Application 2 Consider an object (A) which moves in a uniform rectilinear motion in the negative direction of the x-axis. The speed of (A) is 15 m/s, and its initial abscissa is XoA 30 m. 1. a. Determine the time equation of the motion of (A). b. Draw the (V-1) graph of (A). 2. Another particle (B), whose time equation is XB = 101-70 (S.1), is moving on the same axis. We start timing for (A) and (B) simultaneously. a. Determine the instant at which (A) and (B) meet as well as the position at where they meet. b. Determine the distance separating (A) and (B) at t = 8 s. B- Acceleration rectilinear motion : ARM

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Answer 1

1. For an object (A) moving uniformly in the negative direction of the x-axis with a speed of 15 m/s and initial abscissa of 30 m, the time equation of its motion is X = 30 - 15*t, and its (V-1) graph is a straight line with a slope of -15 and a y-intercept of 30, and 2. When another particle (B) with a time equation of XB = 101-70 (S.1) moves on the same axis, (A) and (B) meet at time t = 2.13 s and position X = -32.7 m. The distance separating (A) and (B) at t = 8 s is 369 m.

1.a. The time equation of the motion of (A) is given by:

X = XoA + Vt

where X is the position of (A) at time t, XoA is the initial position of (A), V is the velocity of (A) and t is the time elapsed since the start of the motion.

Plugging in the given values, we get:

X = 30 - 15t

b. The (V-1) graph of (A) is a straight line with a slope of -15 (since the velocity is constant and negative) and a y-intercept of 30 (since the initial position is 30). The graph looks like this:( below)

2a. To determine the instant at which (A) and (B) meet, we need to find the time t at which their positions are equal. Equating the time equations of (A) and (B), we get:

30 - 15t = 101 - 70t

Solving for t, we get:

t = 2.13 s

To find the position at which they meet, we can plug this value of t into either of the time equations and get:

X = 101 - 70*2.13 = -32.7 m

So (A) and (B) meet at time t = 2.13 s and position X = -32.7 m.

b. To determine the distance separating (A) and (B) at t = 8 s, we need to find their positions at that time. Using the time equation of (A), we get:

Xa = 30 - 158 = -90 m

Using the time equation of (B), we get:

Xb = 101 - 708 = -459 m

The distance separating (A) and (B) at t = 8 s is:

|Xb - Xa| = |-459 - (-90)| = 369 m.

Hence, Two particles moving on the same axis, where one is uniformly moving with an initial abscissa of 30 m and a speed of 15 m/s, and the other is moving with a time equation of XB = 101-70 (S.1), meet at time t = 2.13 s and position X = -32.7 m, while the distance separating them at t = 8 s is 369 m.

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Application 2 Consider An Object (A) Which Moves In A Uniform Rectilinear Motion In The Negative Direction

Related Questions

A closed curve encircles several conductors. The line integral around this curve is ∮B⃗ ⋅dl⃗ = 4.25×10^-4 T⋅m .
A) What is the net current in the conductors?
B) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral?

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a) the net current in the conductors is 3.38 A. b) the line integral in the opposite direction will be: ∮B⃗ ⋅dl⃗ = -4.25×10^-4 T⋅m

To solve this problem, we can use Ampere's Law, which relates the line integral of the magnetic field around a closed loop to the net current passing through the loop.

A) The equation for Ampere's Law is: ∮B⃗ ⋅dl⃗ = μ0I, where μ0 is the permeability of free space and I is the net current passing through the loop. Solving for I, we get:

I = ∮B⃗ ⋅dl⃗ / μ0

Substituting the given values, we get:

I = (4.25×10^-4 T⋅m) / (4π×10^-7 T⋅m/A)

I = 3.38 A

Therefore, the net current in the conductors is 3.38 A.

B) If we integrate around the curve in the opposite direction, the value of the line integral will be negative, since the direction of the magnetic field will be opposite. Specifically, we can use the fact that reversing the direction of the line integral is equivalent to reversing the direction of the loop, which changes the sign of the enclosed current.

