Answer: [tex]4 m/s^{2}[/tex]
Explanation:
First you take your end velocity and subtract the initial velocity ([tex]40-20=20[/tex])
Then you divide the difference by the time ([tex]20/5=4[/tex])
Giving you the result of [tex]4 m/s^{2}[/tex]
This all uses the equation [tex]V_{f}-V_{i}=ta[/tex]
suppose you repeated the experiment but started with all of the mass on the hanger and then de- creased it one increment at a time. would your results be different? why or why not?
If the experiment was repeated but started with all of the mass on the hanger and then decreased it one increment at a time, the results would be different. This is because the mass would be acting as a force, and decreasing the mass would decrease the force applied to the spring.
As a result, the spring would stretch less for each decrease in mass, and the relationship between the mass and the stretch of the spring would be different than in the original experiment. The spring constant of the spring would remain the same, but the data collected would be different due to the change in the force applied.
If you repeated the experiment by starting with all of the mass on the hanger and then decreased it one increment at a time, your results might be slightly different due to potential systematic errors or hysteresis effects in the system.
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An elevator weighing 500kg is to be lifted up at a constant velocity of 0.4m/s. Calculate the power of the motor required for this purpose.
Answer:
2.613 hp
Explanation:
We know that
Power=Work done/time
Power=Work done * Distance/ time ----->eq(1)
where,
Work done=Mg
Distance=0.4m
time=1 s
where M=>mass of the object,
g=>acceleration due to gravity
g=9.8 m/s² and M=500Kg
So,
Work done=500 * 9.8=4900 J
Substituting the values in eq(1) we get
Power = (4900*0.4)/1 = 1960 Watt
750 Watt = 1 hp
1960 Watt = (1*1960)/750
=2.613 hp
how many joules of work are done by a 2.0-horsepower motor that runs for 1.0 hour (assuming 100% efficiency)? round to two significant digits.
A 2.0-horsepower motor that runs for 1.0 hour will do 5.4 x 10^6⁶ joules of work (rounded to two significant digits) assuming 100% efficiency.
To calculate the work done by the motor, we need to use the formula: Work = Power x Time
The power of the motor is given in horsepower, so we need to convert it to watts to use it in the formula. One horsepower is equal to 746 watts, so a 2.0-horsepower motor is equivalent to 1492 watts.
Now we can plug in the values:
Work = 1492 watts x 1 hour
Since 1 hour is equal to 3600 seconds, we can convert it to joules
Work = 1492 watts x 3600 seconds
Work = 5.35 x 10⁶ joules
Finally, we round the answer to two significant digits, which gives us 5.4 x 10⁶ joules. Therefore, a 2.0-horsepower motor that runs for 1.0 hour will do 5.4 x 10⁶ joules of work assuming 100% efficiency.
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D
Question 40
M1
VI
Before
After
M2
V2
1 pts
A railroad car with a mass of 4,500 kg traveling at 3 m/s slams into a stationary railroad car and they couple. After coupling the cars are travelling 1.3 m/s. What is the
mass of the second railroad car, rounded to the nearest whole number? (Please remember that the initial mass you find is both train cars together. You will need to
subtract the mass of the first car from your answer.)
The second car has a mass of
kg.
The mass of the second railroad car, rounded to the nearest whole number, given that the car was initially at rest is 5884 Kg
How do I determine the mass of the second railroad car?The following data were obtained from the question:
Mass of 1st railroad car (m₁) = 4500 KgSpeed of 1st railroad car (u₁) = 3 m/sSpeed of second railroad car (u₂) = 0 m/sSpeed after collision (v) = 1.3 m/sMass of second railroad car (m₂) = ?The mass of second railroad car be obtained as illustrated below:
Momentum before = momentum after
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(4500 × 3) + (m₂ × 0) = 1.3 × (4500 + m₂)
13500 + 0 = 5850 + 1.3m₂
13500 = 5850 + 1.3m₂
Collect like terms
13500 - 5850 = 1.3m₂
7650 = 1.3m₂
Divide both sides by 1.3
m₂ = 7650 / 1.3
m₂ = 5884 Kg
Thus, the mass of the second railroad car is 5884 Kg
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calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun.
The magnitude of the angular momentum of the Earth in a circular orbit around the sun is 1.91 x 10^40 kg m^2/s.
To calculate the magnitude of the angular momentum of the Earth in a circular orbit around the sun, angular momentum, L = I * w where L is the angular momentum, I is the moment of inertia, and w is the angular velocity.
