Please Solve The Following Problems You Must Show All Work

Please Solve The Following Problems You Must Show All Work

Please solve the following problems. You must show all work: 1. One lightyear is defined to be the distance light can travel in one year. a) What is this distance in meters? b) How long does it take for light to get to the Moon 2. Multiply 1.783 x 10 2 * 4.4 x 10 -3 , taking into account significant figures. 3.

What is the surface area of a sphere of diameter 2.4 x10 2 cm? 4. What is larger 4000 L or 4 x 10 5 cm 3 and you must show your reasoning. 5. A car travels 120 meters in one direction in 20 seconds.

Then the car returns ¾ of the way back in 10 seconds. a) Calculate the average speed of the car for the first part of the trip. b) Find the average velocity of the car. 6. A sports car moving at constant speed travels 150 m in 4.00 s. If it then brakes and comes to a stop while decelerating at a rate of 6.0 m/s 2 . How long does it take to stop?

7. Explain a possible situation where you start with a positive velocity that decreases to a negative increasing velocity while there is a constant negative acceleration. 8. A stone is dropped from the roof of a high building. A second stone is dropped 1.25 s later. How long does it take for the stones to be 25.0 meters apart?

Paper For Above instruction

This set of physics problems explores fundamental concepts such as distances in astronomy, significant figure calculations, surface area of geometric shapes, unit conversions, average and instantaneous velocities, acceleration, and free fall physics. Each problem requires methodical calculations grounded in physics principles, emphasizing clear demonstration of work to reinforce understanding and accuracy.

Problem 1: The Distance of a Lightyear and Travel Time to the Moon

A lightyear is the distance that light travels in a year. The speed of light (c) is approximately 299,792,458 meters per second. To determine the distance of a lightyear in meters, we multiply the speed of light by the total seconds in a year:

Total seconds in a year = 365.25 days/year (including leap years) × 24 hours/day × 3600 seconds/hour = 31,557,600 seconds (approximately).

Thus, distance in meters = c × seconds in a year = 299,792,458 m/s × 31,557,600 s ≈ 9.45 × 10^15 meters.

Therefore, a lightyear is approximately 9.45 × 10^15 meters.

To find how long light takes to reach the Moon, we consider the average distance from Earth to the Moon, approximately 384,400 km (or 3.844 × 10^5 km). Converting to meters:

Distance = 3.844 × 10^5 km × 1,000 m/km = 3.844 × 10^8 meters.

Time taken for light to reach the Moon = distance / speed of light = 3.844 × 10^8 m / 2.99792458 × 10^8 m/s ≈ 1.28 seconds.

Problem 2: Multiplication with Significant Figures

Multiply 1.783 × 10^2 by 4.4 × 10^-3. First, multiply the numerical parts: 1.783 × 4.4 ≈ 7.8392.

Combine the powers of ten: 10^2 × 10^-3 = 10^(2 - 3) = 10^-1.

Thus, the product before considering significant figures is approximately 7.8392 × 10^-1.

Determine the number of significant figures: 1.783 has 4 significant figures; 4.4 has 2 significant figures. The result should be rounded to the smallest number of significant figures, which is 2.

Rounded result: 7.8 × 10^-1 or 0.78.

Problem 3: Surface Area of a Sphere

Given the diameter D = 2.4 × 10^2 cm, the radius r = D/2 = 1.2 × 10^2 cm.

Surface area of a sphere = 4πr^2.

Calculate r^2: (1.2 × 10^2)^2 = 1.44 × 10^4 cm^2.

Multiply by 4π: 4 × π ≈ 12.566.

Surface area ≈ 12.566 × 1.44 × 10^4 ≈ 1.81 × 10^5 cm^2.

Problem 4: Comparing 4000 L and 4 × 10^5 cm^3

Recognize that 1 liter (L) equals 1000 cm^3.

Therefore, 4000 L = 4000 × 1000 cm^3 = 4.000 × 10^6 cm^3.

Compare with 4 × 10^5 cm^3: since 4.000 × 10^6 cm^3 > 4 × 10^5 cm^3, the volume of 4000 L is larger.

Problem 5: Car's Travel – Speed and Velocity

a) The first part: speed = distance / time = 120 m / 20 s = 6 m/s.

b) For the second part, the car returns three-fourths of the way back; this distance = (3/4) × 120 m = 90 m.

Time for return trip = 10 s.

Displacement for the entire trip: initial 120 m forward, then 90 m back → net displacement = 120 m - 90 m = 30 m forward.

Total time = 20 s + 10 s = 30 s.

Average velocity = total displacement / total time = 30 m / 30 s = 1 m/s (forward).

Problem 6: Braking of a Car

Initial velocity (v_i) = 150 m / 4.00 s = 37.5 m/s.

Deceleration (a) = -6.0 m/s^2.

Using v_f = v_i + at, where v_f = 0 (comes to stop): 0 = 37.5 m/s + (-6.0 m/s^2) × t.

t = 37.5 m/s / 6.0 m/s^2 ≈ 6.25 seconds.

Problem 7: Positive to Negative Velocity with Constant Negative Acceleration

An example is a car braking while moving forward. Initially, the car has a positive velocity, but as brakes are applied with a constant negative acceleration, its velocity decreases to zero and continues to become negative, indicating motion in reverse. This scenario illustrates how a negative acceleration can reduce positive velocity to zero and then increase the magnitude of negative velocity as the vehicle moves backward.

Problem 8: Dropped Stones and Relative Distance

The first stone is dropped at t=0, hitting the ground at time t_1 = √(2h/g). The second stone is dropped 1.25 seconds later, so its fall time is t_2 = t_1 - 1.25 seconds.

We want to find the time t when the two stones are 25.0 meters apart in vertical distance.

Assuming both stones fall freely under gravity g = 9.8 m/s^2, their heights after time t are:

h_1 = 0.5 × g × t^2 (first stone)

h_2 = 0.5 × g × (t - 1.25)^2 (second stone)

Set the difference in heights equal to 25 meters:

|h_1 - h_2| = 25

Substitute and solve:

h_1 - h_2 = 0.5g (t^2 - (t - 1.25)^2) = 25.

Expanding:

0.5 × 9.8 [t^2 - (t^2 - 2.5t + 1.5625)] = 25.

Simplify:

4.9 [2.5t - 1.5625] = 25.

Calculate:

4.9 × 2.5t - 4.9 × 1.5625 = 25.

12.25t - 7.65625 = 25.

Solve for t:

12.25t = 25 + 7.65625 = 32.65625,

t = 32.65625 / 12.25 ≈ 2.666 seconds.

This is the time after the first stone is dropped. The second stone is released 1.25 seconds later, so the total time from the release of the first stone is approximately 2.67 seconds, at which point they are 25 meters apart.

Conclusion

The problems covered include fundamental physics calculations involving distances, kinematics, surface area, and unit conversions, with an emphasis on calculations involving significant figures and physical reasoning. Understanding these concepts is essential for analyzing motion and understanding physical phenomena accurately.

References

  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers with Modern Physics (10th ed.). Cengage Learning.
  • Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics (10th ed.). Wiley.
  • Giancoli, D. C. (2014). Physics: Principles with Applications (7th ed.). Pearson.
  • Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers (6th ed.). W. H. Freeman.
  • Young, H. D., & Freedman, R. A. (2012). University Physics with Modern Physics (13th ed.). Pearson.
  • NASA. (n.d.). What is a lightyear? NASA Science. https://science.nasa.gov/
  • Wikipedia contributors. (2023). Surface area of a sphere. Wikipedia. https://en.wikipedia.org/wiki/Surface_area_of_a_sphere
  • Khan Academy. (n.d.). Kinematic equations for uniformly accelerated motion. Khan Academy. https://www.khanacademy.org/science/physics
  • HyperPhysics. (n.d.). Speed, velocity, and acceleration. Georgia State University. http://hyperphysics.phy-astr.gsu.edu/
  • Physics Classroom. (n.d.). Free fall acceleration and equations. https://www.physicsclassroom.com/

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