Therefore, the line integral in the opposite direction will be: ∮B⃗ ⋅dl⃗ = -4.25×10^-4 T⋅m

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(1 pt) how would reducing the surface pressure affect the power required to operate the pumps (answer qualitatively)?

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Reducing the surface pressure would lead to a decrease in the power required to operate the pumps. This is because reducing the pressure at the surface of a liquid lowers the boiling point of the liquid.

As a result, less energy is required to move the liquid through the pumps. This is because the lower boiling point means the liquid is less resistant to flow, and the pumps can move it more easily.

Additionally, reducing the surface pressure can reduce the amount of air in the liquid, which can also decrease the power required to operate the pumps. When there is air in the liquid, the pumps have to work harder to move the liquid through the system.

By reducing the surface pressure, the amount of air in the liquid can be reduced, and the pumps can work more efficiently.

Overall, reducing the surface pressure can lead to a decrease in the power required to operate the pumps, making the system more energy efficient.

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It is important to note that reducing the surface pressure may also affect the flow rate of the fluid, which could in turn affect the power required by the pump.

Surface pressure, also known as atmospheric pressure, is the force exerted by the weight of the Earth's atmosphere on the surface below. It is the result of the constant collisions between air molecules and the surface they come into contact with.

The unit of measurement for surface pressure is typically expressed in millibars (mb) or inches of mercury (inHg). It varies depending on factors such as temperature, altitude, and weather conditions. The standard sea-level pressure is around 1013 mb or 29.92 inHg. Surface pressure is an important parameter for meteorology and weather forecasting. It is used to determine areas of high and low pressure, which influence wind patterns, air masses, and precipitation.

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Complete Question:-

How would reducing the surface pressure affect the power required to operate the pumps (answer qualitatively)?

Students measure velocity as a function of changing time for an object moving at a constant rate. The following math model was generated, but the students had to linearize the data first to create this math model. What relationship originally existed between velocity and time? velocity(m/s) = (10(m))/(t(s))

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Inverse proportional relationship between velocity and time originally existed for the measured data of an object moving at a constant rate, which was linearized to obtain the equation: velocity(m/s) = (10(m))/(t(s)).

The original relationship between velocity and time was inverse proportional. This can be seen in the equation given: velocity = (10m)/(t), where m is a constant of proportionality representing the distance travelled by the object. As time increases, velocity decreases, and vice versa. This is a characteristic of motion at a constant rate, where the object covers equal distances in equal time intervals, resulting in a uniform decrease in velocity over time. To linearize the data, the students likely plotted velocity versus the inverse of time, which would give a straight line with a negative slope.

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Sound A has a high pitch and sound B has a low pitch. Which of the following statements about these two sounds are correct? (There could be more than one correct choice.) a. The frequency of A is greater than the frequency of B. The period of A is shorter than the period of c. The amplitude of A is larger than the amplitude of d. Sound B travels faster than sound B through air. e. The wavelength of A is longer than the wavelength of B.

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Answer: a. The frequency of A is greater than the frequency of B.

c. The period of A is shorter than the period of B.

e. The wavelength of A is longer than the wavelength of B.

Explanation:

The frequency of a sound, or the number of waves or cycles per second, determines its pitch. Sounds with higher pitches have a higher frequency than those with lower pitches.

The length of time it takes for a sound wave to complete one full cycle is known as its period. The relationship between the period and frequency is inverse. This implies that the time shortens as the frequency lengthens.

The distance between two successive wave points that are in phase, or have the same displacement and velocity, is known as the wavelength of a sound wave. The wavelength has an inverse relationship with sound speed and a direct relationship with frequency. Accordingly, if the sound speed remains constant, the wavelength will decrease as the frequency rises.

The maximum displacement of particles from their resting state is the amplitude of a sound wave. The pitch and frequency of the sound are unaffected by the amplitude. It solely controls the sound's volume or intensity.

The medium that sound travels through determines its speed. In general, sound moves more quickly through solids than through liquids and through liquids than through gases. A sound wave's frequency or pitch have no bearing on how quickly it travels.

What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell’s internal resistance is 2.00 Ω ?

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Output Voltage of the lithium cell = 2.9994 V

Output Voltage is the maximum voltage that a cell can offer after overcoming its own internal resistance that arises from the construction of the cell.

To find the output voltage of the lithium cell, you need to consider the voltage drop across the internal resistance due to the current draw. You can use Ohm's Law for this calculation: Voltage drop = Current × Resistance.

In this case, the current draw is 0.300 mA, and the internal resistance is 2.00 Ω. First, convert the current to amperes: 0.300 mA = 0.0003 A.

Now, calculate the voltage drop: Voltage drop = 0.0003 A × 2.00 Ω = 0.0006 V.

Finally, subtract the voltage drop from the initial cell voltage: Output voltage = 3.0000 V - 0.0006 V = 2.9994 V.

The output voltage of the lithium cell is approximately 2.9994 V.

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determine the characteristic impedance of two 1-oz cu lands 100 mils in width that are located on opposite sides of a 47-mil glass epoxy board

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The characteristic impedance of the two 1-oz cu lands 100 mils in width that are located on opposite sides of a 47-mil glass epoxy board is approximately 47.4 ohms.

The characteristic impedance (Z0) of a transmission line depends on the geometry of the line and the dielectric constant of the material between the conductors. The formula for the characteristic impedance of a microstrip transmission line is:

Z0 = (87.3 + 100*(w/h)ln(4w/h)) * (h/w)

where w is the width of the trace, h is the height of the substrate, and ln is the natural logarithm.

Assuming a standard FR-4 epoxy substrate with a dielectric constant of 4.5, and using the formula above with w = 100 mils (0.1 inch) and h = 47 mils (0.047 inch), we get:

Z0 = (87.3 + 100*(0.1/0.047)ln(40.1/0.047)) * (0.047/0.1) = 47.4 ohms

Therefore, the characteristic impedance of the two 1-oz cu lands 100 mils in width that are located on opposite sides of a 47-mil glass epoxy board is approximately 47.4 ohms.

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1) Identify a source of interest to you. Provide the bibliographic information for the reader.
2) Summarize the source in at least two well developed paragraphs. Identify the main point of the article as well as the evidence advanced in support of it.
3) Significance. Identify the significance of the source—why is it important?—what practical or theoretical consequences might follow from the main point?—what limitations, objections, or weaknesses might be present that could serve to undermine the significance of the source?
4) Explain what you learned about philosophy as a whole; would you recommend that our class address the themes covered in the source? Why or why not?
5) Recommendation: One a scale of 1-5, with five being the highest, rank the quality and importance of this article. Be sure to explain your ranking.

https://aeon.co/essays/natural-laws-cant-be-broken-but-can-they-be-defined

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The significance of a source in research is crucial as it determines the reliability and validity of the information presented. A credible source is important because it ensures that the information presented is accurate and trustworthy.

Using sources that are not credible or reliable can lead to the spread of misinformation, which can have practical consequences such as wrong decisions and actions based on incorrect information. Theoretical consequences could include flawed research or faulty arguments. It is important to consider the limitations, objections, or weaknesses of a source, as these can undermine its significance. This includes evaluating factors such as bias, sample size, and methodology used in the research.

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--The complete Question is, Identify the significance of the source—why is it important?—what practical or theoretical consequences might follow from the main point?—what limitations, objections, or weaknesses might be present that could serve to undermine the significance of the source?--

A radar gun at point O rotates with the angular velocity of 0.2 rad/s and angular acceleration of 0.040 rad/s^2, at the instant ? = 45 degree, as it follows the motion of the car traveling along the circular road having a radius of r = 220 m. Part A Determine the magnitudes of velocity of the car at this instant. Part B Determine the magnitude of acceleration of the car at this instant.

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(a) the magnitude of the velocity of the car at this instant is 44 m/s.(b) the magnitude of the acceleration of the car at this instant is 8.8 m/s².

Part A:
To determine the velocity of the car at the given instant, we need to use the formula:
v = rω
where v is the velocity of the car, r is the radius of the circular road, and ω is the angular velocity of the radar gun.
At the instant θ = 45 degrees, we can convert this to radians by multiplying by π/180:
θ = 45° × π/180 = 0.7854 radians
We know that the angular velocity of the radar gun is 0.2 rad/s, so we can plug in these values to find the velocity of the car:
v = rω
v = (220 m)(0.2 rad/s)
v = 44 m/s
Therefore, the velocity is 44 m/s.