For a circular orbit, the angular velocity is given by, w = v / r, where v is the speed of the Earth in its orbit and r is the radius of the orbit.
The moment of inertia of a rotating object is given by,
I = 2/5 * m * r^2
where m is the mass of the Earth and r is the radius of the orbit.
We can find the speed of the Earth in its orbit using the formula,
v = 2 * pi * r / T
where T is the period of the Earth's orbit around the sun.
The radius of the Earth's orbit is approximately 1.496 x 10^11 meters, and the period of the Earth's orbit is approximately 365.25 days or 31,557,600 seconds.
Using these values, we can calculate the speed of the Earth in its orbit.
v = 2 * pi * 1.496 x 10^11 / 31,557,600 = 29,783 meters per second
We can also calculate the moment of inertia of the Earth.
I = 2/5 * 5.972 x 10^24 kg * (1.496 x 10^11 meters)^2 = 9.70 x 10^37 kg m^2
L = I * w = (9.70 x 10^37 kg m^2) * (29,783 meters per second / 1.496 x 10^11 meters) = 1.91 x 10^40 kg m^2/s
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1. What do we call the games that involve different manipulative
skills like throwing, tossing, rolling, catching, running, jumping,
hopping, and stretching?
a. Puzzle Games
c. Target Games
b. Simulation Games
d. Role-playing Games
2. Which among the following is NOT a Target Game?
a. Batuhang Bola
c. Tumbang Preso
b. Chinese Garter
d. Shatong
3. How many teams do you need in order to play Batuhang Bola?
d. 2
b. 3
a. 5
C. 4
4. What is the purpose of gaining "life" in Batuhang Bola if a playe
catches a ball?
a. It adds point to the team.
b. It can be used to revive another player or be used to contir
play on if the ball hits him/her.
c. The team will win.
d. It allows the team to hit the opponent intentionally.
hit by the
Role-playing games are those that require players to use a variety of manipulating abilities, such as throwing, rolling, catching, sprinting, jumping, and stretching.
2) TUMBANG PRESO is a very popular game among kids in the entire nation. In the background, in the park, or even in the side streets, it was being played.
3)To play batuhang bola, you need 5 TEAMS.
4) The head is the only portion of the body that is off limits when playing games.
5) In Batuhang Bola, the DEFLECTORS try to avoid being struck by the shooters of the attacking teams until the 5-minute waiting period has passed.
6)TARGET GAMES involve sending an item in the direction of a target while dodging obstacles.
7)TUMBANG PRESO is a common kid's game in the Philippines.
8)Target games need a lot of catching and throwing ability.
9) It can be made a bit more flattened so that it is more difficult to tumble.
10). The abilities needed into using stairs as opposed to an elevator include walking and running.
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Target Games involve throwing, catching, and other physical skills. Chinese Garter is not a Target Game. Batuhang Bola requires two teams. Catching a ball in Batuhang Bola grants a "life" to revive a player or continue to play if hit by the ball.
Games that involve physical skills like throwing, catching, and running are known as Target Games. Chinese Garter is not a Target Game. Batuhang Bola is a Target Game that requires two teams to play. The purpose of gaining "life" in Batuhang Bola when a player catches a ball is to revive another player or continue to play if hit by the ball. This rule makes the game more exciting and strategic, as players must decide whether to risk catching the ball and potentially losing their life or letting it go to avoid being hit. The team with the most remaining lives at the end of the game wins. The game promotes teamwork, coordination, and quick thinking, making it a popular game in the Philippines.
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3. A fire hose is turned on, it exerts a pressure of 10kPa. If the diameter of the
jet is 0.6m, what is the force exerted? (remember there are 1000Pa to a kPa,
and the area of a circle is found by multiplying Pi by the radius squared).
The force exerted by the fire hose is 2.827 kN
force excerted calculation.
First, we need to calculate the area of the jet using the formula for the area of a circle:
Area = π x (radius)^2
We know the diameter of the jet is 0.6m, so the radius is half of that
radius = 0.6m / 2 = 0.3m
Plugging this value into the formula, we get:
Area = π x (0.3m)^2 = 0.2827 m^2
Next, we can use the formula for pressure to calculate the force exerted:
Pressure = Force / Area
Rearranging this formula to solve for force, we get:
Force = Pressure x Area
Plugging in the given values, we get:
Force = 10 kPa x 0.2827 m^2 = 2.827 kN
Therefore, the force exerted by the fire hose is 2.827 kN (kilonewtons).
