth is 6x10^24 kg and that of the moon is 7.4x10^22 kg. If the radius of the moon is 1740 km, then find the value of g if the earth is compressed into the size of moon.
Answers
Answer:
In the surface of the moon, gravitational acceleration is 1.63 m/s*2.
Explanation:
An object of mass M will accelerate gravitationally at a distance R if it is at the following distance:
g = G*M/R^2
Where the gravitational constant is G.
G = 6.67*10^(-11) m^3/(kg*s^2)
At the surface of a moon, the distance between its surface and its center will be equal to its radius, since a moon's mass is concentrated at its center, thus:
R = 1740 km
It's important to remember that we need meters in order to work:
1 km = 1000 m
so:
1740 km = (1740)*1000 m = 1740000 m
R = 1740000 m
Basically, the mass consists of:
M = 7.4x10^22 kg
Incorporating all that into the gravitational acceleration equation, we get:
g = (6.67*10^(-11) m^3 / (kg*s^2))*(7.4x10^22 kg) / ( 1740000 m)^2
g = 1.63 m / s^2
In the surface of the moon, gravitational acceleration is 1.63 m / s*2.
what is the correct tight asymptotic bound for T(n) in 1,2,3 where T(1)=c ?
T(n) = 10T(n/10)+100n
T(n) =T(n/10)+100n
T(n) = T(n/10)+100
Answers
The correct tight asymptotic bound for T(n) in 1, 2, 3 where T(1)=c is Θ(n log n).
To find the tight asymptotic bound of T(n), we will use the Master Theorem. So, let's take a look at each recurrence relation:
1. T(n) = 10T(n/10)+100nApplying the Master Theorem: a = 10, b = 10, f(n) = 100nlogb a = log10 10 = 1 Since f(n) = Θ(n) = Θ(n1), Case 2 of the Master Theorem applies. The solution, therefore, is Θ(n log n).
2. T(n) = T(n/10)+100n Here, a = 1, b = 10, f(n) = 100nlogb a = log10 1 = 0 Since f(n) = Θ(n0) = Θ(1), Case 1 of the Master Theorem applies. The solution, therefore, is Θ(n).
3. T(n) = T(n/10)+100 Here, a = 1, b = 10, f(n) = 100logb a = log10 1 = 0 Since f(n) = Θ(1) = Θ(n0), Case 2 of the Master Theorem applies. The solution, therefore, is Θ(log n).Therefore, the correct tight asymptotic bound for T(n) in 1, 2, 3 where T(1)=c is Θ(n log n).
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Let's use the Master Theorem to find the asymptotic tight bound for each of the three recurrence relations given in the problem statement.
1. T(n) = 10T(n/10) + 100n Here, a = 10, b = 10, and f(n) = 100n. We can calculate the value of logb a as follows: log10 10 = 1 Since f(n) = Θ(n1), we can apply Case 2 of the Master Theorem and get: T(n) = Θ(n log n)Therefore, the correct tight asymptotic bound for T(n) in the first case is Θ(n log n).
2. T(n) = T(n/10) + 100n Here, a = 1, b = 10, and f(n) = 100n. We can calculate the value of logb a as follows: log10 1 = 0 Since f(n) = Θ(n1), we can apply Case 1 of the Master Theorem and get: T(n) = Θ(n)Therefore, the correct tight asymptotic bound for T(n) in the second case is Θ(n).
3. T(n) = T(n/10) + 100 Here, a = 1, b = 10, and f(n) = 100. We can calculate the value of logb a as follows: log10 1 = 0 Since f(n) = Θ(1), we can apply Case 2 of the Master Theorem and get: T(n) = Θ(log n) Therefore, the correct tight asymptotic bound for T(n) in the third case is Θ(log n). Hence, the three asymptotic tight bounds for T(n) are Θ(n log n), Θ(n), and Θ(log n), respectively.
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A specialized part of a cell which has a specific location is called:_______
Answers
A specialized part of a cell which has a specific location is called an organelle.
Organelles are responsible for carrying out the various functions of the cell, and each type of organelle has a specific job. The nucleus is the most important organelle in the cell, as it stores the DNA which contains the cell's genetic material. The nucleus is located in the center of the cell and is surrounded by a membrane.The endoplasmic reticulum is another organelle that has a specific location. It is a network of interconnected sacs and channels that runs throughout the cytoplasm.The mitochondria is an organelle that is responsible for converting energy from food molecules into a form that the cell can use.Overall, organelles are specialized parts of a cell that have a specific location and job. They are responsible for carrying out the various functions of the cell, and each type of organelle has a specific role.
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Please help me I don’t understand this question
Answers
Answer: 225N to the right
Explanation:
The pedaling force is to the right and of 325N.
The air resistance is to the opposite side (in this case to the left) of 100N.
To find Net force of an object with two opposite forces acting on them then we have to subtract the smaller force from the bigger force and but the direction of the bigger force, so
325N-100N=225N
The greater force is towards the right proving the direction of the Net force is to the right.
Therefore the Net force is 225N to the right.
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Two trains sound identical horns of frequency 410 Hz. One train is stationary. The other is moving away from an observer, who heats a beat frequency of 35 Hz. How fast is the moving train going?
Answers
The speed of the moving train is approximately 33.5 m/s.
The beat frequency is given by the difference in frequency between the two horns, which is equal to the Doppler shift in frequency due to the motion of the moving train. Using the formula for the Doppler effect, we can solve for the speed of the train:
\(f_b = f_s\dfrac{(v + v_o)}{(v + v_s)}\)
where \(f_b\) is the beat frequency, \(f_s\) is the horn frequency, v is the speed of sound, \(v_o\) is the observer's speed, and \(v_s\) is the speed of the source.