Part B:
To determine the acceleration of the car at the given instant, we need to use the formula:
a = rα
where a is the acceleration of the car, r is the radius of the circular road, and α is the angular acceleration of the radar gun.
We can plug in the given values to find the acceleration of the car:
a = rα
a = (220 m)(0.040 rad/s²)
a = 8.8 m/s²

Therefore, the acceleration is 8.8 m/s²

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for a particular process at 300 k, δg δ g = -10.0 kj and δh δ h = -7.0 kj. if the process is carried out reversibly, the amount of useful work (in kj) that .

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The amount of useful work that can be obtained for this particular process at 300 K is -3.0 kJ.

For a reversible process, the amount of useful work (in kJ) that can be obtained is given by the equation:

w = -δg = δh - Tδs

where δs is the change in entropy of the system.

Since the process is carried out reversibly, δs can be calculated using the equation:

δs = δqrev / T

where δqrev is the heat absorbed by the system during a reversible process.

Since δh = -7.0 kJ and δg = -10.0 kJ, we know that the process is exothermic (δh < 0) and spontaneous (δg < 0). Therefore, δqrev must be negative, indicating that heat is released from the system.

We can calculate δqrev using the equation:

δqrev = -Tδs = -T(δh / T) = -δh = 7.0 kJ

Substituting this value into the equation for work, we get:

w = -δg = δh - Tδs = -10.0 kJ - (-7.0 kJ) = -3.0 kJ

Therefore, the amount of useful work for this particular process at 300 K is -10.0 kJ.

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Johnny, of mass 65 kg, and Lucy, of mass 45 kg, are facing each other on roller blades. The coefficient of kinetic friction between the roller blades and concrete surface is 0.20. When Johnny pushes Lucy from rest he applies a force for 1.0 s. Lucy then slows down to a stop in another 8.0 s. Calculate:
a. The applied force exerted by Johnny on Lucy.
b. How long it takes Johnny to come to rest.

I tried calculated the force exerted but I would need acceleration which I don't have...any tips on how to solve this one??? help is appreciated!!

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Answer:

John applied a force of approximately [tex]795\; {\rm N}[/tex] (on average, rounded) on Lucy.

John slows down to a stop after approximately another [tex]5.37\; {\rm s}[/tex].

(Assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)

Explanation:

Assuming that the surface is level. The normal force on Johnny will be equal to the weight of Johnny: [tex]N(\text{John}) = m(\text{John})\, g[/tex]. Similarly, the normal force on Lucy will be equal to weight [tex]N(\text{Lucy}) = m(\text{Lucy})\, g[/tex].

Multiply normal force by the coefficient of kinetic friction to find the friction on each person:

[tex]f(\text{John}) = \mu_{k}\, N(\text{John}) = \mu_{k}\, m(\text{John})\, g[/tex].

[tex]f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g[/tex].

Again, because the surface is level, the net force on each person after the first [tex]1.0\; {\rm s}[/tex] will be equal to the friction. Divide that the net force on each person by the mass of that person to find acceleration:
[tex]\displaystyle a(\text{John}) = \frac{\mu_{k}\, m(\text{John})\, g}{m(\text{John})} = \mu_{k}\, g[/tex].

[tex]\displaystyle a(\text{Lucy}) = \frac{\mu_{k}\, m(\text{Lucy})\, g}{m(\text{Lucy})} = \mu_{k}\, g[/tex].

(Note that the magnitude of acceleration is independent of mass and is the same for both John and Lucy.)

[tex]a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}[/tex].

In other words, after the first [tex]1\; {\rm s}[/tex], both John and Lucy will slow down at a rate of [tex]1.962\; {\rm m\cdot s^{-2}}[/tex].