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what are the sources of uncertainty (both random uncertainties and systematic uncertainties) associated with determining the focal length of the 10 cm focal length lens in part 1? review the uncertainty analysis instructions on pilot to help you answer this question
The sources of uncertainty associated with determining the focal length of the 10 cm focal length lens in Part 1 include both random and systematic uncertainties. The random uncertainties include the precision of the measurements taken. The systematic uncertainties include possible errors in the measurement instruments.
In Part 1, the focal length of the 10 cm focal length lens was determined using the lens formula and measurements of object and image distances. Random uncertainties arise due to the inherent variability in measurements, such as the placement of the object and image distances. Systematic uncertainties arise due to factors such as instrument errors and lens deviation from a perfect spherical shape.
These uncertainties can affect the accuracy of the measurement and need to be taken into consideration during the uncertainty analysis. The combination of random and systematic uncertainties contribute to the total uncertainty associated with the determination of the focal length of the lens.
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the maximum force on a force vs. time graph is 400. n and the time interval over which the force acts is from 20 ms - 40 ms. what is the average force exerted on the object?
The average force exerted on the object is equal to the area under the force vs. time graph, which can be calculated as (400 N) x (20 ms) = 8000 Nms. Dividing by the time interval of 20 ms gives an average force of 400 N.
To calculate the average force exerted on the object, we need to find the area under the force vs. time graph during the given time interval. The maximum force on the graph is 400 N and the time interval is from 20 ms to 40 ms. Therefore, the area under the graph can be calculated as the product of the maximum force and the time interval:
Area = (400 N) x (20 ms) = 8000 Nms.
The average force exerted on the object is equal to this area divided by the time interval:
Average Force = Area / Time Interval = 8000 Nms / 20 ms = 400 N
Therefore, the average force exerted on the object during the given time interval is 400 N.
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When there is acceleration, a position vs. time graph is a curve. true or false
Yes it is true that when there is acceleration, a position vs time graph is a curve.
What is acceleration?The concept of acceleration refers to the measure of how quickly an object's velocity changes over time. This physical quantity is denoted by the symbol "a" and is considered a vector quantity since it has both magnitude and direction. To put it simply, acceleration is the rate at which an object's speed changes in a given period of time.
Is it true that when there is acceleration, a position vs. time graph is a curve?Yes it is true. When examining a position vs. time graph, the presence of acceleration is indicated by a curved line rather than a linear one. This is due to the fact that acceleration signifies a modification in velocity over time, and velocity measures how quickly an object's position changes over time. Consequently, as acceleration fluctuates, so too will an object's velocity and position, resulting in a nonlinear curve on the position vs. time graph.
The slope of this curve on said graph represents an object's velocity at any given point in time, while the slope of the tangent line at that same point represents its instantaneous velocity.
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a car is approaching a radio station at a speed of 25.0 m/s. if the radio station broadcasts at a frequency of 74.5 mhz, what change in frequency does the driver observe?
The change in frequency observed by the driver when the car is approaching a radio station at a speed of 25.0 m/s broadcasting at a frequency of 74.5 MHz is 74.59 MHz.
The formula for finding the observed frequency when the source is moving towards the observer is given by;
f′=f (v±v0/c)
Where, f is the frequency of the source, v is the velocity of light, v0 is the velocity of the source observed by the observer, c is the speed of light.
In this case, given that, v0 = 25.0 m/sf = 74.5 MHz, v = 3.0 x 108 m/s, c = 3.0 x 108 m/s
Putting the values in the formula, we get,
f′=74.5×(3.0×10^8+25.0×1000/3.0×10^8)=74.59 MHz (approx)
Hence, the change in frequency observed by the driver when the car is 74.59 MHz (approx).
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Find the center of mass of a uniform L shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 Kg.
Answer: The center of mass of the L shaped lamina is located at (0.458, 0.458) or approximately (0.46, 0.46) units from the lower left corner of the lamina
Explanation:
To find the center of mass of the L shaped lamina, we need to locate the point where the lamina will balance. The center of mass is given by the formula:
x = (M1x1 + M2x2 + M3x3) / (M1 + M2 + M3)
where x is the x-coordinate of the center of mass, M is the mass of the respective part, and x1, x2, and x3 are the x-coordinates of the centers of mass of the respective parts.
In this case, the L shaped lamina can be divided into three parts:
the rectangular part, the small square part, and the triangular part. Each part has the same mass since the lamina is uniform.
The rectangular part has a center of mass at (1/2, 1/4) and a mass of 2 Kg.