We know that \(f_s\) = 410 Hz and \(f_b\) = 35 Hz. The speed of sound in air at standard temperature and pressure is approximately 343 m/s. Since the observer is stationary, \(v_o\) = 0.
Solving for \(v_s\), we get:
\(v_s = \dfrac{(f_s + f_b)}{f_s - 1} \times v\)
\(v_s\) = ((410 Hz + 35 Hz) / 410 Hz - 1) * 343 m/s
\(v_s\) = 33.5 m/s
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a fan is rotating at a constant 339.0 rev/min. what is the magnitude of the acceleration (in m/s2) of a point on one of its blades 13.0 cm from the axis of rotation? m/s2
Answers
Magnitude of the acceleration of fan blades will be approximately 165.9 m/s2
The first step is to convert the rotational speed from revolutions per minute (RPM) to radians per second (rad/s), since acceleration is measured in meters per second squared (m/s^2), which is a linear unit of measurement. We can use the conversion factor:
1 revolution = 2π radians
Therefore, 339.0 rev/min is equal to:
339.0 rev/min * (2π rad/1 rev) * (1 min/60 s) = 35.6 rad/s
The acceleration of a point on one of the fan blades can be found using the formula for centripetal acceleration:
a = rω^2
where:
a is the centripetal acceleration
r is the radius of the fan blade (0.13 m)
ω is the angular velocity in radians per second (35.6 rad/s)
Plugging in the values:
a = (0.13 m)(35.6 rad/s)^2
a = 165.9 m/s^2 (rounded to three significant figures)
Therefore, the magnitude of the acceleration of a point on one of the fan blades is 165.9 m/s^2.
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9.In corrosion fatigue ay the number of cycles for failure increase as the stress is increased b there is always a greater effect of environ-mental factors than mechanical factors c the endurance limit of a material is sharplyreduced d the surface remains bright afiter fracture
Answers
In corrosion fatigue is: "d) the surface remains bright after fracture."
Corrosion fatigue is a phenomenon that occurs when a material is subjected to cyclic loading in a corrosive environment, leading to the degradation of the material. During corrosion fatigue, the surface of the material can exhibit various changes and characteristics.
Option d) states that "the surface remains bright after fracture." This statement is incorrect. In corrosion fatigue, the surface of the material does not remain bright after fracture. Instead, it often exhibits characteristic signs of corrosion, such as pitting, cracking, or discoloration. The combination of cyclic loading and the corrosive environment leads to the formation and propagation of cracks, which ultimately results in failure.
The other options mentioned in the question (a, b, c) are also incorrect or irrelevant to corrosion fatigue. The number of cycles for failure does not necessarily increase as the stress is increased (option a). The effect of environmental factors and mechanical factors can vary depending on the specific situation (option b). The endurance limit of a material is not sharply reduced in corrosion fatigue (option c).
Therefore, the correct statement is option - d.
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Iron has a density of 7.07 g/cm3. What is the mass of 3.51 cm3 of iron?
Answers
The mass of 3.51 cm³ of iron is approximately 24.82 grams.
To calculate the mass of a given volume of iron, we can use the formula:
Mass = Density × Volume
Density of iron = 7.07 g/cm³
Volume of iron = 3.51 cm³
Substituting the values into the formula:
Mass = 7.07 g/cm³ × 3.51 cm³
Calculating the result:
Mass = 24.8157 g
Therefore, the mass of 3.51 cm³ of iron is approximately 24.82 grams.
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A free undamped spring/mass system oscillates with a period of 5 seconds. When 12 N are removed from the spring, the system then has a period of 3 seconds. What was the weight of the original mass on the spring?
Answers
The weight of the original mass on the spring was approximately 6.75 Newtons (N). The period of oscillation of a spring/mass system is determined by the mass and the spring constant.
Let's assume the original mass on the spring is represented by M and the corresponding weight is W.
Given that the original period is 5 seconds and the modified period is 3 seconds, we can set up the following equation using the formula for the period of an oscillating spring/mass system:
T = \(2\pi \sqrt{M/k}\), Where T is the period, M is the mass, and k is the spring constant.
For the original system with a period of 5 seconds, we have:
5 = \(2\pi \sqrt{M/k}\) ...(1)
When 12 N are removed from the spring, the modified system has a period of 3 seconds. This implies that the spring constant has changed, but the mass remains the same. Let's assume the new spring constant is k'.
3 = \(2\pi \sqrt{M/k'}\) ...(2)
Dividing equation (1) by equation (2), we can eliminate the mass M:
5/3 = \(\sqrt{k'/k}\)
Squaring both sides of the equation gives:
25/9 = k'/k.
Rearranging the equation gives:
k' = (25/9)k.
Since the spring constant is directly proportional to the weight of the mass, we can conclude that the weight of the original mass W is also reduced by a factor of (25/9).
Let's assume the weight of the original mass on the spring is W0. Thus, the weight of the modified mass is (W0 - 12 N).
Using the proportionality, we have: (W0 - 12 N) = (25/9)W0.
Simplifying the equation, we find: (9/9)W0 - (12 N) = (25/9)W0, (-16/9)W0 = 12 N.
Multiplying both sides by (-9/16) gives: W0 = (-9/16)(12 N), W0 = -6.75 N
Therefore, the weight of the original mass on the spring was approximately 6.75 Newtons (N).
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The first spacecraft which did not merely fly bya jovian (or giant) planet, but actually went into orbit around it for an extended period of time was
a. Galileo
b. Einstein
c. Voyager
d. the Hubble Space Telescope
e. Cassini
Answers
Answer:The first spacecraft which did not merely fly by a jovian (or giant) planet, but actually went into orbit around it for an extended period of time was option a, Galileo. The Galileo spacecraft was launched in 1989 and orbited Jupiter for almost eight years, from 1995 to 2003.
Explanation: