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12345 [234]
3 years ago
13

Complete each statement about the sign of the work done on a baseball. Carlton catches a baseball and his hand moves backward as

he brings the ball to a halt, doing work on the baseball. Carlton throws a baseball, doing work on the baseball. Carlton carries a baseball in his hand as he rides his bicycle in a straight line at constant speed, doing work on the baseball. Carlton carries a baseball in his hand as he rides his bicycle in a straight line with increasing speed, doing work on the baseball.
Physics
1 answer:
dezoksy [38]3 years ago
8 0

Answer and Explanation:

  • The work done is negative as the force is applied in the forward direction and the displacement of the hand is in the backward direction.
  • The work done is positive as the work is done on the baseball by Carlton. Moreover, both the force applied and the displacement of Carlton are in a straight line in the forward direction.
  • The work done in this case is zero as Carlton rides with constant speed, the Kinetic energy too remains constant and hence net work done is zero.
  • The work done in this case is positive as the work is done on the baseball by Carlton and as the velocity changes its Kinetic energy also varies wit it.
  • Also  the force applied and the displacement of Carlton in this case are in the directed to the same path.
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If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the
koban [17]

Answer:

\displaystyle \frac{480}{7}\approx 68.6\; \rm mph.

Explanation:

The average speed of an object is equal to total distance over total time.

  • Distance traveled: \rm 50 \; mi.

How much time is taken? This trip is divided into two halves, each of distance \displaystyle \frac{50}{2} = 25\;\rm mi.

Time spent on the first half of the trip:

\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours.

Similarly, time spent on the second half of the trip:

\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours.

In total:

\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours.

Average speed:

\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}.

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to \rm 60\; mph rather than \rm 80\; mph.

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If an airplane travels a distance of 500 km in 5 hours, what is its average speed?
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1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential
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<h3>Answer:</h3>

\displaystyle U_s = 0.6 \ J

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: \displaystyle U_s = \frac{1}{2} k \triangle x^2

  • U is energy (in J)
  • k is spring constant (in N/m)
  • Δx is displacement from equilibrium (in m)
<h3>Explanation:</h3>

<u>Step 1: Define</u>

k  = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

  1. Substitute in variables [Elastic Potential Energy]:                                        \displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2
  2. Evaluate exponents:                                                                                      \displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)
  3. Multiply:                                                                                                           \displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)
  4. Multiply:                                                                                                           \displaystyle U_s = 0.6 \ J
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