The small square part has a center of mass at (1/4, 3/4) and a mass of 0.5 Kg.
The triangular part has a center of mass at (1/3, 1/3) and a mass of 0.5 Kg.
Substituting these values into the formula, we get:
x = (2*1/2 + 0.5*1/4 + 0.5*1/3) / (2 + 0.5 + 0.5)
x = 0.458
Therefore, the center of mass of the L shaped lamina is located at (0.458, 0.458) or approximately (0.46, 0.46) units from the lower left corner of the lamina.
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at what distance from earth is the gravitational potentail energy of a spaceship-earth system reduced to half the energy of the system before the launch
The distance from Earth at which the gravitational potential energy of a spaceship-Earth system is reduced to half the energy before launch is approximately 117 million meters (117,000 kilometers).
The gravitational potential energy of a spaceship-Earth system is directly proportional to the distance between them. As the spaceship moves away from the Earth, its potential energy increases. The energy required to move the spaceship away from the Earth against the force of gravity is directly proportional to the mass of the spaceship and the distance between the spaceship and the Earth. To find the distance at which the gravitational potential energy of the spaceship-Earth system is reduced to half the energy before launch, we can use the formula for gravitational potential energy. By solving for the distance using the given values of the masses of the Earth and the spaceship, the gravitational constant, and the initial energy, we can determine that the distance is approximately 117 million meters or 117,000 kilometers.
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A 0.155 kg arrow is shot from ground level, upward at 31.4 m/s. What is its potential energy (PE) when it is 30.0 m above the ground?
Answer:
Kinetic energy (K.E) when it is 30.0 m above the ground is 30.84 J .
Explanation:
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a 190-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,070 a. if the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (use 3.156 107 for the number of seconds in a year.)
Using the given values, the time for one electron to travel the full length of the transmission line is approximately 14 billion years. This calculation is based on the drift velocity of electrons in copper, which is very slow.
The drift velocity of electrons in copper, which is a measurement of how quickly they move in a certain direction under the influence of an electric field, may be used to determine how long it takes for one electron to travel the whole length of the transmission line. The high-voltage transmission line in this instance is 190 km long, 2.00 cm in diameter, and capable of transmitting a constant current of 1,070 A. One electron takes approximately 14 billion years to travel the whole length of the transmission line, according to the free charge density of copper. This is because numerous collisions with other atoms and heat agitation cause the electron drift velocity in copper to be extremely sluggish.
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star a and star b are both standard candles. we are able to determine they are the same luminosity, but star a appears fainter. which star is closer?
The star which appears fainter is farther. Star A and B are both standard candles, but Star A appears fainter, so it is farther.
A standard candle is an object of a known luminosity that astronomers can use to estimate distances based on the difference between its apparent and absolute magnitudes. Astronomers discovered that the luminosity of a standard candle can be calculated using its apparent magnitude because the apparent magnitude of an object is related to its luminosity. The luminosity of the stars determines how bright they are. Star A appears fainter than Star B despite being identical in luminosity because it is farther away from Earth. The farther an object is, the fainter it appears. The relationship between an object's luminosity, distance, and apparent magnitude is described by the inverse-square law, which states that the intensity of light is inversely proportional to the square of the distance from the source.
The formula for the inverse-square law can be used to calculate how much brighter an object appears when it is closer to the observer. If two objects have the same luminosity and one is farther away than the other, the closer object will appear brighter.
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items If you take snapshots of a standing wave on a string, there are certain instants when the string is totally flat v Part A What has happened to the energy of the wave at those instants? The energy is transformed into the potential energy of the string The energy is transformed into the kinetic energy of the string except of the nodes The energy is transformed into the energy of sound waves in the alt Sub Rouest Answer Provide Food
At those instants, the energy of the wave has been transformed into the potential energy of the string.
This is because, during the wave's motion, the peaks and troughs move towards equilibrium. As they move, they store energy in the form of elastic potential energy, which is the energy stored in the string due to its stretching. When the wave is totally flat, the wave has reached equilibrium and the stored energy has been converted into potential energy. This is why the string appears flat at those instants.
The energy is not transformed into the kinetic energy of the string or the energy of sound waves. The nodes of the wave, where the string is flat, are points at which the wave's kinetic energy is zero, and the energy of sound waves is not produced by standing waves.
In conclusion, when a snapshot is taken of a standing wave on a string at certain instants, the energy of the wave is transformed into the potential energy of the string.
