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3 years ago

If the runner completes exactly one lap what is her average speed? her average velocity? Explain your answer.

1 answer:

4 0

Average speed = (total distance covered) / (time to cover the distance).

Runner's average speed = (length of 1 lap) / (her time to run 1 lap).

________________________________

Average velocity =

(straight-line distance between start-point and end-point)
divided by
(time to go from start-point to end-point).

The runner's start-point and end-point are the same place.
The distance between them (her 'displacement') is zero.
Her average velocity = zero .

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If a car with mass 750 kg is lifted to a height of 2.3 m how much gravitational potential energy is added to the car

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3 years ago

Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600

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3 years ago

A 3000-kg crane is supporting a 10,000-kg crate. The crane pivots about point A and is at rest pressed against a support at B. (

Maru [420]

The answers are physically and mathematically reasonable, since both have the <em>same</em> order of magnitude of the <em>external</em> forces shown in the figure.

<h3 /><h3>Procedure - Determination of forces acting on a rigid body</h3>

Let be a system at equilibrium, which mathematically is represented by the following formulas:

$\Sigma F = 0$ (1)

$\Sigma M = 0$ (2)

<h3>a) Force acting on the crane at point A</h3>

We construct equations around points A and B by Newton's Laws and D'Alembert Principle:

<h3>Point A</h3>

$\Sigma M = F_{B}\cdot (1\,m)-(3000\,kgf)\cdot (2\,m)-(10000\,kgf)\cdot (6\,m) = 0$ (3)

<h3 /><h3>Point B</h3>

$\Sigma M = -F_{A,x}\cdot (1\,m) - (3000\,kgf)\cdot (2\,m)-(10000\,kgf)\cdot (6\,m) = 0$ (4)

<h3 /><h3>Entire system</h3>

$\Sigma F_{x} = F_{A,x} + F_{B} = 0$ (5)

$\Sigma F_{y} = F_{A,y} -3000\,kgf-10000\,kgf = 0$ (6)

The solution of the <em>entire</em> system is: $F_{A,x} = -66000\,kgf$ , $F_{B} = 66000\,kgf$ and $F_{A,y} = 13000\,kgf$ .

The magnitude of the force acting on the crane at point A is determined by the Pythagorean theorem:

$F_{A} = \sqrt{(-66000\,kgf)^{2}+(13000\,kgf)^{2}}$

$F_{A} \approx 67268.120\,kgf$

The force acting on the crane at point A has a magnitude of approximately 67268.128 kilograms-force. $\blacksquare$

<h3>b) The force acting on the crane at point B</h3>

The force acting on the crane at point B has a magnitude of approximately 66000 kilograms-force. $\blacksquare$

<h3>c) Order of Magnitude sense-making</h3>

The answers of parts (a) and (b) have an order of magnitude of $10^{3}$ , the same order of magnitude of the external forces shown in the figure. Hence, those answers are physically and mathematically reasonable.

<h3>Remark</h3>

Figure is missing. The statement is incomplete. Complete statement is presented below:

<em>A 3000 kilograms-force is supporting a 10000 kilograms-force crate. The crane pivots about point A and is at rest pressed against a support at B. </em>

<em>(a)</em><em> FInd the force acting on the crane at point A. </em>

<em>(b)</em><em> Find the force acting on the crane at point B. </em>

<em>(c)</em><em> Use order of magnitude sense-making to determine the reasonableness of your answers to parts (a) and (b). Hint: consider how the lever arm to the crate is much different than that to other points. </em>

To learn more on rigid bodies, we kindly invite to check this verified question: brainly.com/question/7031958

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2 years ago