To find the speed of Lucy immediately after the first [tex]1.0\: {\rm s}[/tex], multiply this acceleration by the time [tex]t = 8.0\; {\rm s}[/tex] it took for Lucy to slow down to [tex]0\; {\rm m\cdot s^{-1}}[/tex]:

[tex]\begin{aligned}& (8.0\; {\rm s})\, (1.962\; {\rm m\cdot s^{-2}}) \\ =\; & (8.0)\, (1.962)\; {\rm m\cdot s^{-1}} \\ =\; & 15.696\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Thus, in the first [tex]1.0\; {\rm s}[/tex], Lucy accelerated (from [tex]0\; {\rm m\cdot s^{-1}}[/tex]) to [tex]15.696\; {\rm m\cdot s^{-1}}[/tex].

The average acceleration of Lucy in the first [tex]1.0\; {\rm s}[/tex] would be [tex](15.696) / (1) = 15.696\; {\rm m\cdot s^{-2}}[/tex]. Multiply this average acceleration by the mass of Lucy to find the average net force on Lucy during that [tex]1.0\; {\rm s}[/tex]:

[tex]\begin{aligned}F_{\text{net}}(\text{Lucy}) &= m(\text{Lucy})\, a \\ &= (45)\, (15.696)\; {\rm N} \\ &= 706.320\; {\rm N}\end{aligned}[/tex].

This net force on Lucy during that [tex]1.0\; {\rm s}[/tex] is the combined result of both the push from Johnny and friction:

[tex]F_{\text{net}}(\text{Lucy}) = F(\text{push}) - f(\text{Lucy})[/tex].

Since [tex]f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g[/tex]:

[tex]\begin{aligned}F(\text{push}) &= F_{\text{net}}(\text{Lucy}) + f(\text{Lucy}) \\ &= F_{\text{net}}(\text{Lucy}) + \mu_{k}\, m(\text{Lucy})\, g \\ &= (706.320) \; {\rm N}+ (0.2)\, (45)\, (9.81)\; {\rm N} \\ &= 706.320\; {\rm N} + 88.290\; {\rm N} \\ &=794.610\; {\rm N}\end{aligned}[/tex].

In other words, Johnny would have applied a force of [tex]794.610\; {\rm N}[/tex] on Lucy.

By Newton's Laws of Motion, when Johnny exerts this force on Lucy in that [tex]1.0\; {\rm s}[/tex], Lucy would exert a reaction force on Johnny of the same magnitude: [tex]794.610\; {\rm N}[/tex].

Similar to Lucy, the net force on Johnny during that [tex]1.0\; {\rm s}[/tex] will be the combined effect of the push [tex]F(\text{push})[/tex] and friction [tex]f(\text{John}) = \mu_{k}\, m(\text{John})\, g[/tex]:

[tex]\begin{aligned}F_{\text{net}}(\text{John}) &= F(\text{push}) - f(\text{John}) \\ &= F(\text{push}) - \mu_{k}\, m(\text{John})\, g\\ &= 794.610\; {\rm N} - (0.2)\, (65)\, (9.81)\; {\rm N} \\ &= 667.080\; {\rm N}\end{aligned}[/tex].

Divide net force by mass to find acceleration:

[tex]\begin{aligned}\frac{667.080\; {\rm N}}{65\; {\rm kg}} \approx 10.2628\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].

In other words, Johnny accelerated at a rate of approximately [tex]10.5406\; {\rm m\cdot s^{-2}}[/tex] during that [tex]1.0\; {\rm s}[/tex]. Assuming that Johnny was initially not moving, the velocity of Johnny right after that [tex]1.0\; {\rm s}\![/tex] would be:

[tex](0\; {\rm m\cdot s^{-1}}) + (10.2628\; {\rm m\cdot s^{-2}})\, (1.0\; {\rm s}) = 10.2628\; {\rm m\cdot s^{-1}}[/tex].

After the first [tex]1.0\; {\rm s}[/tex], the acceleration of both John and Lucy (as a result of friction) would both be equal to [tex]a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}[/tex]. Divide initial velocity of Johnny by this acceleration to find the time it took for Johnny to slow down to a stop:

[tex]\displaystyle \frac{10.2628\; {\rm {m\cdot s^{-1}}}}{1.962\; {\rm m\cdot s^{-2}}} \approx 5.23\; {\rm s}[/tex].