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Symmetric versus anti-symmetric problem. If the length of your string it 60 cm and the mass of the bob is 97 gm, what do you expect the period T1 for the mode in Figure 9.2a to be? The straw should be halfway down the string.; What do you predict for the period of T2 of the mode illustrated in Figure 9.2b? Please right your answers with 1 decimal places.
The predicted period for mode T1 is approximately 0.78 seconds and the predicted period for mode T2 is approximately 1.10 seconds.
Assuming the string has negligible mass, the period of oscillation for a simple pendulum is given by,
T = 2π √(L/g)
where L is the length of the pendulum and g is the acceleration due to gravity.
For Figure 9.2a, the straw is at the midpoint of the string, so the effective length of the pendulum is L/2 = 30 cm. The mass of the bob is given as 97 gm.
Using the formula,
T1 = 2π √(L/g)
= 2π √(0.3/9.81)
≈ 0.78 s (to 2 decimal places)
For Figure 9.2b, the straw is at one end of the string, so the effective length of the pendulum is L = 60 cm. The mass of the bob is given as 97 gm.
Using the formula,
T2 = 2π √(L/g)
= 2π √(0.6/9.81)
≈ 1.10 s (to 2 decimal places)
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--The complete question is, Symmetric versus anti-symmetric problem. If the length of your string it 60 cm and the mass of the bob is 97 gm, what do you expect the period T1 for the mode in Figure 9.2a(symmetric) to be? The straw should be halfway down the string.; What do you predict for the period of T2 of the mode illustrated in Figure 9.2b(anti-symmetric)? Please right your answers with 1 decimal places.--
the attraction or repulsion between magnetic poles is called
The attraction or repulsion between magnetic poles is called the Magnetic domain.
A magnetic sphere is a region within a magnetic material in which the magnetization is in a invariant direction. This means that the individual glamorous moments of the tittles are aligned with one another, and they point in the same direction, a magnetic sphere structure is responsible for the magnetic gets of ferromagnetic accoutrements like iron, nickel, cobalt and their blends, and ferrimagnetic accoutrements like ferrite. This includes the conformation of endless attractions and the magnet of ferromagnetic accoutrements to a glamorous field. The regions separating glamorous disciplines are called sphere walls, where the magnetization rotates coherently from the direction in one sphere to that in the coming sphere. The study of magnetic disciplines is called micromagnetics.
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A positive charge of +4.0 μC and a negative charge of –3.0 μC are 6.0 cm apart.
Find the electric potential at a point that is at a distance of 8.0 cm from the negative charge, on a line that makes a 90° angle with the line segment connecting the two charges.
The electric potential at a point that is at a distance of 8.0 cm from the negative charge, on a line that makes a 90° angle with the line segment connecting the two charges, is -1.875 x[tex]10^{7}[/tex] V.
What is Electric Potential?
Electric potential is a measure of the electrical potential energy per unit of charge. It is the amount of work that is needed to move a unit positive charge from a reference point to a specific point in an electric field, without any acceleration. It is measured in volts (V) and is also known as electric potential difference or voltage.
To find the electric potential at a point on a line that makes a 90° angle with the line segment connecting the two charges, we need to find the electric potential due to each charge and then add them algebraically.
The electric potential due to a point charge Q at a distance r from it is given by:
V = kQ/r
where k is the Coulomb's constant (9 x 10^9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex]).
Let's first find the electric potential due to the negative charge at the given point:
V1 = k(-3.0 μC)/(8.0 cm) = -3.375 x [tex]10^{7}[/tex] V
The negative sign indicates that the electric potential is negative, as expected due to the negative charge.
Now, let's find the electric potential due to the positive charge at the given point:
V2 = k(4.0 μC)/(6.0 cm + 8.0 cm) = 1.5 x [tex]10^{7}[/tex] V
The positive sign indicates that the electric potential is positive, as expected due to the positive charge.
Finally, we can find the net electric potential at the given point by adding the electric potentials due to each charge:
V = V1 + V2 = -3.375 x[tex]10^{7}[/tex] V + 1.5 x[tex]10^{7}[/tex] V = -1.875 x [tex]10^{7}[/tex] V
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the speed of sound in water at a temperature of 25°c is 1500 m/s. what is the wavelength of a 300 hz sound wave traveling through water at a temperature of 25°c?
Answer:
The speed of sound in water at a temperature of 25°C is 1500 m/s. We can use the formula:
wavelength = speed of sound / frequency
where frequency is given as 300 Hz.
wavelength = 1500 m/s / 300 Hz
wavelength = 5 meters
Therefore, the wavelength of a 300 Hz sound wave traveling through water at a temperature of 25°C is 5 meters.
Explanation:
Use k = 9 x 10⁹ Nm²/C². 3. Two point charges, q1 and 92, of 4.00 μC each, are placed -16.0 cm and 16.0 cm away from the origin on the x-axis. A charge q3 of -1.00 μC is placed 12.0 cm away from the origin on the y- axis.
a.find thr distsance from q3 to q1 and from q3 to q2.
b.find the magnitude and the direction of the force F13 exerted by q1 on q3.
c.find the magnitude and the direction of the force F23 exerted by q2 on q3.
d.find the magnitude and the direction of the force f12 exerted by q1 on q2.
a. Distance from q₃ to q₁ = 23.0 cm
b. the magnitude and the direction of the force F₁₃ exerted by q₁ on q₃ -3.00 x 10⁻³ N.
c. The magnitude and the direction of the force F23 exerted by q2 on q3 = -3.00 x 10⁻³ N
d. The magnitude and the direction of the force F₂₃ exerted by q₂ on q₁ = -7.5 x 10⁻⁵ N
How to find distance from q₃ to q₁ and from q₃ to q₂?To find the distance from q₃ to q₁ and from q₃ to q₂, we can use the Pythagorean theorem. The distance from q₃ to q₁ is the hypotenuse of a right triangle with legs of 12.0 cm (the distance from q₃ to the origin on the y-axis) and 16.0 cm + 4.00 cm = 20.0 cm (the distance from q₁ to the origin on the x-axis). Thus:
distance from q₃ to q₁ = √(12.0 cm² + 20.0 cm²) = 23.0 cm
Similarly, the distance from q₃ to q₂ is the hypotenuse of a right triangle with legs of 12.0 cm (the distance from q₃ to the origin on the y-axis) and 16.0 cm - 4.00 cm = 12.0 cm (the distance from q₂ to the origin on the x-axis). Thus:
distance from q₃ to q₂ = √(12.0 cm² + 12.0 cm²) = 16.97 cm (to two significant figures)
How to find the magnitude and the direction of the force F₁₃ exerted by q₁ on q₃?
To find the magnitude of the force F₁₃ exerted by q₁ on q₃, we can use Coulomb's law:
F₁₃ = k * q₁ * q₃/ r₁₃²
where k = 9 x 10⁹ Nm²/C² is the Coulomb constant, q₁ and q₃ are the charges in coulombs, and r₁₃ is the distance between the charges in meters. In this case, q₁ = q₃ = 4.00 μC = 4.00 x 10⁻⁶ C and r₁₃ = 12.0 cm = 0.12 m. Thus:
F₁₃ = (9 x 10⁹ Nm²/C²) * (4.00 x 10⁻⁶ C) * (-1.00 x 10⁻⁶ C) / (0.12 m)²
= -3.00 x 10⁻³ N
The negative sign indicates that the force is attractive, since q₁ and q₃ have opposite signs.
c. q₂ = q₁, so The magnitude and the direction of the force F₂₃ exerted by q₂ on q₃ = -3.00 x 10⁻³ N
d. q₃ = 1/4q₂, so -3.00 x 10⁻³/4 N
= −0.00075 = -7.5 x 10⁻⁵
The magnitude and the direction of the force F₂₃ exerted by q₂ on q₁ = -7.5 x 10⁻⁵ N
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a man has a mass of 85 kg. how far above the surface of the earth would he have to go to lose 20% of his body weight?
6,371 km above the surface of the earth he has to go to lose 20% of his body weight
let's calculate as follows.
Step 1: We know that the weight of an object is given as;
Weight = Mass x Acceleration due to gravity
So, man's weight on earth can be given as;`
Weight = Mass x Acceleration due to gravity
Weight = 85 kg x 9.81 m/s²
Weight = 834.85 N
Step 2: To find the height the man needs to go to lose 20% of his body weight;
Let h be the height in meters the man needs to go to lose 20% of his body weight.`20% of the man's weight = 20/100 x 834.85 = 166.97 N
Since weight decreases as one moves away from the surface of the earth. Therefore, let's consider the man's weight at a height h
Weight = Mass x Acceleration due to gravity = 0.8 x 834.85``
0.8 x 834.85 = 667.88 N
At height h, the weight of the man is 667.88 N
So, let's calculate h;
Weight = Mass x Acceleration due to gravity
667.88 = 85 x 9.81 x (Earth radius / (Earth radius + h)²)
(Earth radius + h)² = 9.81 x 85 / 667.88(6371000 + h)² = 7180.63
(6371000 + h) = ±84.81
h = 6371000 + 84.81
Since h is the distance above the surface of the earth the man must go to lose 20% of his body weight, he needs to go 6,371,084.81 meters (approx. 6,371 km).
Therefore, to lose 20% of his body weight, the man needs to go 6,371 km above the surface of the earth.
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3. What average net force is required to stop a 7 kg shopping cart in 2 s if it's initially
traveling at 3. 5 m/s?
An average net force of -12.25 N is required to stop a 7 kg shopping cart that is initially moving at 3.5 m/s in 2 seconds, acting in the opposite direction to the cart's initial velocity.
To determine the average net force required to stop a 7 kg shopping cart in 2 s, we can use the equation:
Δv = aΔt
where Δv is the change in velocity, a is the acceleration, and Δt is the time interval.
Initially, the shopping cart is traveling at a velocity of 3.5 m/s, and it comes to a stop in 2 s, so Δv = -3.5 m/s. We can rearrange the equation above to solve for the acceleration:
a = Δv / Δt = (-3.5 m/s) / (2 s) = -1.75 m/s²
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which is necessary to stop the cart. Finally, we can use Newton's second law, F = ma, to calculate the average net force required:
F = ma = (7 kg) x (-1.75 m/s²) = -12.25 N
The negative sign indicates that the force is in the opposite direction to the initial velocity of the cart. Therefore, an average net force of 12.25 N is required to stop the 7 kg shopping cart in 2 s, assuming constant acceleration.
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what conclusions can you draw about the best wavelength for measuring a given sample with beer's law
In order to determine the best wavelength for measuring a given sample with Beer's Law, one must first identify the components of the sample and the maximum absorption of each component. This can be done by measuring the absorption of each component at different wavelengths. Once this data is collected, the wavelength with the highest absorption for the sample can be determined.
According to Beer's Law, the absorption of light by a solution is proportional to its concentration and the path length of the light. The best wavelength for measuring a given sample with Beer's Law is determined by the absorbance of the sample.
The conclusion that can be drawn about the best wavelength for measuring a given sample with Beer's Law is that it is determined by the sample's absorbance.
Absorbance is directly proportional to concentration and path length, as determined by Beer's Law.
Therefore, the wavelength at which a sample has the highest absorbance is the best wavelength for measuring that particular sample. It's worth noting that the best wavelength for measuring a given sample may differ from that of another sample. This is because different samples may have different molecular structures and therefore absorb different wavelengths of light.
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A wave has a speed of 20 m/s and a wavelength of 5 meters. if the same wave was created in the same medium, with half the original frequency, how would the wavelength change?
Answer:
The wavelength of the wave would increase to 10 meters.
Explanation:
We can use the formula:
velocity = frequency × wavelength
to relate the velocity, frequency, and wavelength of a wave.
Given that the wave has a speed of 20 m/s and a wavelength of 5 meters, we can solve for its frequency as follows:
frequency = velocity ÷ wavelength = 20 m/s ÷ 5 meters = 4 Hz
Now, if the same wave is created in the same medium, but with half the original frequency, its new frequency will be:
new frequency = 4 Hz ÷ 2 = 2 Hz
To find the new wavelength of the wave, we can rearrange the formula above to solve for wavelength:
wavelength = velocity ÷ frequency
Using the new frequency of 2 Hz, we get:
new wavelength = 20 m/s ÷ 2 Hz = 10 meters
Therefore, if the same wave was created in the same medium, with half the original frequency, the wavelength of the wave would increase to 10 meters.
g what is the power p supplied to a resistor whose resistance is r when it is known that it has a voltage v across it? express the power p in terms of r and v .
The power p supplied to a resistor whose resistance is r when it is known that it has a voltage v across it can be expressed as P = v²/r.
What is power?Power is the rate at which energy is transferred. The power P can be represented as P = W/t, where W is the work done, and t is the time required to complete the work.
What is resistance?Resistance is the ratio of voltage to current in an electrical circuit. It is a measure of how difficult it is to transfer a current through a component.
What is the power supplied to a resistor whose resistance is r when it is known that it has a voltage v across it?The power P provided to a resistor whose resistance is r when it is known that it has a voltage v across it is given by:
P = v²/r
Therefore, power can be expressed in terms of resistance and voltage as P = v²/r, where v is the voltage across the resistor and r is its resistance.
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v = Circular motion constant speed
a = Centripetal acceleration
f = Centripetal force
In circular motion with constant speed, the object moves in a circular path at a constant speed, while experiencing centripetal acceleration and centripetal force directed towards the center of the circle.
Using a circular image show the circular motion constant speed, centripetal acceleration, centripetal force?In circular motion with constant speed, the object moves in a circular path at a constant speed. This means that the object covers equal distance in equal time intervals. The velocity of the object is tangential to the circle at every point.
Centripetal acceleration is the acceleration of an object moving in a circular path, directed towards the center of the circle. It is always perpendicular to the velocity of the object and is given by the formula a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.
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you fire a 250-g arrow at a 2.5-kg box resting on the table. the box slides 1.34 meters across the table before it stops. the coefficient of kinetic friction between the box and the table is 0.30. how much energy did friction dissipate? at what velocity is the arrow/box moving right after the collision? what was the initial arrow's speed right before it hit the box?
The friction dissipated 9.82 J of energy, the velocity of the arrow/box system right after the collision is 1.12 m/s to the right, and the initial velocity of the arrow was 11.2 m/s to the right.
To solve this problem, we'll use the conservation of momentum and the work-energy principle.
First, let's find the initial velocity of the arrow before it hits the box. We can use the conservation of momentum equation:
m1v1 = (m1 + m2)vf
where m1 is the mass of the arrow,
v1 is the initial velocity of the arrow,
m2 is the mass of the box,
and vf is the final velocity of the arrow and box system after the collision.
Plugging in the values we get:
(0.25 kg)(v1) = (0.25 kg + 2.5 kg)(vf)
Solving for v1, we get:
v1 = 10vf
Next, we need to find the velocity of the arrow/box system after the collision.
We can use the work-energy principle:
W friction = ΔK
where W friction is the work done by friction and ΔK is the change in kinetic energy of the arrow/box system.
Since the arrow and box start at rest, the initial kinetic energy is zero.
The work done by friction is:
W friction = f friction x d
where f friction is the force of friction and d is the distance the box slides.
The force of friction is:
ffriction = μk x Fn
where μk is the coefficient of kinetic friction and Fn is the normal force. Since the box is on a horizontal surface, the normal force is equal to the weight of the box:
Fn = m2g
where g is the acceleration due to gravity.
Plugging in the values we get:
[tex]f friction = (0.30)(2.5 kg)(9.81 m/s^2) = 7.34 N[/tex]
Now we can find the work done by friction:
W friction = (7.34 N)(1.34 m) = 9.82 J
The change in kinetic energy is:
Δ[tex]K = (1/2)(m1 + m2)vf^2 - 0[/tex]
where vf is the final velocity of the arrow and box system after the collision.
Plugging in the values we get:
Δ[tex]K = (1/2)(0.25 kg + 2.5 kg)vf^2[/tex]
Setting W friction equal to ΔK and solving for vf, we get:
vf = [tex]\sqrt{(2Wfriction/(m1+m2))}[/tex]
[tex]\sqrt{ (2(9.82 J)/(0.25 kg + 2.5 kg)) }[/tex]
= 1.12 m/s
Finally, we can find the initial velocity of the arrow:
v1 = 10vf
= 10(1.12 m/s)
= 11.2 m/s.
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a projectile is launched upward with a velocity of 128 feet per second from the top of a 50-foot platform. what is the maximum height attained by the projectile?
We can solve this problem using the kinematic equations of motion for a projectile under constant acceleration due to gravity. In particular,
we can use the equation:
h = y + viy×t - (1/2)gt²
where h is the maximum height attained by the projectile,
y is the initial height of the projectile (in this case, 50 feet), viy is the initial vertical velocity of the projectile (in this case, 128 feet per second),
t is the time it takes for the projectile to reach its maximum height,
and g is the acceleration due to gravity (which we take to be 32.2 feet per second squared).
To find the time it takes for the projectile to reach its maximum height, we can use the fact that the projectile will reach its maximum height when its vertical velocity becomes zero.
At this point, the projectile will have traveled halfway through its trajectory, so the time it takes to reach the maximum height is given by:
tmax = viy/g
Plugging in the given values, we get:
tmax = 128/32.2 seconds
tmax ≈ 3.98 seconds
Now, we can use the equation for the maximum height:
h = y + viy×t - (1/2)gt²
Plugging in the values we have calculated, we get:
h = 50 + 128×3.98 - (1/2)32.2(3.98)²
h ≈ 403.2 feet
Therefore, the maximum height attained by the projectile is approximately 403.2 feet.
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